Resolução de exercícios - Física 1 - Young Freedman, Exercícios de Física

Baixe Resolução de exercícios - Física 1 - Young Freedman e outras Exercícios em PDF para Física, somente na Docsity! 1-1 UNITS, PHYSICAL QUANTITIES AND VECTORS 1.1. IDENTIFY: Convert units from mi to km and from km to ft. SET UP: 1 in. 2.54 cm= , 1 km = 1000 m , 12 in. 1 ft= , 1 mi = 5280 ft . EXECUTE: (a) 2 3 5280 ft 12 in. 2.54 cm 1 m 1 km1.00 mi (1.00 mi) 1.61 km 1 mi 1 ft 1 in. 10 cm 10 m ⎛ ⎞⎛ ⎞⎛ ⎞⎛ ⎞⎛ ⎞= =⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠⎝ ⎠⎝ ⎠⎝ ⎠ (b) 3 2 310 m 10 cm 1 in. 1 ft1.00 km (1.00 km) 3.28 10 ft 1 km 1 m 2.54 cm 12 in. ⎛ ⎞⎛ ⎞⎛ ⎞⎛ ⎞= = ×⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠⎝ ⎠⎝ ⎠ EVALUATE: A mile is a greater distance than a kilometer. There are 5280 ft in a mile but only 3280 ft in a km. 1.2. IDENTIFY: Convert volume units from L to 3in. . SET UP: 31 L 1000 cm= . 1 in. 2.54 cm= EXECUTE: 33 31000 cm 1 in.0.473 L 28.9 in. . 1 L 2.54 cm ⎛ ⎞ ⎛ ⎞× × =⎜ ⎟ ⎜ ⎟ ⎝ ⎠⎝ ⎠ EVALUATE: 31 in. is greater than 31 cm , so the volume in 3in. is a smaller number than the volume in 3cm , which is 3473 cm . 1.3. IDENTIFY: We know the speed of light in m/s. /t d v= . Convert 1.00 ft to m and t from s to ns. SET UP: The speed of light is 83.00 10 m/sv = × . 1 ft 0.3048 m= . 91 s 10 ns= . EXECUTE: 9 8 0.3048 m 1.02 10 s 1.02 ns 3.00 10 m/s t −= = × = × EVALUATE: In 1.00 s light travels 8 5 53.00 10 m 3.00 10 km 1.86 10 mi× = × = × . 1.4. IDENTIFY: Convert the units from g to kg and from 3cm to 3m . SET UP: 1 kg 1000 g= . 1 m 1000 cm= . EXECUTE: 3 4 3 3 g 1 kg 100 cm kg11.3 1.13 10 cm 1000 g 1 m m ⎛ ⎞ ⎛ ⎞ × × = ×⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ EVALUATE: The ratio that converts cm to m is cubed, because we need to convert 3cm to 3m . 1.5. IDENTIFY: Convert volume units from 3in. to L. SET UP: 31 L 1000 cm= . 1 in. 2.54 cm= . EXECUTE: ( ) ( ) ( ) 33 3327 in. 2.54 cm in. 1 L 1000 cm 5.36 L× × = EVALUATE: The volume is 35360 cm . 31 cm is less than 31 in. , so the volume in 3cm is a larger number than the volume in 3in. . 1.6. IDENTIFY: Convert 2ft to 2m and then to hectares. SET UP: 4 21.00 hectare 1.00 10 m= × . 1 ft 0.3048 m= . EXECUTE: The area is 22 4 2 43,600 ft 0.3048 m 1.00 hectare(12.0 acres) 4.86 hectares 1 acre 1.00 ft 1.00 10 m ⎛ ⎞⎛ ⎞ ⎛ ⎞ =⎜ ⎟⎜ ⎟ ⎜ ⎟×⎝ ⎠ ⎝ ⎠⎝ ⎠ . EVALUATE: Since 1 ft 0.3048 m= , 2 2 21 ft (0.3048) m= . 1.7. IDENTIFY: Convert seconds to years. SET UP: 91 billion seconds 1 10 s= × . 1 day 24 h= . 1 h 3600 s= . EXECUTE: ( )9 1 h 1 day 1 y1.00 billion seconds 1.00 10 s 31.7 y 3600 s 24 h 365 days ⎛ ⎞⎛ ⎞⎛ ⎞= × =⎜ ⎟⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠⎝ ⎠ . 1 1-2 Chapter 1 EVALUATE: The conversion 71 y 3.156 10 s= × assumes 1 y 365.24 d= , which is the average for one extra day every four years, in leap years. The problem says instead to assume a 365-day year. 1.8. IDENTIFY: Apply the given conversion factors. SET UP: 1 furlong 0.1250 mi and 1 fortnight 14 days.= = 1 day 24 h.= EXECUTE: ( ) 0.125 mi 1 fortnight 1 day180,000 furlongs fortnight 67 mi/h 1 furlong 14 days 24 h ⎛ ⎞⎛ ⎞⎛ ⎞ =⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠⎝ ⎠ EVALUATE: A furlong is less than a mile and a fortnight is many hours, so the speed limit in mph is a much smaller number. 1.9. IDENTIFY: Convert miles/gallon to km/L. SET UP: 1 mi 1.609 km= . 1 gallon 3.788 L.= EXECUTE: (a) 1.609 km 1 gallon55.0 miles/gallon (55.0 miles/gallon) 23.4 km/L 1 mi 3.788 L ⎛ ⎞⎛ ⎞= =⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠ . (b) The volume of gas required is 1500 km 64.1 L 23.4 km/L = . 64.1 L 1.4 tanks 45 L/tank = . EVALUATE: 1 mi/gal 0.425 km/L = . A km is very roughly half a mile and there are roughly 4 liters in a gallon, so 2 41 mi/gal km/L∼ , which is roughly our result. 1.10. IDENTIFY: Convert units. SET UP: Use the unit conversions given in the problem. Also, 100 cm 1 m= and 1000 g 1 kg= . EXECUTE: (a) mi 1h 5280 ft ft60 88 h 3600s 1mi s ⎛ ⎞ ⎛ ⎞⎛ ⎞ =⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ (b) 2 2 ft 30.48cm 1 m m32 9.8 s 1ft 100 cm s ⎛ ⎞⎛ ⎞ ⎛ ⎞ =⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠⎝ ⎠ (c) 3 3 3 3 g 100 cm 1 kg kg1.0 10 cm 1 m 1000 g m ⎛ ⎞⎛ ⎞ ⎛ ⎞ =⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ EVALUATE: The relations 60 mi/h 88 ft/s= and 3 3 31 g/cm 10 kg/m= are exact. The relation 2 232 ft/s 9.8 m/s= is accurate to only two significant figures. 1.11. IDENTIFY: We know the density and mass; thus we can find the volume using the relation density mass/volume /m V= = . The radius is then found from the volume equation for a sphere and the result for the volume. SET UP: 3Density 19.5 g/cm= and critical 60.0 kg.m = For a sphere 34 3V rπ= . EXECUTE: 3 critical 3 60.0 kg 1000 g/ density 3080 cm 19.5 g/cm 1.0 kg V m ⎛ ⎞⎛ ⎞= = =⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠ . ( )33 3 3 3 3080 cm 9.0 cm 4 4 Vr π π = = = . EVALUATE: The density is very large, so the 130 pound sphere is small in size. 1.12. IDENTIFY: Use your calculator to display 710π × . Compare that number to the number of seconds in a year. SET UP: 1 yr 365.24 days,= 1 day 24 h,= and 1 h 3600 s.= EXECUTE: 724 h 3600 s(365.24 days/1 yr) 3.15567... 10 s 1 day 1 h ⎛ ⎞⎛ ⎞ = ×⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠ ; 7 710 s 3.14159... 10 sπ × = × The approximate expression is accurate to two significant figures. EVALUATE: The close agreement is a numerical accident. 1.13. IDENTIFY: The percent error is the error divided by the quantity. SET UP: The distance from Berlin to Paris is given to the nearest 10 km. EXECUTE: (a) 3 3 10 m 1.1 10 %. 890 10 m −= × × (b) Since the distance was given as 890 km, the total distance should be 890,000 meters. We know the total distance to only three significant figures. EVALUATE: In this case a very small percentage error has disastrous consequences. 1.14. IDENTIFY: When numbers are multiplied or divided, the number of significant figures in the result can be no greater than in the factor with the fewest significant figures. When we add or subtract numbers it is the location of the decimal that matters. Units, Physical Quantities and Vectors 1-5 1.25. IDENTIFY: Estimation problem SET UP: Estimate that the pile is 18 in. 18 in. 5 ft 8 in..× × Use the density of gold to calculate the mass of gold in the pile and from this calculate the dollar value. EXECUTE: The volume of gold in the pile is 318 in. 18 in. 68 in. 22,000 in. .V = × × = Convert to 3cm : 3 3 3 5 322,000 in. (1000 cm / 61.02 in. ) 3.6 10 cm .V = = × The density of gold is 319.3 g/cm , so the mass of this volume of gold is 3 5 3 6(19.3 g/cm )(3.6 10 cm ) 7 10 g.m = × = × The monetary value of one gram is $10, so the gold has a value of 6 7($10/ gram)(7 10 grams) $7 10 ,× = × or about 6$100 10× (one hundred million dollars). EVALUATE: This is quite a large pile of gold, so such a large monetary value is reasonable. 1.26. IDENTIFY: Estimate the diameter of a drop and from that calculate the volume of a drop, in 3m . Convert 3m to L. SET UP: Estimate the diameter of a drop to be 2 mmd = . The volume of a spherical drop is 3 34 1 3 6V r dπ π= = . 3 310 cm 1 L= . EXECUTE: 3 3 31 6 (0.2 cm) 4 10 cmV π −= = × . The number of drops in 1.0 L is 3 5 3 3 1000 cm 2 10 4 10 cm− = × × EVALUATE: Since 3V d∼ , if our estimate of the diameter of a drop is off by a factor of 2 then our estimate of the number of drops is off by a factor of 8. 1.27. IDENTIFY: Estimate the number of students and the average number of pizzas eaten by each student in a school year. SET UP: Assume a school of thousand students, each of whom averages ten pizzas a year (perhaps an underestimate) EXECUTE: They eat a total of 104 pizzas. EVALUATE: The same answer applies to a school of 250 students averaging 40 pizzas a year each. 1.28. IDENTIFY: The number of bills is the distance to the moon divided by the thickness of one bill. SET UP: Estimate the thickness of a dollar bills by measuring a short stack, say ten, and dividing the measurement by the total number of bills. I obtain a thickness of roughly 1 mm. From Appendix F, the distance from the earth to the moon is 83.8 10 m.× EXECUTE: 8 3 12 12 bills 3.8 10 m 10 mm 3.8 10 bills 4 10 bills 0.1 mm/bill 1 m N ⎛ ⎞⎛ ⎞× = = × ≈ ×⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠ EVALUATE: This answer represents 4 trillion dollars! The cost of a single space shuttle mission in 2005 is significantly less – roughly 1 billion dollars. 1.29. IDENTIFY: The cost would equal the number of dollar bills required; the surface area of the U.S. divided by the surface area of a single dollar bill. SET UP: By drawing a rectangle on a map of the U.S., the approximate area is 2600 mi by 1300 mi or 3,380,000 2mi . This estimate is within 10 percent of the actual area, 3,794,083 2mi . The population is roughly 83.0 10× while the area of a dollar bill, as measured with a ruler, is approximately 1 86 in. by 5 82 in. EXECUTE: ( ) ( ) ( )[ ] ( ) ( ) 222 16 2 U.S. 3,380,000 mi 5280 ft / 1 mi 12 in. 1 ft 1.4 10 in.A ⎡ ⎤= = ×⎣ ⎦ ( )( ) 2 bill 6.125 in. 2.625 in. 16.1 in.A = = ( ) ( )16 2 2 14 bills U.S. billTotal cost 1.4 10 in. 16.1 in. / bill 9 10 billsN A A= = = × = × 14 8 6Cost per person (9 10 dollars) /(3.0 10 persons) 3 10 dollars/person= × × = × EVALUATE: The actual cost would be somewhat larger, because the land isn’t flat. 1.30. IDENTIFY: The displacements must be added as vectors and the magnitude of the sum depends on the relative orientation of the two displacements. SET UP: The sum with the largest magnitude is when the two displacements are parallel and the sum with the smallest magnitude is when the two displacements are antiparallel. EXECUTE: The orientations of the displacements that give the desired sum are shown in Figure 1.30. EVALUATE: The orientations of the two displacements can be chosen such that the sum has any value between 0.6 m and 4.2 m. Figure 1.30 1-6 Chapter 1 1.31. IDENTIFY: Draw each subsequent displacement tail to head with the previous displacement. The resultant displacement is the single vector that points from the starting point to the stopping point. SET UP: Call the three displacements A " , B " , and C " . The resultant displacement R " is given by R = A + B + C " "" " . EXECUTE: The vector addition diagram is given in Figure 1.31. Careful measurement gives that R " is 7.8 km, 38 north of east# . EVALUATE: The magnitude of the resultant displacement, 7.8 km, is less than the sum of the magnitudes of the individual displacements, 2.6 km 4.0 km 3.1 km+ + . Figure 1.31 1.32. IDENTIFY: Draw the vector addition diagram, so scale. SET UP: The two vectors A " and B " are specified in the figure that accompanies the problem. EXECUTE: (a) The diagram for C = A + B " " " is given in Figure 1.32a. Measuring the length and angle of C " gives 9.0 mC = and an angle of 34θ = ° . (b) The diagram for −D = A B "" " is given in Figure 1.32b. Measuring the length and angle of D " gives 22 mD = and an angle of 250θ = ° . (c) ( )− − −A B = A + B " " , so − −A B " " has a magnitude of 9.0 m (the same as A + B " " ) and an angle with the x+ axis of 214° (opposite to the direction of A + B " " ). (d) ( )− − −B A = A B " "" " , so −B A "" has a magnitude of 22 m and an angle with the x+ axis of 70° (opposite to the direction of −A B " " ). EVALUATE: The vector −A " is equal in magnitude and opposite in direction to the vector A " . Figure 1.32 1.33. IDENTIFY: Since she returns to the starting point, the vectors sum of the four displacements must be zero. SET UP: Call the three given displacements A " , B " , and C " , and call the fourth displacement D " . 0A + B + C + D = " "" " . EXECUTE: The vector addition diagram is sketched in Figure 1.33. Careful measurement gives that D " is144 m, 41 south of west.# Units, Physical Quantities and Vectors 1-7 EVALUATE: D " is equal in magnitude and opposite in direction to the sum A + B + C " "" . Figure 1.33 1.34. IDENTIFY and SET UP: Use a ruler and protractor to draw the vectors described. Then draw the corresponding horizontal and vertical components. EXECUTE: (a) Figure 1.34 gives components 4.7 m, 8.1 m. (b) Figure 1.34 gives components 15.6 km,15.6 km− . (c) Figure 1.34 gives components 3.82 cm, 5.07 cm− . EVALUATE: The signs of the components depend on the quadrant in which the vector lies. Figure 1.34 1.35. IDENTIFY: For each vector V " , use that cosxV V θ= and sinyV V θ= , when θ is the angle V " makes with the x+ axis, measured counterclockwise from the axis. SET UP: For A " , 270.0θ = ° . For B " , 60.0θ = ° . For C " , 205.0θ = ° . For D " , 143.0θ = ° . EXECUTE: 0xA = , 8.00 myA = − . 7.50 mxB = , 13.0 myB = . 10.9 mxC = − , 5.07 myC = − . 7.99 mxD = − , 6.02 myD = . EVALUATE: The signs of the components correspond to the quadrant in which the vector lies. 1.36. IDENTIFY: tan y x A A θ = , for θ measured counterclockwise from the x+ -axis. SET UP: A sketch of xA , yA and A " tells us the quadrant in which A " lies. EXECUTE: (a) 1.00 mtan 0.500 2.00 m y X A θ A − = = = − . ( )1tan 0.500 360 26.6 333θ −= − = ° − ° = ° . (b) 1.00 mtan 0.500 2.00 m y x A θ A = = = . ( )1tan 0.500 26.6θ −= = ° . (c) 1.00 mtan 0.500 2.00 m y x A θ A = = = − − . ( )1tan 0.500 180 26.6 153θ −= − = ° − ° = ° . (d) 1.00 mtan 0.500 2.00 m y x A θ A − = = = − . ( )1tan 0.500 180 26.6 207θ −= = ° + ° = ° EVALUATE: The angles 26.6° and 207° have the same tangent. Our sketch tells us which is the correct value of θ . 1.37. IDENTIFY: Find the vector sum of the two forces. SET UP: Use components to add the two forces. Take the -directionx+ to be forward and the -directiony+ to be upward. 1-10 Chapter 1 (c) Similarly, ( )4 10 cm 1 30 cm 2 80 cm,. − . = . ( )3 75 cm 2 25 cm 6 00 cm. − . = . .2 2 (d) 2 2(2.80cm) ( 6.00cm) 6.62+ − = cm, 6.00arctan 295 2.80 −⎛ ⎞ = °⎜ ⎟ ⎝ ⎠ (which is 360 65° − ° ). EVALUATE: We can draw the vector addition diagram in each case and verify that our results are qualitatively correct. 1.43. IDENTIFY: Vector addition problem. ( ).− −A B = A + B " "" " SET UP: Find the x- and y-components of A " and .B " Then the x- and y-components of the vector sum are calculated from the x- and y-components of A " and .B " EXECUTE: cos(60.0 )xA A= ° (2.80 cm)cos(60.0 ) 1.40 cmxA = ° = + sin(60.0 )yA A= ° (2.80 cm)sin(60.0 ) 2.425 cmyA = ° = + cos( 60.0 )xB B= − ° (1.90 cm)cos( 60.0 ) 0.95 cmxB = − ° = + sin( 60.0 )yB B= − ° (1.90 cm)sin( 60.0 ) 1.645 cmyB = − ° = − Note that the signs of the components correspond to the directions of the component vectors. Figure 1.43a (a) Now let .= +R A B "" " 1.40 cm 0.95 cm 2.35 cm.x x xR A B= + = + + = + 2.425 cm 1.645 cm 0.78 cm.y y yR A B= + = + − = + 2 2 2 2(2.35 cm) (0.78 cm)x yR R R= + = + 2.48 cmR = 0.78 cmtan 0.3319 2.35 cm y x R R θ + = = = + + 18.4θ = ° Figure 1.43b EVALUATE: The vector addition diagram for = +R A B "" " is R " is in the 1st quadrant, with ,y xR R< in agreement with our calculation. Figure 1.43c Units, Physical Quantities and Vectors 1-11 (b) EXECUTE: Now let .−=R A B "" " 1.40 cm 0.95 cm 0.45 cm.x x xR A B= − = + − = + 2.425 cm 1.645 cm 4.070 cm.y y yR A B= − = + + = + 2 2 2 2(0.45 cm) (4.070 cm)x yR R R= + = + 4.09 cmR = 4.070 cmtan 9.044 0.45 cm y x R R θ = = = + 83.7θ = ° Figure 1.43d EVALUATE: The vector addition diagram for ( )−= +R A B "" " is R " is in the 1st quadrant, with ,x yR R< in agreement with our calculation. Figure 1.43e (c) EXECUTE: ( )− − −B A = A B " "" " −B A "" and −A B " " are equal in magnitude and opposite in direction. 4.09 cmR = and 83.7 180 264θ = ° + ° = ° Figure 1.43f 1-12 Chapter 1 EVALUATE: The vector addition diagram for ( )−= +R B A "" " is R " is in the 3rd quadrant, with ,x yR R< in agreement with our calculation. Figure 1.43g 1.44. IDENTIFY: The velocity of the boat relative to the earth, B/Ev" , the velocity of the water relative to the earth, W/Ev" , and the velocity of the boat relative to the water, B/Wv" , are related by B/E B/W W/Ev = v + v" " " . SET UP: W/E 5.0 km/h=v" , north and B/W 7.0 km/h=v" , west. The vector addition diagram is sketched in Figure 1.44. EXECUTE: 2 2 2 B/E W/E B/Wv v v= + and 2 2 B/E (5.0 km/h) (7.0 km/h) 8.6 km/hv = + = . W/E B/W 5.0 km/htan 7.0 km/h v v φ = = and 36φ = ° , north of west. EVALUATE: Since the two vectors we are adding are perpendicular we can use the Pythagorean theorem directly to find the magnitude of their vector sum. Figure 1.44 1.45. IDENTIFY: Let 625 NA = and 875 NB = . We are asked to find the vector C " such that 0A + B = C = " "" . SET UP: 0xA = , 625 NyA = − . (875 N)cos30 758 NxB = =° , (875 N)sin30 438 NyB = =° . EXECUTE: ( ) (0 758 N) 758 Nx x xC A B= − + = − + = − . ( ) ( 625 N 438 N) 187 Ny y yC A B= − + = − − + = + . Vector C " and its components are sketched in Figure 1.45. 2 2 781 Nx yC C C= + = . 187 Ntan 758 N y x C C φ = = and 13.9φ = ° . C " is at an angle of 13.9° above the x− -axis and therefore at an angle 180 13.9 166.1− =° ° ° counterclockwise from the -axisx+ . EVALUATE: A vector addition diagram for A + B + C " "" verifies that their sum is zero. Figure 1.45 Units, Physical Quantities and Vectors 1-15 EXECUTE: (a) ˆ ˆ4.00 3.00 ,= +A i j " ˆ ˆ5.00 2.00 ;= −B i j " 5.00,A = 5.39B = ˆ ˆ ˆ ˆ(4.00 3.00 ) (5.00 2.00 ) (4.00)(5.00) (3.00)( 2.00)⋅ = + ⋅ − = + − =A B i j i j " " 20.0 6.0 14.0.− = + (b) 14.0cos 0.519; (5.00)(5.39)AB φ ⋅ = = = A B " " 58.7 .φ = ° EVALUATE: The component of B " along A " is in the same direction as ,A " so the scalar product is positive and the angle φ is less than 90 .° 1.55. IDENTIFY: For all of these pairs of vectors, the angle is found from combining Equations (1.18) and (1.21), to give the angleφ as arccos arccos x x y yA B A B AB AB φ +⎛ ⎞ ⎛ ⎞⋅ = =⎜ ⎟ ⎜ ⎟ ⎝ ⎠⎝ ⎠ A B " " . SET UP: Eq.(1.14) shows how to obtain the components for a vector written in terms of unit vectors. EXECUTE: (a) 22, 40, 13,A B⋅ = − = =A B " " and so 22arccos 165 40 13 φ ⎛ ⎞− = = °⎜ ⎟ ⎝ ⎠ . (b) 60, 34, 136,A B⋅ = = =A B " " 60arccos 28 34 136 φ ⎛ ⎞= = °⎜ ⎟ ⎝ ⎠ . (c) 0⋅ =A B " " and 90φ = ° . EVALUATE: If 0⋅ >A B " " , 0 90φ≤ < ° . If 0⋅ <A B " " , 90 180φ ≤°< ° . If 0⋅ =A B " " , 90φ = ° and the two vectors are perpendicular. 1.56. IDENTIFY: cosAB φ⋅ =A B " " and sinAB φ=A× B " " , where φ is the angle between A " and B " . SET UP: Figure 1.56 shows A " and B " . The components A% of A " along B " and A⊥ of A " perpendicular to B " are shown in Figure 1.56a. The components of B% of B " along A " and B⊥ of B " perpendicular to A " are shown in Figure 1.56b. EXECUTE: (a) From Figures 1.56a and b, cosA A φ=% and cosB B φ=% . cosAB BA ABφ⋅ = = =A B % % " " . (b) sinA A φ⊥ = and sinB B φ⊥ = . sinAB BA ABφ ⊥ ⊥= = =A× B " " . EVALUATE: When A " and B " are perpendicular, A " has no component along B " and B " has no component along A " and 0⋅ =A B " " . When A " and B " are parallel, A " has no component perpendicular to B " and B " has no component perpendicular to A " and 0=A× B " " . Figure 1.56 1.57. IDENTIFY: A× D " " has magnitude sinAD φ . Its direction is given by the right-hand rule. SET UP: 180 53 127φ = − =° ° ° EXECUTE: 2(8.00 m)(10.0 m)sin127 63.9 m= =A× D " " ° . The right-hand rule says A× D " " is in the -directionz− (into the page). EVALUATE: The component of D " perpendicular to A " is sin53.0 7.00 mD D⊥ = =° . 263.9 mAD⊥= =A× D " " , which agrees with our previous result. 1.58. IDENTIFY: Target variable is the vector ,A× B " " expressed in terms of unit vectors. SET UP: We are given A " and B " in unit vector form and can take the vector product using Eq.(1.24). EXECUTE: ˆ ˆ4.00 3.00 ,= +A i j " ˆ ˆ5.00 2.00= −B i j " 1-16 Chapter 1 ( ) ( )ˆ ˆ ˆ ˆ4.00 3.00 5.00 2.00× = + × − =A B i j i j " " ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ20.0 8.00 15.0 6.00× − × + × − ×i i i j j i j j But ˆ ˆ ˆ ˆ 0× = × =i i j j and ˆ ˆ ˆ,× =i j k ˆ ˆ ˆ,× = −j i k so ( )ˆ ˆ ˆ8.00 15.0 23.0 .× = − + − = −A B k k k " " The magnitude of ×A B " " is 23.0. EVALUATE: Sketch the vectors A " and B " in a coordinate system where the xy-plane is in the plane of the paper and the z-axis is directed out toward you. Figure 1.58 By the right-hand rule ×A B " " is directed into the plane of the paper, in the -direction.z− This agrees with the above calculation that used unit vectors. 1.59. IDENTIFY: The right-hand rule gives the direction and Eq.(1.22) gives the magnitude. SET UP: 120.0φ = ° . EXECUTE: (a) The direction of A×B " " is into the page (the -directionz− ). The magnitude of the vector product is ( )( ) 2sin 2.80 cm 1.90 cm sin120 4.61 cmAB φ = =# . (b) Rather than repeat the calculations, Eq. (1.23) may be used to see that B × A "" has magnitude 4.61 cm2 and is in the -directionz+ (out of the page). EVALUATE: For part (a) we could use Eq. (1.27) and note that the only non-vanishing component is ( ) ( )2.80 cm cos60.0 1.90 cm sin 60z x y y xC A B A B= − = ° − ° ( ) ( ) 2 2.80 cm sin 60.0 1.90 cm cos60.0 4.61 cm− ° ° = − . This gives the same result. 1.60. IDENTIFY: Area is length times width. Do unit conversions. SET UP: 1 mi 5280 ft= . 31 ft 7.477 gal= . EXECUTE: (a) The area of one acre is 21 1 1 8 80 640mi mi mi ,× = so there are 640 acres to a square mile. (b) ( ) 22 21 mi 5280 ft1 acre 43,560 ft 640 acre 1 mi ⎛ ⎞ ⎛ ⎞× × =⎜ ⎟ ⎜ ⎟ ⎝ ⎠⎝ ⎠ (all of the above conversions are exact). (c) (1 acre-foot) ( )3 5 3 7.477 gal43,560 ft 3.26 10 gal, 1 ft ⎛ ⎞= × = ×⎜ ⎟ ⎝ ⎠ which is rounded to three significant figures. EVALUATE: An acre is much larger than a square foot but less than a square mile. A volume of 1 acre-foot is much larger than a gallon. 1.61. IDENTIFY: The density relates mass and volume. Use the given mass and density to find the volume and from this the radius. SET UP: The earth has mass 24 E 5.97 10 kgm = × and radius 6 E 6.38 10 mr = × . The volume of a sphere is 34 3V rπ= . 3 31.76 g/cm 1760 km/mρ = = . EXECUTE: (a) The planet has mass 25 E5.5 3.28 10 kgm m= = × . 25 22 3 3 3.28 10 kg 1.86 10 m 1760 kg/m mV ρ × = = = × . 1/ 31/ 3 22 3 7 43 3[1.86 10 m ] 1.64 10 m 1.64 10 km 4 4 Vr π π ⎛ ⎞×⎛ ⎞= = = × = ×⎜ ⎟⎜ ⎟ ⎝ ⎠ ⎝ ⎠ (b) E2.57r r= EVALUATE: Volume V is proportional to mass and radius r is proportional to 1/ 3V , so r is proportional to 1/ 3m . If the planet and earth had the same density its radius would be 1/ 3 E E(5.5) 1.8r r= . The radius of the planet is greater than this, so its density must be less than that of the earth. Units, Physical Quantities and Vectors 1-17 1.62. IDENTIFY and SET UP: Unit conversion. EXECUTE: (a) 91.420 10 cycles/s,f = × so 10 9 1 s 7.04 10 s 1.420 10 −= × × for one cycle. (b) 12 10 3600 s/h 5.11 10 cycles/h 7.04 10 s/cycle− = × × (c) Calculate the number of seconds in 4600 million 9years 4.6 10 y= × and divide by the time for 1 cycle: 9 7 6 10 (4.6 10 y)(3.156 10 s/y) 2.1 10 cycles 7.04 10 s/cycle 2 − × × = × × (d) The clock is off by 1 s in 5100,000 y 1 10 y,= × so in 94.60 10 y× it is off by 9 4 5 4.60 10(1 s) 4.6 10 s 1 10 ⎛ ⎞× = ×⎜ ⎟×⎝ ⎠ (about 13 h). EVALUATE: In each case the units in the calculation combine algebraically to give the correct units for the answer. 1.63. IDENTIFY: The number of atoms is your mass divided by the mass of one atom. SET UP: Assume a 70-kg person and that the human body is mostly water. Use Appendix D to find the mass of one H2O molecule: 27 2618 015 u 1 661 10 kg/u 2 992 10 kg/molecule. × . × = . × .2 2 EXECUTE: ( ) ( )26 2770 kg / 2 992 10 kg/molecule 2 34 10. × = . ×2 molecules. Each 2H O molecule has 3 atoms, so there are about 276 10× atoms. EVALUATE: Assuming carbon to be the most common atom gives 273 10× molecules, which is a result of the same order of magnitude. 1.64. IDENTIFY: Estimate the volume of each object. The mass m is the density times the volume. SET UP: The volume of a sphere of radius r is 34 3V rπ= . The volume of a cylinder of radius r and length l is 2V r lπ= . The density of water is 31000 kg m . EXECUTE: (a) Estimate the volume as that of a sphere of diameter 10 cm: 4 35.2 10 mV −= × . ( )( )( )3 4 30.98 1000 kg m 5.2 10 m 0.5 kgm −= × = . (b) Approximate as a sphere of radius 0.25 mr μ= (probably an over estimate): 20 36.5 10 mV −= × . ( )( )( )3 20 3 17 140.98 1000 kg m 6.5 10 m 6 10 kg 6 10 gm − − −= × = × = × . (c) Estimate the volume as that of a cylinder of length 1 cm and radius 3 mm: 2 7 32.8 10 mV r lπ −= = × . ( )( )( )3 7 3 40.98 1000 kg m 2.8 10 m 3 10 kg 0.3 gm − −= × = × = . EVALUATE: The mass is directly proportional to the volume. 1.65. IDENTIFY: Use the volume V and density ρ to calculate the mass M: ,so M Mρ V V ρ = = . SET UP: The volume of a cube with sides of length x is 3x . The volume of a sphere with radius R is 34 3 Rπ . EXECUTE: (a) 3 5 3 3 3 0.200 kg 2.54 10 m 7.86 10 kg/m x −= = × × . 22.94 10 m 2.94 cmx −= × = . (b) 3 5 34 2.54 10 m 3 Rπ −= × . 21.82 10 m 1.82 cmR −= × = . EVALUATE: 4 3 4.2π = , so a sphere with radius R has a greater volume than a cube whose sides have length R. 1.66. IDENTIFY: Estimate the volume of sand in all the beaches on the earth. The diameter of a grain of sand determines its volume. From the volume of one grain and the total volume of sand we can calculate the number of grains. SET UP: The volume of a sphere of diameter d is 31 6V dπ= . Consulting an atlas, we estimate that the continents have about 51.45 10 km× of coastline. Add another 25% of this for rivers and lakes, giving 51.82 10 km× of coastline. Assume that a beach extends 50 m beyond the water and that the sand is 2 m deep. 91 billion 1 10= × . EXECUTE: (a) The volume of sand is 8 10 3(1.82 10 m)(50 m)(2 m) 2 10 m× = × . The volume of a grain is 3 3 12 31 6 (0.2 10 m) 4 10 mV π − −= × = × . The number of grains is 10 3 21 12 3 2 10 m 5 10 4 10 m− × = × × . The number of grains of sand is about 2210 . (b) The number of stars is 9 9 22(100 10 )(100 10 ) 10× × = . The two estimates result in comparable numbers for these two quantities. 1-20 Chapter 1 (c) 2 2 x yA A A= + 2 2(3.03 cm) (8.10 cm) 8.65 cmA = + = 8.10 cmtan 2.67 3.03 cm y x A A θ = = = 69.5θ = ° Figure 1.71b EVALUATE: The A " we calculated agrees qualitatively with vector A " in the vector addition diagram in part (a). 1.72. IDENTIFY: Add the vectors using the method of components. SET UP: 0xA = , 8.00 myA = − . 7.50 mxB = , 13.0 myB = . 10.9 mxC = − , 5.07 myC = − . EXECUTE: (a) 3.4 mx x x xR A B C= + + = − . 0.07 my y y yR A B C= + + = − . 3.4 mR = . 0.07 mtan 3.4 m θ − = − . 1.2θ = ° below the -axisx− . (b) 18.4 mx x x xS C A B= − − = − . 10.1 my y y yS C A B= − − = − . 21.0 mS = . 10.1 mtan 18.4 m y x S S θ − = = − . 28.8θ = ° below the -axisx− . EVALUATE: The magnitude and direction we calculated for R " and S " agree with our vector diagrams. Figure 1.72 1.73. IDENTIFY: Vector addition. Target variable is the 4th displacement. SET UP: Use a coordinate system where east is in the -directionx+ and north is in the -direction.y+ Let ,A " ,B " and C " be the three displacements that are given and let D " be the fourth unmeasured displacement. Then the resultant displacement is .= + + +R A B C D " "" " " And since she ends up back where she started, 0.=R " 0 ,= + + +A B C D " "" " so ( )= − + +D A B C " "" " ( )x x x xD A B C= − + + and ( )y y y yD A B C= − + + EXECUTE: 180 m,xA = − 0yA = cos315 (210 m)cos315 148.5 mxB B= ° = ° = + sin315 (210 m)sin315 148.5 myB B= ° = ° = − cos60 (280 m)cos60 140 mxC C= ° = ° = + sin 60 (280 m)sin 60 242.5 myC C= ° = ° = + Figure 1.73a ( ) ( 180 m 148.5 m 140 m) 108.5 mx x x xD A B C= − + + = − − + + = − Units, Physical Quantities and Vectors 1-21 ( ) (0 148.5 m 242.5 m) 94.0 my y y yD A B C= − + + = − − + = − 2 2 x yD D D= + 2 2( 108.5 m) ( 94.0 m) 144 mD = − + − = 94.0 mtan 0.8664 108.5 m y x D D θ − = = = − 180 40.9 220.9θ = ° + ° = ° ( D " is in the third quadrant since both xD and yD are negative.) Figure 1.73b The direction of D " can also be specified in terms of 180 40.9 ;φ θ= − ° = ° D " is 41° south of west. EVALUATE: The vector addition diagram, approximately to scale, is Vector D " in this diagram agrees qualitatively with our calculation using components. Figure 1.73c 1.74. IDENTIFY: Solve for one of the vectors in the vector sum. Use components. SET UP: Use coordinates for which x+ is east and y+ is north. The vector displacements are: 2.00 km, 0 of east; 3.50 m, 45 south of east;= ° = °A B $ $ and 5.80 m, 0 east= °R $ EXECUTE: ( ) ( )( )5.80 km 2.00 km 3.50 km cos45 1.33 kmx x x xC R A B= − − = − − ° = ; y y y yC R A B= − − ( )( )0 km 0 km 3.50 km sin 45 2.47 km= − − − ° = ; ( ) ( )2 21.33 km 2.47 km 2.81 kmC = + = ; ( ) ( )1tan 2.47 km 1.33 km 61.7 north of east.θ − ⎡ ⎤= = °⎣ ⎦ The vector addition diagram in Figure 1.74 shows good qualitative agreement with these values. EVALUATE: The third leg lies in the first quadrant since its x and y components are both positive. Figure 1.74 1.75. IDENTIFY: The sum of the vector forces on the beam sum to zero, so their x components and their y components sum to zero. Solve for the components of F " . SET UP: The forces on the beam are sketched in Figure 1.75a. Choose coordinates as shown in the sketch. The 100-N pull makes an angle of 30.0 40.0 70.0+ =° ° ° with the horizontal. F " and the 100-N pull have been replaced by their x and y components. EXECUTE: (a) The sum of the x-components is equal to zero gives (100 N)cos70.0 0xF + =° and 34.2 NxF = − . The sum of the y-components is equal to zero gives (100 N)sin70.0 124 N 0yF + − =° and 30.0 NyF = + . F " and its components are sketched in Figure 1.75b. 2 2 45.5 Nx yF F F= + = . 30.0 Ntan 34.2 N y x F F φ = = and 41.3φ = ° . F " is directed at 41.3° above the x− -axis in Figure 1.75a. (b) The vector addition diagram is given in Figure 1.75c. F " determined from the diagram agrees with F " calculated in part (a) using components. 1-22 Chapter 1 EVALUATE: The vertical component of the 100 N pull is less than the 124 N weight so F " must have an upward component if all three forces balance. Figure 1.75 1.76. IDENTIFY: The four displacements return her to her starting point, so ( )−D = A + B + C " "" " , where A " , B " and C " are in the three given displacements and D " is the displacement for her return. START UP: Let x+ be east and y+ be north. EXECUTE: (a) ( ) ( ) ( )[ 147 km sin85 106 km sin167 166 km sin 235 ] 34.3 kmxD = − ° + ° + ° = − . ( ) ( ) ( )[ 147 km cos85 106 km cos167 166 km cos235 ] 185.7 kmyD = − ° + ° + ° = + . 2 2( 34.3 km) (185.7 km) 189 kmD = − + = . (b) The direction relative to north is 34.3 kmarctan 10.5 185.7 km φ ⎛ ⎞ = = °⎜ ⎟ ⎝ ⎠ . Since 0xD < and 0yD > , the direction of D " is 10.5° west of north. EVALUATE: The four displacements add to zero. 1.77. IDENTIFY and SET UP: The vector A " that connects points 1 1( , )x y and 2 2( , )x y has components 2 1xA x x= − and 2 1yA y y= − . EXECUTE: (a) Angle of first line is 1 200 20tan 42 . 210 10 θ − −⎛ ⎞= = °⎜ ⎟−⎝ ⎠ Angle of second line is 42 30 72 .° + ° = ° Therefore 10 250 cos 72 87X = + ° = , 20 250 sin 72 258Y = + ° = for a final point of (87,258). (b) The computer screen now looks something like Figure 1.77. The length of the bottom line is ( ) ( )2 2210 87 200 258 136− + − = and its direction is 1 258 200tan 25 210 87 − −⎛ ⎞ = °⎜ ⎟−⎝ ⎠ below straight left. EVALUATE: Figure 1.77 is a vector addition diagram. The vector first line plus the vector arrow gives the vector for the second line. Figure 1.77 Units, Physical Quantities and Vectors 1-25 (b) The vector diagram is sketched in Figure 1.83c. The final displacement D " from this diagram agrees with the vector D " calculated in part (a) using components. EVALUATE: Note that D " is the negative of the sum of A " , B " , and C " . Figure 1.83 1.84. IDENTIFY: If the vector from your tent to Joe’s is A " and from your tent to Karl’s is B " , then the vector from Joe’s tent to Karl’s is −B A "" . SET UP: Take your tent's position as the origin. Let x+ be east and y+ be north. EXECUTE: The position vector for Joe’s tent is ( ) ( )ˆ ˆ ˆ ˆ[21.0 m]cos 23 [21.0 m]sin 23 (19.33 m) (8.205 m) .° − ° −i j = i j The position vector for Karl's tent is ( ) ( )ˆ ˆ ˆ ˆ[32.0 m]cos 37 [32.0 m]sin 37 (25.56 m) (19.26 m) .° °i + j = i + j The difference between the two positions is ( ) ( )ˆ ˆ ˆ ˆ19.33 m 25.56 m 8.205 m 19.25 m (6.23 m) (27.46 m) .− − − − −i + j = i j The magnitude of this vector is the distance between the two tents: ( ) ( )2 26.23 m 27.46 m 28.2 mD = − + − = EVALUATE: If both tents were due east of yours, the distance between them would be 32.0 m 21.0 m 17.0 m− = . If Joe’s was due north of yours and Karl’s was due south of yours, then the distance between them would be 32.0 m 21.0 m 53.0 m+ = . The actual distance between them lies between these limiting values. 1.85. IDENTIFY: In Eqs.(1.21) and (1.27) write the components of A " and B " in terms of A, B, Aθ and Bθ . SET UP: From Appendix B, cos( ) cos cos sin sina b a b a b− = + and sin( ) sin cos cos sina b a b a b− = − . EXECUTE: (a) With 0z zA B= = , Eq.(1.21) becomes ( )( ) ( )( ) cos cos sin sinx x y y A B A BA B A B A θ B θ A θ B θ+ = + ( ) ( )cos cos sin sin cos cos x x y y A B A B A BA B A B AB θ θ θ θ AB θ θ AB φ+ = + = − = , where the expression for the cosine of the difference between two angles has been used. (b) With 0z zA B= = , ˆ zCC = k " and zC C= . From Eq.(1.27), ( )( ) ( )( ) cos sin sin cos x y y x A B A AC A B A B A θ B θ A θ B θ= − = − ( )cos sin sin cos sin sinA B A B B AC AB θ θ θ θ AB θ θ AB φ= − = − = , where the expression for the sine of the difference between two angles has been used. EVALUATE: Since they are equivalent, we may use either Eq.(1.18) or (1.21) for the scalar product and either (1.22) or (1.27) for the vector product, depending on which is the more convenient in a given application. 1.86. IDENTIFY: Apply Eqs.(1.18) and (1.22). SET UP: The angle between the vectors is 20 90 0 140 .° + ° = °°+3 EXECUTE: (a) Eq. (1.18) gives ( )( ) 23.60 m 2.40 m cos 140 6.62 m .⋅ = ° = −A B " " (b) From Eq.(1.22), the magnitude of the cross product is ( )( ) 23.60 m 2.40 m sin 140 5.55 m° = and the direction, from the right-hand rule, is out of the page (the -directionz+ ). EVALUATE: We could also use Eqs.(1.21) and (1.27), with the components of A " and B " . 1-26 Chapter 1 1.87. IDENTIFY: Compare the magnitude of the cross product, sinAB φ , to the area of the parallelogram. SET UP: The two sides of the parallelogram have lengths A and B. φ is the angle between A " and B " . EXECUTE: (a) The length of the base is B and the height of the parallelogram is sinA φ , so the area is sinAB φ . This equals the magnitude of the cross product. (b) The cross product A× B " " is perpendicular to the plane formed by A " and B " , so the angle is 90° . EVALUATE: It is useful to consider the special cases 0φ = ° , where the area is zero, and 90φ = ° , where the parallelogram becomes a rectangle and the area is AB. 1.88. IDENTIFY: Use Eq.(1.27) for the components of the vector product. SET UP: Use coordinates with the -axisx+ to the right, -axisy+ toward the top of the page, and -axisz+ out of the page. 0xA = , 0yA = and 3.50 cmzA = − . The page is 20 cm by 35 cm, so 20 cmxB = and 35 cmyB = . EXECUTE: ( ) ( ) ( )2 2122 cm , 70 cm , 0. x y z = = − =A × B A × B A × B " " "" " " EVALUATE: From the components we calculated the magnitude of the vector product is 2141 cm . 40.3 cmB = and 90φ = ° , so 2sin 141 cmAB φ = , which agrees. 1.89. IDENTIFY: A " and B " are given in unit vector form. Find A, B and the vector difference .−A B " " SET UP: 2.00 3.00 4.00 ,= − + +A i j k " " " " 3.00 1.00 3.00= + −B i j k " " "" Use Eq.(1.8) to find the magnitudes of the vectors. EXECUTE: (a) 2 2 2 2 2 2( 2.00) (3.00) (4.00) 5.38x y zA A A A= + + = − + + = 2 2 2 2 2 2(3.00) (1.00) ( 3.00) 4.36x y zB B B B= + + = + + − = (b) ˆ ˆ ˆ ˆ ˆ ˆ( 2.00 3.00 4.00 ) (3.00 1.00 3.00 )− = − + + − + −A B i j k i j k " " ˆ ˆ ˆ ˆ ˆ ˆ( 2.00 3.00) (3.00 1.00) (4.00 ( 3.00)) 5.00 2.00 7.00 .− = − − + − + − − = − + +A B i j k i j k " " (c) Let ,= −C A B " " " so 5.00,xC = − 2.00,yC = + 7.00zC = + 2 2 2 2 2 2( 5.00) (2.00) (7.00) 8.83x y zC C C C= + + = − + + = ( ),− = − −B A A B " "" " so −A B " " and −B A "" have the same magnitude but opposite directions. EVALUATE: A, B and C are each larger than any of their components. 1.90. IDENTIFY: Calculate the scalar product and use Eq.(1.18) to determine φ . SET UP: The unit vectors are perpendicular to each other. EXECUTE: The direction vectors each have magnitude 3 , and their scalar product is ( )( ) ( )( ) ( )( )1 1 1 1 1 1 1,+ − + − =2 so from Eq. (1.18) the angle between the bonds is 1 1arccos arccos 109 . 33 3 −⎛ ⎞ ⎛ ⎞= − = °⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠ EVALUATE: The angle between the two vectors in the bond directions is greater than 90° . 1.91. IDENTIFY: Use the relation derived in part (a) of Problem 1.92: 2 2 2 2 cos ,C A B AB φ= + + where φ is the angle between A " and B " . SET UP: cos 0φ = for 90φ = ° . cos 0φ < for 90 180φ< <° ° and cos 0φ > for 0 90φ< <° ° . EXECUTE: (a) If 2 2 2 , cos 0,C A B φ= + = and the angle between A " and B " is 90° (the vectors are perpendicular). (b) If 2 2 2, cos 0,C A B φ< + < and the angle between A " and B " is greater than 90° . (c) If 2 2 2 , cos 0,C A B φ> + > and the angle between A " and B " is less than 90 .° EVALUATE: It is easy to verify the expression from Problem 1.92 for the special cases 0φ = , where C A B= + , and for 180φ = ° , where C A B= − . 1.92. IDENTIFY: Let C = A + B " " " and calculate the scalar product ⋅C C " " . SET UP: For any vector V " , 2V⋅ =V V " " . cosAB φ⋅ =A B " " . EXECUTE: (a) Use the linearity of the dot product to show that the square of the magnitude of the sum A + B " " is ( ) ( ) 2 2 2 2 2 2 2 cos A B A B AB φ ⋅ = ⋅ + ⋅ + ⋅ + ⋅ = ⋅ + ⋅ + ⋅ = + + ⋅ = + + A+ B A+ B A A A B B A B B A A B B A B A B " " " " " " " " " "" " " " " " " " " " Units, Physical Quantities and Vectors 1-27 (b) Using the result of part (a), with ,A B= the condition is that 2 2 2 22 cos A A A A φ= + + , which solves for 1 2 2cos ,φ= + 1 2cos ,φ = − and 120 .φ = ° EVALUATE: The expression 2 2 2 2 cosC A B AB φ= + + is called the law of cosines. 1.93. IDENTIFY: Find the angle between specified pairs of vectors. SET UP: Use cos AB φ ⋅ = A B " " EXECUTE: (a) ˆA = k " (along line ab) ˆ ˆ ˆB = i + j + k " (along line ad ) 1,A = 2 2 21 1 1 3B = + + = ( )ˆ ˆ ˆ ˆ 1⋅ ⋅ =A B = k i + j + k " " So cos 1/ 3; AB φ ⋅ = = A B " " 54.7φ = ° (b) ˆ ˆ ˆA = i + j + k " (along line ad ) ˆ ˆB = j + k " (along line ac) 2 2 21 1 1 3;A = + + = 2 21 1 2B = + = ( ) ( )ˆ ˆ ˆ ˆ ˆ 1 1 2⋅ ⋅ = + =A B = i + j + k i + j " " So 2 2cos ; 3 2 6AB φ ⋅ = = = A B " " 35.3φ = ° EVALUATE: Each angle is computed to be less than 90 ,° in agreement with what is deduced from Fig. 1.43 in the textbook. 1.94. IDENTIFY: The cross product A× B " " is perpendicular to both A " and B " . SET UP: Use Eq.(1.27) to calculate the components of A× B " " . EXECUTE: The cross product is 6.00 11.00ˆ ˆ ˆ ˆ ˆ ˆ( 13.00) (6.00) ( 11.00) 13 (1.00) 13.00 13.00 ⎡ ⎤⎛ ⎞− − − −⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦ i + j+ k = i + j k . The magnitude of the vector in square brackets is 1.93, and so a unit vector in this direction is ˆ ˆ ˆ(1.00) (6.00 /13.00) (11.00/13.00) 1.93 ⎡ ⎤− − ⎢ ⎥ ⎢ ⎥⎣ ⎦ i + j k . The negative of this vector, ˆ ˆ ˆ(1.00) (6.00/13.00) (11.00 /13.00) 1.93 ⎡ ⎤− ⎢ ⎥ ⎢ ⎥⎣ ⎦ i j+ k , is also a unit vector perpendicular to A " and B " . EVALUATE: Any two vectors that are not parallel or antiparallel form a plane and a vector perpendicular to both vectors is perpendicular to this plane. 1.95. IDENTIFY and SET UP: The target variables are the components of .C " We are given A " and .B " We also know ⋅A C " " and ,⋅B C "" and this gives us two equations in the two unknowns xC and .yC EXECUTE: A " and C " are perpendicular, so 0.⋅ =A C " " 0,x x y yA C A C+ = which gives 5.0 6.5 0.x yC C− = 15.0,⋅ =B C "" so 3.5 7.0 15.0x yC C− + = We have two equations in two unknowns xC and .yC Solving gives 8.0xC = and 6.1yC = EVALUATE: We can check that our result does give us a vector C " that satisfies the two equations 0⋅ =A C " " and 15.0.⋅ =B C "" 1.96. IDENTIFY: Calculate the magnitude of the vector product and then use Eq.(1.22). SET UP: The magnitude of a vector is related to its components by Eq.(1.12). 1-30 Chapter 1 1.102. IDENTIFY: Define ˆ ˆ ˆA B CS = i + j+ k " . Show that 0⋅r S = "" if 0Ax By Cz+ + = . SET UP: Use Eq.(1.21) to calculate the scalar product. EXECUTE: ˆ ˆ ˆ ˆ ˆ ˆ( ) ( )x y z A B C Ax By Cz⋅ = + + ⋅ + + = + +r S i j k i j k "" If the points satisfy 0,Ax By Cz+ + = then 0⋅ =r S "" and all points r" are perpendicular to S " . The vector and plane are sketched in Figure 1.102. EVALUATE: If two vectors are perpendicular their scalar product is zero. Figure 1.102 2-1 MOTION ALONG A STRAIGHT LINE 2.1. IDENTIFY: The average velocity is av-x xv t Δ = Δ . SET UP: Let x+ be upward. EXECUTE: (a) av- 1000 m 63 m 197 m/s 4.75 sxv − = = (b) av- 1000 m 0 169 m/s 5.90 sxv − = = EVALUATE: For the first 1.15 s of the flight, av- 63 m 0 54.8 m/s 1.15 sxv − = = . When the velocity isn’t constant the average velocity depends on the time interval chosen. In this motion the velocity is increasing. 2.2. IDENTIFY: av-x xv t Δ = Δ SET UP: 513.5 days 1.166 10 s= × . At the release point, 65.150 10 mx = + × . EXECUTE: (a) 6 2 1 av- 6 5.150 10 m 4.42 m/s 1.166 10 sx x xv t − × = = = − Δ × (b) For the round trip, 2 1x x= and 0xΔ = . The average velocity is zero. EVALUATE: The average velocity for the trip from the nest to the release point is positive. 2.3. IDENTIFY: Target variable is the time tΔ it takes to make the trip in heavy traffic. Use Eq.(2.2) that relates the average velocity to the displacement and average time. SET UP: av-x xv t Δ = Δ so av-xx v tΔ = Δ and av- . x xt v Δ Δ = EXECUTE: Use the information given for normal driving conditions to calculate the distance between the two cities: av- (105 km/h)(1 h/60 min)(140 min) 245 km.xx v tΔ = Δ = = Now use av-xv for heavy traffic to calculate ;tΔ xΔ is the same as before: av- 245 km 3.50 h 3 h 70 km/hx xt v Δ Δ = = = = and 30 min. The trip takes an additional 1 hour and 10 minutes. EVALUATE: The time is inversely proportional to the average speed, so the time in traffic is (105/ 70)(140 m) 210 min.= 2.4. IDENTIFY: The average velocity is av-x xv t Δ = Δ . Use the average speed for each segment to find the time traveled in that segment. The average speed is the distance traveled by the time. SET UP: The post is 80 m west of the pillar. The total distance traveled is 200 m 280 m 480 m+ = . EXECUTE: (a) The eastward run takes time 200 m 40.0 s 5.0 m/s = and the westward run takes 280 m 70.0 s 4.0 m/s = . The average speed for the entire trip is 480 m 4.4 m/s 110.0 s = . (b) av- 80 m 0.73 m/s 110.0 sx xv t Δ − = = = − Δ . The average velocity is directed westward. 2 2-2 Chapter 2 EVALUATE: The displacement is much less than the distance traveled and the magnitude of the average velocity is much less than the average speed. The average speed for the entire trip has a value that lies between the average speed for the two segments. 2.5. IDENTIFY: When they first meet the sum of the distances they have run is 200 m. SET UP: Each runs with constant speed and continues around the track in the same direction, so the distance each runs is given by d vt= . Let the two runners be objects A and B. EXECUTE: (a) 200 mA Bd d+ = , so (6.20 m/s) (5.50 m/s) 200 mt t+ = and 200 m 17.1 s 11.70 m/s t = = . (b) (6.20 m/s)(17.1 s) 106 mA Ad v t= = = . (5.50 m/s)(17.1 s) 94 mB Bd v t= = = . The faster runner will be 106 m from the starting point and the slower runner will be 94 m from the starting point. These distances are measured around the circular track and are not straight-line distances. EVALUATE: The faster runner runs farther. 2.6. IDENTIFY: To overtake the slower runner the first time the fast runner must run 200 m farther. To overtake the slower runner the second time the faster runner must run 400 m farther. SET UP: t and 0x are the same for the two runners. EXECUTE: (a) Apply 0 0xx x v t− = to each runner: 0 f( ) (6.20 m/s)x x t− = and 0 s( ) (5.50 m/s)x x t− = . 0 f 0 s( ) ( ) 200 mx x x x− = − + gives (6.20 m/s) (5.50 m/s) 200 mt t= + and 200 m 286 s 6.20 m/s 5.50 m/s t = = − . 0 f( ) 1770 mx x− = and 0 s( ) 1570 mx x− = . (b) Repeat the calculation but now 0 f 0 s( ) ( ) 400 mx x x x− = − + . 572 st = . The fast runner has traveled 3540 m. He has made 17 full laps for 3400 m and 140 m past the starting line in this 18th lap. EVALUATE: In part (a) the fast runner will have run 8 laps for 1600 m and will be 170 m past the starting line in his 9th lap. 2.7. IDENTIFY: In time St the S-waves travel a distance S Sd v t= and in time Pt the P-waves travel a distance P Pd v t= . SET UP: S P 33 st t= + EXECUTE: S P 33 sd d v v = + . 1 1 33 s 3.5 km/s 6.5 km/s d ⎛ ⎞− =⎜ ⎟ ⎝ ⎠ and 250 kmd = . EVALUATE: The times of travel for each wave are S 71 st = and P 38 st = . 2.8. IDENTIFY: The average velocity is av-x xv t Δ = Δ . Use ( )x t to find x for each t. SET UP: (0) 0x = , (2.00 s) 5.60 mx = , and (4.00 s) 20.8 mx = EXECUTE: (a) av- 5.60 m 0 2.80 m/s 2.00 sxv − = = + (b) av- 20.8 m 0 5.20 m/s 4.00 sxv − = = + (c) av- 20.8 m 5.60 m 7.60 m/s 2.00 sxv − = = + EVALUATE: The average velocity depends on the time interval being considered. 2.9. (a) IDENTIFY: Calculate the average velocity using Eq.(2.2). SET UP: av-x xv t Δ = Δ so use ( )x t to find the displacement xΔ for this time interval. EXECUTE: 0 :t = 0x = 10.0 s:t = 2 2 3 3(2.40 m/s )(10.0 s) (0.120 m/s )(10.0 s) 240 m 120 m 120 m.x = − = − = Then av- 120 m 12.0 m/s. 10.0 sx xv t Δ = = = Δ (b) IDENTIFY: Use Eq.(2.3) to calculate ( )xv t and evaluate this expression at each specified t. SET UP: 22 3 .x dxv bt ct dt = = − EXECUTE: (i) 0 :t = 0xv = (ii) 5.0 s:t = 2 3 22(2.40 m/s )(5.0 s) 3(0.120 m/s )(5.0 s) 24.0 m/s 9.0 m/s 15.0 m/s.xv = − = − = (iii) 10.0 s:t = 2 3 22(2.40 m/s )(10.0 s) 3(0.120 m/s )(10.0 s) 48.0 m/s 36.0 m/s 12.0 m/s.xv = − = − = Motion Along a Straight Line 2-5 2.14. IDENTIFY: av- x x va t Δ = Δ . ( )xa t is the slope of the xv versus t graph. SET UP: 60 km/h 16.7 m/s= EXECUTE: (a) (i) 2 av- 16.7 m/s 0 1.7 m/s 10 sxa − = = . (ii) 2 av- 0 16.7 m/s 1.7 m/s 10 sxa − = = − . (iii) 0xvΔ = and av- 0xa = . (iv) 0xvΔ = and av- 0xa = . (b) At 20 st = , xv is constant and 0xa = . At 35 st = , the graph of xv versus t is a straight line and 2 av- 1.7 m/sx xa a= = − . EVALUATE: When av-xa and xv have the same sign the speed is increasing. When they have opposite sign the speed is decreasing. 2.15. IDENTIFY and SET UP: Use x dxv dt = and x x dva dt = to calculate ( )xv t and ( ).xa t EXECUTE: 22.00 cm/s (0.125 cm/s )x dxv t dt = = − 20.125 cm/sx x dva dt = = − (a) At 0,t = 50.0 cm,x = 2.00 cm/s,xv = 20.125 cm/s .xa = − (b) Set 0xv = and solve for t: 16.0 s.t = (c) Set 50.0 cmx = and solve for t. This gives 0t = and 32.0 s.t = The turtle returns to the starting point after 32.0 s. (d) Turtle is 10.0 cm from starting point when 60.0 cmx = or 40.0 cm.x = Set 60.0 cmx = and solve for t: 6.20 st = and 25.8 s.t = At 6.20 s,t = 1.23 cm/s.xv = + At 25.8 s,t = 1.23 cm/s.xv = − Set 40.0 cmx = and solve for t: 36.4 st = (other root to the quadratic equation is negative and hence nonphysical). At 36.4 s,t = 2.55 cm/s.xv = − (e) The graphs are sketched in Figure 2.15. Figure 2.15 EVALUATE: The acceleration is constant and negative. xv is linear in time. It is initially positive, decreases to zero, and then becomes negative with increasing magnitude. The turtle initially moves farther away from the origin but then stops and moves in the -direction.x− 2.16. IDENTIFY: Use Eq.(2.4), with 10 stΔ = in all cases. SET UP: xv is negative if the motion is to the right. EXECUTE: (a) ( ) ( )( ) ( ) 25.0 m/s 15.0 m/s / 10 s 1.0 m/s− = − (b) ( ) ( )( ) ( ) 215.0 m/s 5.0 m/s / 10 s 1.0 m/s− − − = − (c) ( ) ( )( ) ( ) 215.0 m/s 15.0 m/s / 10 s 3.0 m/s− − + = − EVALUATE: In all cases, the negative acceleration indicates an acceleration to the left. 2.17. IDENTIFY: The average acceleration is av- x x va t Δ = Δ SET UP: Assume the car goes from rest to 65 mi/h (29 m/s) in 10 s. In braking, assume the car goes from 65 mi/h to zero in 4.0 s. Let x+ be in the direction the car is traveling. EXECUTE: (a) 2 av- 29 m/s 0 2.9 m/s 10 sxa − = = (b) 2 av- 0 29 m/s 7.2 m/s 4.0 sxa − = = − 2-6 Chapter 2 (c) In part (a) the speed increases so the acceleration is in the same direction as the velocity. If the velocity direction is positive, then the acceleration is positive. In part (b) the speed decreases so the acceleration is in the direction opposite to the direction of the velocity. If the velocity direction is positive then the acceleration is negative, and if the velocity direction is negative then the acceleration direction is positive. EVALUATE: The sign of the velocity and of the acceleration indicate their direction. 2.18. IDENTIFY: The average acceleration is av- x x va t Δ = Δ . Use ( )xv t to find xv at each t. The instantaneous acceleration is x x dva dt = . SET UP: (0) 3.00 m/sxv = and (5.00 s) 5.50 m/sxv = . EXECUTE: (a) 2 av- 5.50 m/s 3.00 m/s 0.500 m/s 5.00 s x x va t Δ − = = = Δ (b) 3 3(0.100 m/s )(2 ) (0.200 m/s )x x dva t t dt = = = . At 0t = , 0xa = . At 5.00 st = , 21.00 m/sxa = . (c) Graphs of ( )xv t and ( )xa t are given in Figure 2.18. EVALUATE: ( )xa t is the slope of ( )xv t and increases at t increases. The average acceleration for 0t = to 5.00 st = equals the instantaneous acceleration at the midpoint of the time interval, 2.50 st = , since ( )xa t is a linear function of t. Figure 2.18 2.19. (a) IDENTIFY and SET UP: xv is the slope of the x versus t curve and xa is the slope of the xv versus t curve. EXECUTE: 0t = to 5 st = : x versus t is a parabola so xa is a constant. The curvature is positive so xa is positive. xv versus t is a straight line with positive slope. 0 0.xv = 5 st = to 15 st = : x versus t is a straight line so xv is constant and 0.xa = The slope of x versus t is positive so xv is positive. 15 st = to 25 s:t = x versus t is a parabola with negative curvature, so xa is constant and negative. xv versus t is a straight line with negative slope. The velocity is zero at 20 s, positive for 15 s to 20 s, and negative for 20 s to 25 s. 25 st = to 35 s:t = x versus t is a straight line so xv is constant and 0.xa = The slope of x versus t is negative so xv is negative. 35 st = to 40 s:t = x versus t is a parabola with positive curvature, so xa is constant and positive. xv versus t is a straight line with positive slope. The velocity reaches zero at 40 s.t = Motion Along a Straight Line 2-7 The graphs of ( )xv t and ( )xa t are sketched in Figure 2.19a. Figure 2.19a (b) The motions diagrams are sketched in Figure 2.19b. Figure 2.19b EVALUATE: The spider speeds up for the first 5 s, since xv and xa are both positive. Starting at 15 st = the spider starts to slow down, stops momentarily at 20 s,t = and then moves in the opposite direction. At 35 st = the spider starts to slow down again and stops at 40 s.t = 2.20. IDENTIFY: ( )x dxv t dt = and ( ) x x dva t dt = SET UP: 1( )n nd t nt dt −= for 1n ≥ . EXECUTE: (a) 2 6 5( ) (9.60 m/s ) (0.600 m/s )xv t t t= − and 2 6 4( ) 9.60 m/s (3.00 m/s )xa t t= − . Setting 0xv = gives 0t = and 2.00 st = . At 0t = , 2.17 mx = and 29.60 m/sxa = . At 2.00 st = , 15.0 mx = and 238.4 m/sxa = − . (b) The graphs are given in Figure 2.20. 2-10 Chapter 2 (b) 2 0 0 (6.67 ft/s )(19.9 s) 133 ft/s 90.5 mphx x xv v a t= + = + = = (c) 0 2 0 88.0 ft/s 3.32 s 26.5 ft/s x x x v vt a − − = = = − EVALUATE: The magnitude of the acceleration while braking is much larger than when speeding up. That is why it takes much longer to go from 0 to 60 mph than to go from 60 mph to 0. 2.29. IDENTIFY: The acceleration xa is the slope of the graph of xv versus t. SET UP: The signs of xv and of xa indicate their directions. EXECUTE: (a) Reading from the graph, at 4.0 st = , 2.7 cm/sxv = , to the right and at 7.0 st = , 1.3 cm/sxv = , to the left. (b) xv versus t is a straight line with slope 28.0 cm/s 1.3 cm/s 6.0 s − = − . The acceleration is constant and equal to 21.3 cm/s , to the left. It has this value at all times. (c) Since the acceleration is constant, 21 0 0 2x xx x v t a t− = + . For 0t = to 4.5 s, 2 21 0 2(8.0 cm/s)(4.5 s) ( 1.3 cm/s )(4.5 s) 22.8 cmx x− = + − = . For 0t = to 7.5 s, 2 21 0 2(8.0 cm/s)(7.5 s) ( 1.3 cm/s )(7.5 s) 23.4 cmx x− = + − = (d) The graphs of xa and x versus t are given in Fig. 2.29. EVALUATE: In part (c) we could have instead used 0 0 2 x xv vx x t+⎛ ⎞− = ⎜ ⎟ ⎝ ⎠ . Figure 2.29 2.30. IDENTIFY: Use the constant acceleration equations to find x, 0xv , xv and xa for each constant-acceleration segment of the motion. SET UP: Let x+ be the direction of motion of the car and let 0x = at the first traffic light. EXECUTE: (a) For 0t = to 8 st = : 0 0 20 m/s (8 s) 80 m 2 2 x xv vx t+ +⎛ ⎞ ⎛ ⎞= = =⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ . 20 20 m/s 2.50 m/s 8 s x x x v va t − = = = + . The car moves from 0x = to 80 mx = . The velocity xv increases linearly from zero to 20 m/s. The acceleration is a constant 22.50 m/s . Constant speed for 60 m: The car moves from 80 mx = to 140 mx = . xv is a constant 20 m/s. 0xa = . This interval starts at 8 st = and continues until 60 m 8 s 11 s 20 m/s t = + = . Slowing from 20 m/s until stopped: The car moves from 140 mx = to 180 mx = . The velocity decreases linearly from 20 m/s to zero. 0 0 2 x xv vx x t+⎛ ⎞− = ⎜ ⎟ ⎝ ⎠ gives 2(40 m) 4 s 20 m/s 0 t = = + . 2 2 0 02 ( )x x xv v a x x= + − gives 2 2(20.0 m/s) 5.00 m/s 2(40 m)xa − = = − This segment is from 11 st = to 15 st = . The acceleration is a constant 25.00 m/s− . The graphs are drawn in Figure 2.30a. (b) The motion diagram is sketched in Figure 2.30b. Motion Along a Straight Line 2-11 EVALUATE: When a! and v! are in the same direction, the speed increases ( 0t = to 8 st = ). When a! and v! are in opposite directions, the speed decreases ( 11 st = to 15 st = ). When 0a = the speed is constant 8 st = to 11 st = . Figure 2.30a-b 2.31. (a) IDENTIFY and SET UP: The acceleration xa at time t is the slope of the tangent to the xv versus t curve at time t. EXECUTE: At 3 s,t = the xv versus t curve is a horizontal straight line, with zero slope. Thus 0.xa = At 7 s,t = the xv versus t curve is a straight-line segment with slope 245 m/s 20 m/s 6.3 m/s . 9 s 5 s − = − Thus 26.3 m/s .xa = At 11 st = the curve is again a straight-line segment, now with slope 20 45 m/s 11.2 m/s . 13 s 9 s − − = − − Thus 211.2 m/s .xa = − EVALUATE: 0xa = when xv is constant, 0xa > when xv is positive and the speed is increasing, and 0xa < when xv is positive and the speed is decreasing. (b) IDENTIFY: Calculate the displacement during the specified time interval. SET UP: We can use the constant acceleration equations only for time intervals during which the acceleration is constant. If necessary, break the motion up into constant acceleration segments and apply the constant acceleration equations for each segment. For the time interval 0t = to 5 st = the acceleration is constant and equal to zero. For the time interval 5 st = to 9 st = the acceleration is constant and equal to 26.25 m/s . For the interval 9 st = to 13 st = the acceleration is constant and equal to 211.2 m/s .− EXECUTE: During the first 5 seconds the acceleration is constant, so the constant acceleration kinematic formulas can be used. 0 20 m/sxv = 0xa = 5 st = 0 ?x x− = 0 0xx x v t− = ( 0xa = so no 21 2 xa t term) 0 (20 m/s)(5 s) 100 m;x x− = = this is the distance the officer travels in the first 5 seconds. During the interval 5 st = to 9 s the acceleration is again constant. The constant acceleration formulas can be applied to this 4 second interval. It is convenient to restart our clock so the interval starts at time 0t = and ends at time 5 s.t = (Note that the acceleration is not constant over the entire 0t = to 9 st = interval.) 0 20 m/sxv = 26.25 m/sxa = 4 st = 0 100 mx = 0 ?x x− = 21 0 0 2x xx x v t a t− = + 2 21 0 2(20 m/s)(4 s) (6.25 m/s )(4 s) 80 m 50 m 130 m.x x− = + = + = Thus 0 130 m 100 m 130 m 230 m.x x− + = + = 2-12 Chapter 2 At 9 st = the officer is at 230 m,x = so she has traveled 230 m in the first 9 seconds. During the interval 9 st = to 13 st = the acceleration is again constant. The constant acceleration formulas can be applied for this 4 second interval but not for the whole 0t = to 13 st = interval. To use the equations restart our clock so this interval begins at time 0t = and ends at time 4 s.t = 0 45 m/sxv = (at the start of this time interval) 211.2 m/sxa = − 4 st = 0 230 mx = 0 ?x x− = 21 0 0 2x xx x v t a t− = + 2 21 0 2(45 m/s)(4 s) ( 11.2 m/s )(4 s) 180 m 89.6 m 90.4 m.x x− = + − = − = Thus 0 90.4 m 230 m 90.4 m 320 m.x x= + = + = At 13 st = the officer is at 320 m,x = so she has traveled 320 m in the first 13 seconds. EVALUATE: The velocity xv is always positive so the displacement is always positive and displacement and distance traveled are the same. The average velocity for time interval tΔ is av- / .xv x t= Δ Δ For 0t = to 5 s, av- 20 m/s.xv = For 0t = to 9 s, av- 26 m/s.xv = For 0t = to 13 s, av- 25 m/s.xv = These results are consistent with Fig. 2.33. 2.32. IDENTIFY: In each constant acceleration interval, the constant acceleration equations apply. SET UP: When xa is constant, the graph of xv versus t is a straight line and the graph of x versus t is a parabola. When 0xa = , xv is constant and x versus t is a straight line. EXECUTE: The graphs are given in Figure 2.32. EVALUATE: The slope of the x versus t graph is ( )xv t and the slope of the xv versus t graph is ( )xa t . Figure 2.32 2.33. (a) IDENTIFY: The maximum speed occurs at the end of the initial acceleration period. SET UP: 220.0 m/sxa = 15.0 min 900 st = = 0 0xv = ?xv = 0x x xv v a t= + EXECUTE: 2 40 (20.0 m/s )(900 s) 1.80 10 m/sxv = + = × (b) IDENTIFY: Use constant acceleration formulas to find the displacement .xΔ The motion consists of three constant acceleration intervals. In the middle segment of the trip 0xa = and 41.80 10 m/s,xv = × but we can’t directly find the distance traveled during this part of the trip because we don’t know the time. Instead, find the distance traveled in the first part of the trip (where 220.0 m/sxa = + ) and in the last part of the trip (where 220.0 m/sxa = − ). Subtract these two distances from the total distance of 83.84 10 m× to find the distance traveled in the middle part of the trip (where 0).xa = first segment SET UP: 0 ?x x− = 15.0 min 900 st = = 220.0 m/sxa = + 0 0xv = 21 0 0 2x xx x v t a t− = + EXECUTE: 2 2 6 31 0 20 (20.0 m/s )(900 s) 8.10 10 m 8.10 10 kmx x− = + = × = × second segment SET UP: 0 ?x x− = 15.0 min 900 st = = 220.0 m/sxa = − 4 0 1.80 10 m/sxv = × 21 0 0 2x xx x v t a t− = + EXECUTE: 2 2 6 31 0 2(1.80 10 s)(900 s) ( 20.0 m/s )(900 s) 8.10 10 m 8.10 10 kmx x 4− = × + − = × = × (The same distance as traveled as in the first segment.) Motion Along a Straight Line 2-15 EXECUTE: (a) 0 200 my y− = , 29.80 m/sya = , 0 0yv = . 2 2 0 02 ( )y y yv v a y y= + − gives 22(9.80 m/s )(200 m) 60 m/s 200 km/h 140 mi/hyv = = = = . (b) Raindrops actually have a speed of about 1 m/s as they strike the ground. (c) The actual speed at the ground is much less than the speed calculated assuming free-fall, so neglect of air resistance is a very poor approximation for falling raindrops. EVALUATE: In the absence of air resistance raindrops would land with speeds that would make them very dangerous. 2.39. IDENTIFY: Apply the constant acceleration equations to the motion of the flea. After the flea leaves the ground, ,ya g= downward. Take the origin at the ground and the positive direction to be upward. (a) SET UP: At the maximum height 0.yv = 0yv = 0 0.440 my y− = 29.80 m/sya = − 0 ?yv = 2 2 0 02 ( )y y yv v a y y= + − EXECUTE: 2 0 02 ( ) 2( 9.80 m/s )(0.440 m) 2.94 m/sy yv a y y= − − = − − = (b) SET UP: When the flea has returned to the ground 0 0.y y− = 0 0y y− = 0 2.94 m/syv = + 29.80 m/sya = − ?t = 21 0 0 2y yy y v t a t− = + EXECUTE: With 0 0y y− = this gives 0 2 2 2(2.94 m/s) 0.600 s. 9.80 m/s y y v t a = − = − = − EVALUATE: We can use 0y y yv v a t= + to show that with 0 2.94 m/s,yv = 0yv = after 0.300 s. 2.40. IDENTIFY: Apply constant acceleration equations to the motion of the lander. SET UP: Let y+ be positive. Since the lander is in free-fall, 21.6 m/sya = + . EXECUTE: 0 0.8 m/syv = , 0 5.0 my y− = , 21.6 m/sya = + in 2 2 0 02 ( )y y yv v a y y= + − gives 2 2 2 0 02 ( ) (0.8 m/s) 2(1.6 m/s )(5.0 m) 4.1 m/sy y yv v a y y= + − = + = . EVALUATE: The same descent on earth would result in a final speed of 9.9 m/s, since the acceleration due to gravity on earth is much larger than on the moon. 2.41. IDENTIFY: Apply constant acceleration equations to the motion of the meterstick. The time the meterstick falls is your reaction time. SET UP: Let y+ be downward. The meter stick has 0 0yv = and 29.80 m/sya = . Let d be the distance the meterstick falls. EXECUTE: (a) 21 0 0 2y yy y v t a t− = + gives 2 2(4.90 m/s )d t= and 24.90 m/s dt = . (b) 2 0.176 m 0.190 s 4.90 m/s t = = EVALUATE: The reaction time is proportional to the square of the distance the stick falls. 2.42. IDENTIFY: Apply constant acceleration equations to the vertical motion of the brick. SET UP: Let y+ be downward. 29.80 m/sya = EXECUTE: (a) 0 0yv = , 2.50 st = , 29.80 m/sya = . 2 2 21 1 0 0 2 2 (9.80 m/s )(2.50 s) 30.6 my yy y v t a t− = + = = . The building is 30.6 m tall. (b) 2 0 0 (9.80 m/s )(2.50 s) 24.5 m/sy y yv v a t= + = + = (c) The graphs of ya , yv and y versus t are given in Fig. 2.42. Take 0y = at the ground. 2-16 Chapter 2 EVALUATE: We could use either 0 0 2 y yv v y y t +⎛ ⎞ − = ⎜ ⎟ ⎝ ⎠ or 2 2 0 02 ( )y y yv v a y y= + − to check our results. Figure 2.42 2.43. IDENTIFY: When the only force is gravity the acceleration is 29.80 m/s , downward. There are two intervals of constant acceleration and the constant acceleration equations apply during each of these intervals. SET UP: Let y+ be upward. Let 0y = at the launch pad. The final velocity for the first phase of the motion is the initial velocity for the free-fall phase. EXECUTE: (a) Find the velocity when the engines cut off. 0 525 my y− = , 22.25 m/sya = + , 0 0yv = . 2 2 0 02 ( )y y yv v a y y= + − gives 22(2.25 m/s )(525 m) 48.6 m/syv = = . Now consider the motion from engine cut off to maximum height: 0 525 my = , 0 48.6 m/syv = + , 0yv = (at the maximum height), 29.80 m/sya = − . 2 2 0 02 ( )y y yv v a y y= + − gives 2 2 2 0 0 2 0 (48.6 m/s) 121 m 2 2( 9.80 m/s ) y y y v v y y a − − − = = = − and 121 m 525 m 646 my = + = . (b) Consider the motion from engine failure until just before the rocket strikes the ground: 0 525 my y− = − , 29.80 m/sya = − , 0 48.6 m/syv = + . 2 2 0 02 ( )y y yv v a y y= + − gives 2 2(48.6 m/s) 2( 9.80 m/s )( 525 m) 112 m/syv = − + − − = − . Then 0y y yv v a t= + gives 0 2 112 m/s 48.6 m/s 16.4 s 9.80 m/s y y y v v t a − − − = = = − . (c) Find the time from blast-off until engine failure: 0 525 my y− = , 0 0yv = , 22.25 m/sya = + . 21 0 0 2y yy y v t a t− = + gives 0 2 2( ) 2(525 m) 21.6 s 2.25 m/sy y yt a − = = = . The rocket strikes the launch pad 21.6 s 16.4 s 38.0 s+ = after blast off. The acceleration ya is 22.25 m/s+ from 0t = to 21.6 st = . It is 29.80 m/s− from 21.6 st = to 38.0 s . 0y y yv v a t= + applies during each constant acceleration segment, so the graph of yv versus t is a straight line with positive slope of 22.25 m/s during the blast-off phase and with negative slope of 29.80 m/s− after engine failure. During each phase 21 0 0 2y yy y v t a t− = + . The sign of ya determines the curvature of ( )y t . At 38.0 st = the rocket has returned to 0y = . The graphs are sketched in Figure 2.43. EVALUATE: In part (b) we could have found the time from 21 0 0 2y yy y v t a t− = + , finding yv first allows us to avoid solving for t from a quadratic equation. Figure 2.43 Motion Along a Straight Line 2-17 2.44. IDENTIFY: Apply constant acceleration equations to the vertical motion of the sandbag. SET UP: Take y+ upward. 29.80 m/sya = − . The initial velocity of the sandbag equals the velocity of the balloon, so 0 5.00 m/syv = + . When the balloon reaches the ground, 0 40.0 my y− = − . At its maximum height the sandbag has 0yv = . EXECUTE: (a) 0.250 st = : 2 2 21 1 0 0 2 2(5.00 m/s)(0.250 s) ( 9.80 m/s )(0.250 s) 0.94 my yy y v t a t− = + = + − = . The sandbag is 40.9 m above the ground. 2 0 5.00 m/s ( 9.80 m/s )(0.250 s) 2.55 m/sy y yv v a t= + = + + − = . 1.00 st = : 2 21 0 2(5.00 m/s)(1.00 s) ( 9.80 m/s )(1.00 s) 0.10 my y− = + − = . The sandbag is 40.1 m above the ground. 2 0 5.00 m/s ( 9.80 m/s )(1.00 s) 4.80 m/sy y yv v a t= + = + + − = − . (b) 0 40.0 my y− = − , 0 5.00 m/syv = , 29.80 m/sya = − . 21 0 0 2y yy y v t a t− = + gives 2 240.0 m (5.00 m/s) (4.90 m/s )t t− = − . 2 2(4.90 m/s ) (5.00 m/s) 40.0 m 0t t− − = and ( )21 5.00 ( 5.00) 4(4.90)( 40.0) s (0.51 2.90) s 9.80 t = ± − − − = ± . t must be positive, so 3.41 st = . (c) 2 0 5.00 m/s ( 9.80 m/s )(3.41 s) 28.4 m/sy y yv v a t= + = + + − = − (d) 0 5.00 m/syv = , 29.80 m/sya = − , 0yv = . 2 2 0 02 ( )y y yv v a y y= + − gives 2 2 2 0 0 2 0 (5.00 m/s) 1.28 m 2 2( 9.80 m/s ) y y y v v y y a − − − = = = − . The maximum height is 41.3 m above the ground. (e) The graphs of ya , yv , and y versus t are given in Fig. 2.44. Take 0y = at the ground . EVALUATE: The sandbag initially travels upward with decreasing velocity and then moves downward with increasing speed. Figure 2.44 2.45. IDENTIFY: The balloon has constant acceleration ,ya g= downward. (a) SET UP: Take the y+ direction to be upward. 2.00 s,t = 0 6.00 m/s,yv = − 29.80 m/s ,ya = − ?yv = EXECUTE: 2 0 6.00 m/s ( 9.80 m/s )(2.00 s) 25.5 m/sy y yv v a t= + = − + − = − (b) SET UP: 0 ?y y− = EXECUTE: 2 21 1 0 0 2 2( 6.00 m/s)(2.00 s) ( 9.80 m/s )(2.00 s) 31.6 my yy y v t a t 2− = + = − + − = − (c) SET UP: 0 10.0 m,y y− = − 0 6.00 m/s,yv = − 29.80 m/s ,ya = − ?yv = 2 2 0 02 ( )y y yv v a y y= + − EXECUTE: 2 2 2 0 02 ( ) ( 6.00 m/s) 2( 9.80 m/s )( 10.0 m) 15.2 m/sy y yv v a y y= − + − = − − + − − = − (d) The graphs are sketched in Figure 2.45. Figure 2.45 EVALUATE: The speed of the balloon increases steadily since the acceleration and velocity are in the same direction. 25.5 m/syv = when 0 31.6 m,y y− = so yv is less than this (15.2 m/s) when 0y y− is less (10.0 m). 2-20 Chapter 2 EVALUATE: We can verify that x x dva dt = and x dxv dt = . Figure 2.50 2.51. 2 xa At Bt= − with 31.50 m/sA = and 40.120 m/sB = (a) IDENTIFY: Integrate ( )xa t to find ( )xv t and then integrate ( )xv t to find ( ).x t SET UP: 0 0 t x x xv v a dt= + ∫ EXECUTE: 2 2 31 1 0 0 2 30 ( ) t x x xv v At Bt dt v At Bt= + − = + −∫ At rest at 0t = says that 0 0,xv = so 2 3 3 2 41 1 1 1 2 3 2 3(1.50 m/s ) (0.120 m/s )xv At Bt t t3= − = − 3 2 4 3(0.75 m/s ) (0.040 m/s )xv t t= − SET UP: 0 0 t xx x v dt− + ∫ EXECUTE: ( )2 3 3 41 1 1 1 0 02 3 6 120 t x x At Bt dt x At Bt= + − = + −∫ At the origin at 0t = says that 0 0,x = so 3 4 3 3 4 41 1 1 1 6 12 6 12(1.50 m/s ) (0.120 m/s )x At Bt t t= − = − 3 3 4 4(0.25 m/s ) (0.010 m/s )x t t= − EVALUATE: We can check our results by using them to verify that ( )x dxv t dt = and ( ) .x x dva t dt = (b) IDENTIFY and SET UP: At time t, when xv is a maximum, 0.xdv dt = (Since ,x x dva dt = the maximum velocity is when 0.xa = For earlier times xa is positive so xv is still increasing. For later times xa is negative and xv is decreasing.) EXECUTE: 0x x dva dt = = so 2 0At Bt− = One root is 0,t = but at this time 0xv = and not a maximum. The other root is 3 4 1.50 m/s 12.5 s 0.120 m/s At B = = = At this time 3 2 4 3(0.75 m/s ) (0.040 m/s )xv t t= − gives 3 2 4 3(0.75 m/s )(12.5 s) (0.040 m/s )(12.5 s) 117.2 m/s 78.1 m/s 39.1 m/s.xv = − = − = EVALUATE: For 12.5 s,t < 0xa > and xv is increasing. For 12.5 s,t > 0xa < and xv is decreasing. 2.52. IDENTIFY: ( )a t is the slope of the v versus t graph and the distance traveled is the area under the v versus t graph. SET UP: The v versus t graph can be approximated by the graph sketched in Figure 2.52. EXECUTE: (a) Slope 0 for 1.3 msa t= = ≥ . (b) max Area under - graphh v t= Triangle RectangleA A≈ + ( )1 (1.3 ms) 133 cm/s (2.5 ms 1.3 ms)(133 cm s) 2 ≈ + − 0.25 cm≈ (c) slopea = of v-t graph. 25133cm s(0.5 ms) (1.0 ms) 1.0 10 cm s 1.3ms a a≈ ≈ = × . (1.5 ms) 0 because the slope is zero.a = Motion Along a Straight Line 2-21 (d) areah = under v-t graph. ( ) 3 Triangle 1(0.5 ms) (0.5 ms) 33 cm/s 8.3 10 cm 2 h A −≈ = = × . 2 Triangle 1(1.0 ms) (1.0 ms)(100 cm s) 5.0 10 cm 2 h A −≈ = = × . ( )Triangle Rectangle 1(1.5 ms) (1.3 ms) 133 cm/s (0.2 ms)(1.33) 0.11 cm 2 h A A≈ + = = EVALUATE: The acceleration is constant until 1.3 mst = , and then it is zero. 2980 cm/sg = . The acceleration during the first 1.3 ms is much larger than this and gravity can be neglected for the portion of the jump that we are considering. Figure 2.52 2.53. (a) IDENTIFY and SET UP: The change in speed is the area under the xa versus t curve between vertical lines at 2.5 st = and 7.5 s.t = EXECUTE: This area is 2 21 2 (4.00 cm/s 8.00 cm/s )(7.5 s 2.5 s) 30.0 cm/s+ − = This acceleration is positive so the change in velocity is positive. (b) Slope of xv versus t is positive and increasing with t. The graph is sketched in Figure 2.53. Figure 2.53 EVALUATE: The calculation in part (a) is equivalent to av-( ) .x xv a tΔ = Δ Since xa is linear in t, av- 0( ) / 2.x x xa a a= + Thus 2 21 av- 2 (4.00 cm/s 8.00 cm/s )xa = + for the time interval 2.5 st = to 7.5 s.t = 2.54. IDENTIFY: The average speed is the total distance traveled divided by the total time. The elapsed time is the distance traveled divided by the average speed. SET UP: The total distance traveled is 20 mi. With an average speed of 8 mi/h for 10 mi, the time for that first 10 miles is 10 mi 1.25 h 8 mi/h = . EXECUTE: (a) An average speed of 4 mi/h for 20 mi gives a total time of 20 mi 5.0 h 4 mi/h = . The second 10 mi must be covered in 5.0 h 1.25 h 3.75 h− = . This corresponds to an average speed of 10 mi 2.7 mi/h 3.75 h = . (b) An average speed of 12 mi/h for 20 mi gives a total time of 20 mi 1.67 h 12 mi/h = . The second 10 mi must be covered in 1.67 h 1.25 h 0.42 h− = . This corresponds to an average speed of 10 mi 24 mi/h 0.42 h = . (c) An average speed of 16 mi/h for 20 mi gives a total time of 20 mi 1.25 h 16 mi/h = . But 1.25 h was already spent during the first 10 miles and the second 10 miles would have to be covered in zero time. This is not possible and an average speed of 16 mi/h for the 20-mile ride is not possible. EVALUATE: The average speed for the total trip is not the average of the average speeds for each 10-mile segment. The rider spends a different amount of time traveling at each of the two average speeds. 2.55. IDENTIFY: ( )x dxv t dt = and x x dva dt = . SET UP: 1( )n nd t nt dt −= , for 1n ≥ . EXECUTE: (a) 3 2 2( ) (9.00 m/s ) (20.0 m/s ) 9.00 m/sxv t t t= − + . 3 2( ) (18.0 m/s ) 20.0 m/sxa t t= − . The graphs are sketched in Figure 2.55. 2-22 Chapter 2 (b) The particle is instantaneously at rest when ( ) 0xv t = . 0 0xv = and the quadratic formula gives 21 (20.0 (20.0) 4(9.00)(9.00)) s 1.11 s 0.48 s 18.0 t = ± − = ± . 0.63 st = and 1.59 st = . These results agree with the -xv t graphs in part (a). (c) For 0.63 st = , 3 2 2(18.0 m/s )(0.63 s) 20.0 m/s 8.7 m/sxa = − = − . For 1.59 st = , 28.6 m/sxa = + . At 0.63 st = the slope of the -xv t graph is negative and at 1.59 st = it is positive, so the same answer is deduced from the ( )xv t graph as from the expression for ( )xa t . (d) ( )xv t is instantaneously not changing when 0xa = . This occurs at 2 3 20.0 m/s 1.11 s 18.0m/s t = = . (e) When the particle is at its greatest distance from the origin, 0xv = and 0xa < (so the particle is starting to move back toward the origin). This is the case for 0.63 st = , which agrees with the x-t graph in part (a) . At 0.63 st = , 2.45 mx = . (f) The particle’s speed is changing at its greatest rate when xa has its maximum magnitude. The -xa t graph in part (a) shows this occurs at 0t = and at 2.00 st = . Since xv is always positive in this time interval, the particle is speeding up at its greatest rate when xa is positive, and this is for 2.00 st = . The particle is slowing down at its greatest rate when xa is negative and this is for 0t = . EVALUATE: Since ( )xa t is linear in t, ( )xv t is a parabola and is symmetric around the point where ( )xv t has its minimum value ( 1.11 st = ). For this reason, the answer to part (d) is midway between the two times in part (c). Figure 2.55 2.56. IDENTIFY: The average velocity is av-x xv t Δ = Δ . The average speed is the distance traveled divided by the elapsed time. SET UP: Let x+ be in the direction of the first leg of the race. For the round trip, 0xΔ ≥ and the total distance traveled is 50.0 m. For each leg of the race both the magnitude of the displacement and the distance traveled are 25.0 m. EXECUTE: (a) av- 25.0 m 1.25 m/s 20.0 sx xv t Δ = = = Δ . This is the same as the average speed for this leg of the race. (b) av- 25.0 m 1.67 m/s 15.0 sx xv t Δ = = = Δ . This is the same as the average speed for this leg of the race. (c) 0xΔ = so av- 0xv = . (d) The average speed is 50.0 m 1.43 m/s 35.0 s = . EVALUATE: Note that the average speed for the round trip is not equal to the arithmetic average of the average speeds for each leg. 2.57. IDENTIFY: Use information about displacement and time to calculate average speed and average velocity. Take the origin to be at Seward and the positive direction to be west. (a) SET UP: distance traveledaverage speed time = EXECUTE: The distance traveled (different from the net displacement 0( )x x− ) is 76 km 34 km 110 km.+ = Find the total elapsed time by using 0 av-x x x xv t t Δ − = = Δ to find t for each leg of the journey. Seward to Auora: 0 av- 76 km 0.8636 h 88 km/hx x xt v − = = = Motion Along a Straight Line 2-25 EVALUATE: When xv and xa are both positive, the speed increases. When xv is positive and xa is negative, the speed decreases. Figure 2.61 2.62. IDENTIFY: Since light travels at constant speed, d ct= SET UP: The distance from the earth to the sun is 111.50 10 m× . The distance from the earth to the moon is 83.84 10 m× . 186,000 mi/sc = . EXECUTE: (a) 1 8 154365 d 24 h 3600 s(3.0 10 m/s)(1 y) 9.5 10 m 1 y 1 d 1 h d ct ⎛ ⎞⎛ ⎞⎛ ⎞= = × = ×⎜ ⎟⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠⎝ ⎠ (b) 8 9(3.0 10 m/s)(10 s) 0.30 md ct −= = × = (c) 11 8 1.5 10 m 500 s 8 33 min 3.0 10 m s dt . c × = = = = × (d) 8 8 2(3.84 10 m) 2.6 s 3.0 10 m s dt c × = = = × (e) 93 10 mi 16,100 s 4 5 h 186,000 mi s dt . c × = = = = EVALUATE: The speed of light is very large but it still takes light a measurable length of time to travel a large distance. 2.63. IDENTIFY: Speed is distance d divided by time t. The distance around a circular path is 2d Rπ= , where R is the radius of the circular path. SET UP: The radius of the earth is 6 E 6.38 10 mR = × . The earth rotates once in 1 day 86,400 s= . The radius of the earth’s orbit around the sun is 111.50 10 m× and the earth completes this orbit in 71 year 3.156 10 s= × . The speed of light in vacuum is 83.00 10 m/sc = × . EXECUTE: (a) 6 E2 2 (6.38 10 m) 464 m/s 86,400 s d Rv t t π π × = = = = . (b) 11 4 7 2 2 (1.50 10 m) 2.99 10 m/s 3.156 10 s Rv t π π × = = = × × . (c) The time for light to go around once is 6 E 8 2 2 (6.38 10 m) 0.1336 s c 3.00 10 m/s d Rt c π π × = = = = × . In 1.00 s light would go around the earth 1.00 s 7.49 times 0.1336 s = . EVALUATE: All these speeds are large compared to speeds of objects in our everyday experience. 2.64. IDENTIFY: When the graph of xv versus t is a straight line the acceleration is constant, so this motion consists of two constant acceleration segments and the constant acceleration equations can be used for each segment. For 0t = to 5.0 s, xv is positive and the ball moves in the x+ direction. For 5.0 st = to 20.0 s, xv is negative and the ball moves in the x− direction. The acceleration xa is the slope of the xv versus t graph. SET UP: For the 0t = to 5.0 st = segment, 0 0xv = and 30.0 m/sxv = . For the 5.0 st = to 20.0 st = segment, 0 20.0 m/sxv = − and 0xv = . 2-26 Chapter 2 EXECUTE: (a) For 0t = to 5.0 s, 0 0 0 30.0 m/s (5.0 m/s) 75.0 m 2 2 x xv vx x t+ +⎛ ⎞ ⎛ ⎞− = = =⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ . The ball travels a distance of 75.0 m. For 5.0 st = to 20.0 s, 0 20.0 m/s 0 (15.0 m/s) 150.0 m 2 x x − +⎛ ⎞− = = −⎜ ⎟ ⎝ ⎠ . The total distance traveled is 75.0 m 150.0 m 225.0 m+ = . (b) The total displacement is 0 75.0 m +( 150.0 m) 75.0 mx x− = − = − . The ball ends up 75.0 m in the negative x- direction from where it started. (c) For 0t = to 5.0 s, 230.0 m/s 0 6.00 m/s 5.0 sxa − = = . For 5.0 st = to 20.0 s, 20 ( 20.0 m/s) 1.33 m/s 15.0 sxa − − = = + . The graph of xa versus t is given in Figure 2.64. (d) The ball is in contact with the floor for a small but nonzero period of time and the direction of the velocity doesn't change instantaneously. So, no, the actual graph of ( )xv t is not really vertical at 5.00 s. EVALUATE: For 0t = to 5.0 s, both xv and xa are positive and the speed increases. For 5.0 st = to 20.0 s, xv is negative and xa is positive and the speed decreases. Since the direction of motion is not the same throughout, the displacement is not equal to the distance traveled. Figure 2.64 2.65. IDENTIFY and SET UP: Apply constant acceleration equations. Find the velocity at the start of the second 5.0 s; this is the velocity at the end of the first 5.0 s. Then find 0x x− for the first 5.0 s. EXECUTE: For the first 5.0 s of the motion, 0 0,xv = 5.0 s.t = 0x x xv v a t= + gives (5.0 s).x xv a= This is the initial speed for the second 5.0 s of the motion. For the second 5.0 s: 0 (5.0 s),x xv a= 5.0 s,t = 0 150 m.x x− = 21 0 0 2x xx x v t a t− = + gives 2 2150 m (25 s ) (12.5 s )x xa a= + and 24.0 m/sxa = Use this xa and consider the first 5.0 s of the motion: 2 2 21 1 0 0 2 20 (4.0 m/s )(5.0 s) 50.0 m.x xx x v t a t− = + = + = EVALUATE: The ball is speeding up so it travels farther in the second 5.0 s interval than in the first. In fact, 0x x− is proportional to t 2 since it starts from rest. If it goes 50.0 m in 5.0 s, in twice the time (10.0 s) it should go four times as far. In 10.0 s we calculated it went 50 m 150 m 200 m,+ = which is four times 50 m. 2.66. IDENTIFY: Apply 21 0 0 2x xx x v t a t− = + to the motion of each train. A collision means the front of the passenger train is at the same location as the caboose of the freight train at some common time. SET UP: Let P be the passenger train and F be the freight train. For the front of the passenger train 0 0x = and for the caboose of the freight train 0 200 mx = . For the freight train F 15.0 m/sv = and F 0a = . For the passenger train P 25.0 m/sv = and 2 P 0.100 m/sa = − . EXECUTE: (a) 21 0 0 2x xx x v t a t− = + for each object gives 21 P P P2x v t a t= + and F F200 mx v t= + . Setting P Fx x= gives 21 P P F2 200 mv t a t v t+ = + . 2 2(0.0500 m/s ) (10.0 m/s) 200 m 0t t− + = . The quadratic formula gives ( )21 10.0 (10.0) 4(0.0500)(200) s (100 77.5) s 0.100 t = + ± − = ± . The collision occurs at 100 s 77.5 s 22.5 st = − = . The equations that specify a collision have a physical solution (real, positive t), so a collision does occur. Motion Along a Straight Line 2-27 (b) 2 21 P 2(25.0 m/s)(22.5 s) ( 0.100 m/s )(22.5 s) 537 mx = + − = . The passenger train moves 537 m before the collision. The freight train moves (15.0 m/s)(22.5 s) 337 m= . (c) The graphs of Fx and Px versus t are sketched in Figure 2.66. EVALUATE: The second root for the equation for t, 177.5 st = is the time the trains would meet again if they were on parallel tracks and continued their motion after the first meeting. Figure 2.66 2.67. IDENTIFY: Apply constant acceleration equations to the motion of the two objects, you and the cockroach. You catch up with the roach when both objects are at the same place at the same time. Let T be the time when you catch up with the cockroach. SET UP: Take 0x = to be at the 0t = location of the roach and positive x to be in the direction of motion of the two objects. roach: 0 1.50 m/s,xv = 0,xa = 0 0,x = 1.20 m,x = t T= you: 0 0.80 m/s,xv = 0 0.90 m,x = − 1.20 m,x = ,t T= ?xa = Apply 21 0 0 2x xx x v t a t− = + to both objects: EXECUTE: roach: 1.20 m (1.50 m/s) ,T= so 0.800 s.T = you: 21 21.20 m ( 0.90 m) (0.80 m/s) xT a T− − = + 21 22.10 m (0.80 m/s)(0.800 s) (0.800 s)xa= + 22.10 m 0.64 m (0.320 s ) xa= + 24.6 m/s .xa = EVALUATE: Your final velocity is 0 4.48 m/s.x x xv v a t= + = Then 0 0 2.10 m, 2 x xv vx x t+⎛ ⎞− = =⎜ ⎟ ⎝ ⎠ which checks. You have to accelerate to a speed greater than that of the roach so you will travel the extra 0.90 m you are initially behind. 2.68. IDENTIFY: The insect has constant speed 15 m/s during the time it takes the cars to come together. SET UP: Each car has moved 100 m when they hit. EXECUTE: The time until the cars hit is 100 m 10 s 10 m/s = . During this time the grasshopper travels a distance of (15 m/s)(10 s) 150 m= . EVALUATE: The grasshopper ends up 100 m from where it started, so the magnitude of his final displacement is 100 m. This is less than the total distance he travels since he spends part of the time moving in the opposite direction. 2.69. IDENTIFY: Apply constant acceleration equations to each object. Take the origin of coordinates to be at the initial position of the truck, as shown in Figure 2.69a Let d be the distance that the auto initially is behind the truck, so 0(auto)x d= − and 0(truck) 0.x = Let T be the time it takes the auto to catch the truck. Thus at time T the truck has undergone a displacement 0 40.0 m,x x− = so is at 0 40.0 m 40.0 m.x x= + = The auto has caught the truck so at time T is also at 40.0 m.x = Figure 2.69a 2-30 Chapter 2 The car goes from 0 24.0 mx = − to 51.5 m.x = So 0 75.5 mx x− = for the car. Calculate the time it takes the car to travel this distance: 20.600 m/s ,xa = 0 0,xv = 0 75.5 m,x x− = ?t = 21 0 0 2x xx x v t a t− = + 0 2 2( ) 2(75.5 m) 15.86 s 0.600 m/sx x xt a − = = = It takes the car 15.9 s to pass the truck. (b) Need how far the car travels relative to the earth, so go now to coordinates fixed to the earth. In these coordinates 0 20.0 m/sxv = for the car. Take the origin to be at the initial position of the car. 0 20.0 m/s,xv = 20.600 m/s ,xa = 15.86 s,t = 0 ?x x− = 2 2 21 1 0 0 2 2(20.0 m/s)(15.86 s) (0.600 m/s )(15.86 s)x xx x v t a t− = + = + 0 317.2 m 75.5 m 393 m.x x− = + = (c) In coordinates fixed to the earth: 2 0 20.0 m/s (0.600 m/s )(15.86 s) 29.5 m/sx x xv v a t= + = + = EVALUATE: In 15.9 s the truck travels 0 (20.0 m/s)(15.86 s) 317.2 m.x x− = = The car travels 392.7 m 317.2 m 75 m− = farther than the truck, which checks with part (a). In coordinates attached to the truck, for the car 0 0,xv = 9.5 m/sxv = and in 15.86 s the car travels 0 0 75 m, 2 x xv vx x t+⎛ ⎞− = =⎜ ⎟ ⎝ ⎠ which checks with part (a). 2.74. IDENTIFY: The acceleration is not constant so the constant acceleration equations cannot be used. Instead, use ( ) x x dva t dt = and 0 0 ( ) t xx x v t dt= + ∫ . SET UP: 11 1 n nt dt t n += +∫ for 0n ≥ . EXECUTE: (a) 2 31 0 0 30 ( ) [ ] t x t x t dt x t tα β α β= + − = + −∫ . 0x = at 0t = gives 0 0x = and 3 3 31 3( ) (4.00 m/s) (0.667 m/s )x t t t t tα β= − = − . 3( ) 2 (4.00 m/s )x x dva t t t dt β= = − = − . (b) The maximum positive x is when 0xv = and 0xa < . 0xv = gives 2 0tα β− = and 3 4.00 m/s 1.41 s 2.00 m/s t α β = = = . At this t, xa is negative. For 1.41 st = , 3 3(4.00 m/s)(1.41 s) (0.667 m/s )(1.41 s) 3.77 mx = − = . EVALUATE: After 1.41 st = the object starts to move in the x− direction and goes to x = −∞ as t →∞ . 2.75. ( ) ,a t tα β= + with 22.00 m/sα = − and 33.00 m/sβ = (a) IDENTIFY and SET UP: Integrage ( )xa t to find ( )xv t and then integrate ( )xv t to find ( ).x t EXECUTE: 21 0 0 0 20 0 ( ) t t x x x x xv v a dt v dt v t tα β α β= + = + + = + +∫ ∫ 2 2 31 1 1 0 0 0 0 02 2 60 0 ( ) t t x x xx x v dt x v t t dt x v t t tα β α β= + = + + + = + + +∫ ∫ At 0,t = 0.x x= To have 0x x= at 1 4.00 st = requires that 2 31 1 0 1 1 12 6 0.xv t t tα β+ + = Thus 2 3 2 21 1 1 1 0 1 16 2 6 2(3.00 m/s )(4.00 s) ( 2.00 m/s )(4.00 s) 4.00 m/s.xv t tβ α= − − = − − − = − (b) With 0xv as calculated in part (a) and 4.00 s,t = 2 2 3 21 1 0 0 2 24.00 s ( 2.00 m/s )(4.00 s) (3.00 m/s )(4.00 s) 12.0 m/s.xv v t tα β= + + = − + − + = + EVALUATE: 0xa = at 0.67 s.t = For 0.67 s,t > 0.xa > At 0,t = the particle is moving in the -directionx− and is speeding up. After 0.67 s,t = when the acceleration is positive, the object slows down and then starts to move in the -directionx+ with increasing speed. Motion Along a Straight Line 2-31 2.76. IDENTIFY: Find the distance the professor walks during the time t it takes the egg to fall to the height of his head. SET UP: Let y+ be downward. The egg has 0 0yv = and 29.80 m/sya = . At the height of the professor’s head, the egg has 0 44.2 my y− = . EXECUTE: 21 0 0 2y yy y v t a t− = + gives 0 2 2( ) 2(44.2 m) 3.00 s 9.80 m/sy y yt a − = = = . The professor walks a distance 0 0 (1.20 m/s)(3.00 s) 3.60 mxx x v t− = = = . Release the egg when your professor is 3.60 m from the point directly below you. EVALUATE: Just before the egg lands its speed is 2(9.80 m/s )(3.00s) 29.4 m/s= . It is traveling much faster than the professor. 2.77. IDENTIFY: Use the constant acceleration equations to establish a relationship between maximum height and acceleration due to gravity and between time in the air and acceleration due to gravity. SET UP: Let y+ be upward. At the maximum height, 0yv = . When the rock returns to the surface, 0 0y y− = . EXECUTE: (a) 2 2 0 02 ( )y y yv v a y y= + − gives 21 02y ya H v= − , which is constant, so E E M Ma H a H= . 2 E M E 2 M 9.80 m/s 2.64 3.71 m/s aH H H H a ⎛ ⎞ ⎛ ⎞ = = =⎜ ⎟ ⎜ ⎟ ⎝ ⎠⎝ ⎠ . (b) 21 0 0 2y yy y v t a t− = + with 0 0y y− = gives 02y ya t v= − , which is constant, so E E M Ma T a T= . E M E M 2.64aT T T a ⎡ ⎤ = =⎢ ⎥ ⎣ ⎦ . EVALUATE: On Mars, where the acceleration due to gravity is smaller, the rocks reach a greater height and are in the air for a longer time. 2.78. IDENTIFY: Calculate the time it takes her to run to the table and return. This is the time in the air for the thrown ball. The thrown ball is in free-fall after it is thrown. Assume air resistance can be neglected. SET UP: For the thrown ball, let y+ be upward. 29.80 m/sya = − . 0 0y y− = when the ball returns to its original position. EXECUTE: (a) It takes her 5.50 m 2.20 s 2.50 m/s = to reach the table and an equal time to return. For the ball, 0 0y y− = , 4.40 st = and 29.80 m/sya = − . 21 0 0 2y yy y v t a t− = + gives 21 1 0 2 2 ( 9.80 m/s )(4.40 s) 21.6 m/sy yv a t= − = − − = . (b) Find 0y y− when 2.20 st = . 2 2 21 1 0 0 2 2(21.6 m/s)(2.20 s) ( 9.80 m/s )(2.20 s) 23.8 my yy y v t a t− = + = + − = EVALUATE: It takes the ball the same amount of time to reach its maximum height as to return from its maximum height, so when she is at the table the ball is at its maximum height. Note that this large maximum height requires that the act either be done outdoors, or in a building with a very high ceiling. 2.79. (a) IDENTIFY: Use constant acceleration equations, with ,ya g= downward, to calculate the speed of the diver when she reaches the water. SET UP: Take the origin of coordinates to be at the platform, and take the -directiony+ to be downward. 0 21.3 m,y y− = + 29.80 m/s ,ya = + 0 0yv = (since diver just steps off), ?yv = 2 2 0 02 ( )y y yv v a y y= + − EXECUTE: 2 02 ( ) 2(9.80 m/s )(31.3 m) 20.4 m/s.y yv a y y= + − = + = + We know that yv is positive because the diver is traveling downward when she reaches the water. The announcer has exaggerated the speed of the diver. EVALUATE: We could also use 21 0 0 2y yy y v t a t− = + to find 2.085 s.t = The diver gains 9.80 m/s of speed each second, so has 2(9.80 m/s )(2.085 s) 20.4 m/syv = = when she reaches the water, which checks. (b) IDENTIFY: Calculate the initial upward velocity needed to give the diver a speed of 25.0 m/s when she reaches the water. Use the same coordinates as in part (a). SET UP: 0 ?,yv = 25.0 m/s,yv = + 29.80 m/s ,ya = + 0 21.3 my y− = + 2 2 0 02 ( )y y yv v a y y= + − 2-32 Chapter 2 EXECUTE: 2 2 2 0 02 ( ) (25.0 m/s) 2(9.80 m/s )(21.3 m) 14.4 m/sy y yv v a y y= − − − = − − = − 0( yv is negative since the direction of the initial velocity is upward.) EVALUATE: One way to decide if this speed is reasonable is to calculate the maximum height above the platform it would produce: 0 14.4 m/s,yv = − 0yv = (at maximum height), 29.80 m/s ,ya = + 0 ?y y− = 2 2 0 02 ( )y y yv v a y y= + − 2 2 2 0 0 0 ( 14.4 s) 10.6 m 2 2( 9.80 m/s) y y y v v y y a − − − − = = = − + This is not physically attainable; a vertical leap of 10.6 m upward is not possible. 2.80. IDENTIFY: The flowerpot is in free-fall. Apply the constant acceleration equations. Use the motion past the window to find the speed of the flowerpot as it reaches the top of the window. Then consider the motion from the windowsill to the top of the window. SET UP: Let y+ be downward. Throughout the motion 29.80 m/sya = + . EXECUTE: Motion past the window: 0 1.90 my y− = , 0.420 st = , 29.80 m/sya = + . 21 0 0 2y yy y v t a t− = + gives 20 1 1 0 2 2 1.90 m (9.80 m/s )(0.420 s) 2.466 m/s 0.420 sy y y yv a t t − = − = − = . This is the velocity of the flowerpot when it is at the top of the window. Motion from the windowsill to the top of the window: 0 0yv = , 2.466 m/syv = , 29.80 m/sya = + . 2 2 0 02 ( )y y yv v a y y= + − gives 2 2 2 0 0 2 (2.466 m/s) 0 0.310 m 2 2(9.80 m/s ) y y y v v y y a − − − = = = . The top of the window is 0.310 m below the windowsill. EVALUATE: It takes the flowerpot 0 2 2.466 m/s 0.252 s 9.80 m/s y y y v v t a − = = = to fall from the sill to the top of the window. Our result says that from the windowsill the pot falls 0.310 m 1.90 m 2.21 m+ = in 0.252 s 0.420 s 0.672 s+ = . 2 2 21 1 0 0 2 2 (9.80 m/s )(0.672 s) 2.21 my yy y v t a t− = + = = , which checks. 2.81. IDENTIFY: For parts (a) and (b) apply the constant acceleration equations to the motion of the bullet. In part (c) neglect air resistance, so the bullet is free-fall. Use the constant acceleration equations to establish a relation between initial speed 0v and maximum height H. SET UP: For parts (a) and (b) let x+ be in the direction of motion of the bullet. For part (c) let y+ be upward, so ya g= − . At the maximum height, 0yv = . EXECUTE: (a) 0 0.700 mx x− = , 0 0xv = , 965 m/sxv = . 2 2 0 02 ( )x x xv v a x x= + − gives 2 2 2 5 20 0 (965 m/s) 0 6.65 10 m/s 2( ) 2(0.700 m) x x x v va x x − − = = = × − . 46.79 10xa g = × , so 4(6.79 10 )xa g= × . (b) 0 0 2 x xv vx x t+⎛ ⎞− = ⎜ ⎟ ⎝ ⎠ gives 0 0 2( ) 2(0.700 m) 1.45 ms 0 965 m/sx x x xt v v − = = = + + . (c) 2 2 0 02 ( )y y yv v a y y= + − and 0yv = gives 2 0 0 2y y v a y y = − − , which is constant. 2 2 01 02 1 2 v v H H = . 22 1 0102 2 2 1 2 01 01 / 4 vvH H H H v v ⎛ ⎞ ⎛ ⎞ = = =⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ . EVALUATE: 2 2 2 0 2 (965 m/s) 47.5 km 2 2( 9.80 m/s ) y y y v v H a − − = = = − . Rifle bullets fired vertically don't actually reach such a large height; it is not an accurate approximation to ignore air resistance. 2.82. IDENTIFY: Assume the firing of the second stage lasts a very short time, so the rocket is in free-fall after 25.0 s. The motion consists of two constant acceleration segments. SET UP: Let y+ be upward. After 25.0 st = , 29.80 m/sya = − . EXECUTE: (a) Find the height of the rocket at 25.0 st = : 0 0yv = , 23.50 m/sya = + , 25.0 st = . 2 2 31 1 0 0 2 2 (3.50 m/s)(25.0 s) 1.0938 10 my yy y v t a t− = + = = × . Find the displacement of the rocket from firing of the Motion Along a Straight Line 2-35 3.834 m 4.90 m 4 h h= + . Let 2h u= and solve for u. 21 4 3.834 m 4.90 m 0u u− − = . ( )22 3.834 ( 3.834) 4.90 mu = ± − + . Only the positive root is physical, so 16.52 mu = and 2 273 mh u= = , which rounds to 270 m. The building is 270 m tall. EVALUATE: With 273 mh = the total time of fall is 2 7.46 s y ht a = = . In 7.47 s 1.00 s 6.46 s− = Spider-Man falls a distance 2 21 0 2 (9.80 m/s )(6.46 s) 204 my y− = = . This leaves 69 m for the last 1.0 s of fall, which is / 4h . 2.88. IDENTIFY: Apply constant acceleration equations to the motion of the rock. Sound travels at constant speed. SET UP: Let fallt be the time for the rock to fall to the ground and let st be the time it takes the sound to travel from the impact point back to you. fall s 10.0 st t+ = . Both the rock and sound travel a distance d that is equal to the height of the cliff. Take y+ downward for the motion of the rock. The rock has 0 0yv = and 29.80 m/sya = . EXECUTE: (a) For the rock, 21 0 0 2y yy y v t a t− = + gives fall 2 2 9.80 m/s dt = . For the sound, s 10.0 s 330 m/s dt = = . Let 2 dα = . 20.00303 0.4518 10.0 0α α+ − = . 19.6α = and 384 md = . (b) You would have calculated 2 21 2 (9.80 m/s )(10.0 s) 490 md = = . You would have overestimated the height of the cliff. It actually takes the rock less time than 10.0 s to fall to the ground. EVALUATE: Once we know d we can calculate that fall 8.8 st = and s 1.2 st = . The time for the sound of impact to travel back to you is 12% of the total time and cannot be neglected. The rock has speed 86 m/s just before it strikes the ground. 2.89. (a) IDENTIFY: Let y+ be upward. The can has constant acceleration .ya g= − The initial upward velocity of the can equals the upward velocity of the scaffolding; first find this speed. SET UP: 0 15.0 m,y y− = − 3.25 s,t = 29.80 m/s ,ya = − 0 ?yv = EXECUTE: 21 0 0 2y yy y v t a t− = + gives 0 11.31 m/syv = Use this 0 yv in 0y y yv v a t= + to solve for :yv 20.5 m/syv = − (b) IDENTIFY: Find the maximum height of the can, above the point where it falls from the scaffolding: SET UP: 0,yv = 0 11.31 m/s,yv = + 29.80 m/s ,ya = − 0 ?y y− = EXECUTE: 2 2 0 02 ( )y y yv v a y y= + − gives 0 6.53 my y− = The can will pass the location of the other painter. Yes, he gets a chance. EVALUATE: Relative to the ground the can is initially traveling upward, so it moves upward before stopping momentarily and starting to fall back down. 2.90. IDENTIFY: Both objects are in free-fall. Apply the constant acceleration equations to the motion of each person. SET UP: Let y+ be downward, so 29.80 m/sya = + for each object. EXECUTE: (a) Find the time it takes the student to reach the ground: 0 180 my y− = , 0 0yv = , 29.80 m/sya = . 21 0 0 2y yy y v t a t− = + gives 0 2 2( ) 2(180 m) 6.06 s 9.80 m/sy y yt a − = = = . Superman must reach the ground in 6.06 s 5.00 s 1.06 s− = : 1.06 st = , 0 180 my y− = , 29.80 m/sya = + . 21 0 0 2y yy y v t a t− = + gives 20 1 1 0 2 2 180 m (9.80 m/s )(1.06 s) 165 m/s 1.06 sy y y yv a t t − = − = − = . Superman must have initial speed 0 165 m/sv = . (b) The graphs of y-t for Superman and for the student are sketched in Figure 2.90. (c) The minimum height of the building is the height for which the student reaches the ground in 5.00 s, before Superman jumps. 2 2 21 1 0 0 2 2 (9.80 m/s )(5.00 s) 122 my yy y v t a t− = + = = . The skyscraper must be at least 122 m high. 2-36 Chapter 2 EVALUATE: 165 m/s 369 mi/h= , so only Superman could jump downward with this initial speed. Figure 2.90 2.91. IDENTIFY: Apply constant acceleration equations to the motion of the rocket and to the motion of the canister after it is released. Find the time it takes the canister to reach the ground after it is released and find the height of the rocket after this time has elapsed. The canister travels up to its maximum height and then returns to the ground. SET UP: Let y+ be upward. At the instant that the canister is released, it has the same velocity as the rocket. After it is released, the canister has 29.80 m/sya = − . At its maximum height the canister has 0yv = . EXECUTE: (a) Find the speed of the rocket when the canister is released: 0 0yv = , 23.30 m/sya = , 0 235 my y− = . 2 2 0 02 ( )y y yv v a y y= + − gives 2 02 ( ) 2(3.30 m/s )(235 m) 39.4 m/sy yv a y y= − = = . For the motion of the canister after it is released, 0 39.4 m/syv = + , 29.80 m/sya = − , 0 235 my y− = − . 21 0 0 2y yy y v t a t− = + gives 2 2235 m (39.4 m/s) (4.90 m/s )t t− = − . The quadratic formula gives 12.0 st = as the positive solution. Then for the motion of the rocket during this 12.0 s, 2 2 21 1 0 0 2 2235 m (39.4 m/s)(12.0 s) (3.30 m/s )(12.0 s) 945 my yy y v t a t− = + = + + = . (b) Find the maximum height of the canister above its release point: 0 39.4 m/syv = + , 0yv = , 29.80 m/sya = − . 2 2 0 02 ( )y y yv v a y y= + − gives 2 2 2 0 0 2 0 (39.4 m/s) 79.2 m 2 2( 9.80 m/s ) y y y v v y y a − − − = = = − . After its release the canister travels upward 79.2 m to its maximum height and then back down 79.2 m 235 m+ to the ground. The total distance it travels is 393 m. EVALUATE: The speed of the rocket at the instant that the canister returns to the launch pad is 2 0 39.4 m/s (3.30 m/s )(12.0 s) 79.0 m/sy y yv v a t= + = + = . We can calculate its height at this instant by 2 2 0 02 ( )y y yv v a y y= + − with 0 0yv = and 79.0 m/syv = . 2 2 2 0 0 2 (79.0 m/s) 946 m 2 2(3.30 m/s ) y y y v v y y a − − = = = , which agrees with our previous calculation. 2.92. IDENTIFY: Both objects are in free-fall and move with constant acceleration 29.80 m/s , downward. The two balls collide when they are at the same height at the same time. SET UP: Let y+ be upward, so 29.80 m/sya = − for each ball. Let 0y = at the ground. Let ball A be the one thrown straight up and ball B be the one dropped from rest at height H. 0 0Ay = , 0By H= . EXECUTE: (a) 21 0 0 2y yy y v t a t− = + applied to each ball gives 21 0 2Ay v t gt= − and 21 2By H gt= − . A By y= gives 2 21 1 0 2 2v t gt H gt− = − and 0 Ht v = . (b) For ball A at its highest point, 0yAv = and 0y y yv v a t= + gives 0vt g = . Setting this equal to the time in part (a) gives 0 0 H v v g = and 2 0vH g = . EVALUATE: In part (a), using 0 Ht v = in the expressions for Ay and By gives 2 0 1 2A B gHy y H v ⎛ ⎞ = = −⎜ ⎟ ⎝ ⎠ . H must be less than 2 02v g in order for the balls to collide before ball A returns to the ground. This is because it takes ball A Motion Along a Straight Line 2-37 time 02vt g = to return to the ground and ball B falls a distance 2 2 01 2 2vgt g = during this time. When 2 02vH g = the two balls collide just as ball A reaches the ground and for H greater than this ball A reaches the ground before they collide. 2.93. IDENTIFY and SET UP: Use /xv dx dt= and /x xa dv dt= to calculate ( )xv t and ( )xa t for each car. Use these equations to answer the questions about the motion. EXECUTE: 2 ,Ax t tα β= + 2 ,A Ax dxv t dt α β= = + 2Ax Ax dva dt β= = 2 3,Bx t tγ δ= − 22 3 ,B Bx dxv t t dt γ δ= = − 2 6Bx Bx dva t dt γ δ= − − (a) IDENTIFY and SET UP: The car that initially moves ahead is the one that has the larger 0 .xv EXECUTE: At 0,t = Axv α= and 0.Bxv = So initially car A moves ahead. (b) IDENTIFY and SET UP: Cars at the same point implies .A Bx x= 2 2 3t t t tα β γ δ+ = − EXECUTE: One solution is 0,t = which says that they start from the same point. To find the other solutions, divide by t: 2t t tα β γ δ+ = − 2 ( ) 0t tδ β γ α+ − + = ( ) ( )2 21 1( ) ( ) 4 1.60 (1.60) 4(0.20)(2.60) 4.00 s 1.73 s 2 0.40 t β γ β γ δα δ = − − ± − − = + ± − = ± So A Bx x= for 0,t = 2.27 st = and 5.73 s.t = EVALUATE: Car A has constant, positive .xa Its xv is positive and increasing. Car B has 0 0xv = and xa that is initially positive but then becomes negative. Car B initially moves in the -directionx+ but then slows down and finally reverses direction. At 2.27 st = car B has overtaken car A and then passes it. At 5.73 s,t = car B is moving in the -directionx− as it passes car A again. (c) IDENTIFY: The distance from A to B is .B Ax x− The rate of change of this distance is ( ) .B Ad x x dt − If this distance is not changing, ( ) 0.B Ad x x dt − = But this says 0.Bx Axv v− = (The distance between A and B is neither decreasing nor increasing at the instant when they have the same velocity.) SET UP: Ax Bxv v= requires 22 2 3t t tα β γ δ+ = − EXECUTE: 23 2( ) 0t tδ β γ α+ − + = ( ) ( )2 21 12( ) 4( ) 12 3.20 4( 1.60) 12(0.20)(2.60) 6 1.20 t β γ β γ δα δ = − − ± − − = ± − − 2.667 s 1.667 s,t = ± so Ax Bxv v= for 1.00 st = and 4.33 s.t = EVALUATE: At 1.00 s,t = 5.00 m/s.Ax Bxv v= = At 4.33 s,t = 13.0 m/s.Ax Bxv v= = Now car B is slowing down while A continues to speed up, so their velocities aren’t ever equal again. (d) IDENTIFY and SET UP: Ax Bxa a= requires 2 2 6 tβ γ δ= − EXECUTE: 2 2 3 2.80 m/s 1.20 m/s 2.67 s. 3 3(0.20 m/s ) t γ β δ − − = = = EVALUATE: At 0,t = ,Bx Axa a> but Bxa is decreasing while Axa is constant. They are equal at 2.67 st = but for all times after that .Bx Axa a< 2.94. IDENTIFY: The apple has two segments of motion with constant acceleration. For the motion from the tree to the top of the grass the acceleration is g, downward and the apple falls a distance H h− . For the motion from the top of the grass to the ground the acceleration is a, upward, the apple travels downward a distance h, and the final speed is zero. SET UP: Let y+ be upward and let 0y = at the ground. The apple is initially a height H h+ above the ground. EXECUTE: (a) Motion from 0y H h= + to y H= : 0y y H− = − , 0 0yv = , ya g= − . 2 2 0 02 ( )y y yv v a y y= + − gives 2yv gH= − . The speed of the apple is 2gH as it enters the grass. 2-40 Chapter 2 EXECUTE: (a) Let the height be h and denote the 1.30-s interval as ;tΔ the simultaneous equations 2 21 2 1 2 3 2, ( )h gt h g t t= = − Δ can be solved for t. Eliminating h and taking the square root, 3 , 2 t t t = − Δ and , 1 2/3 tt Δ = − and substitution into 21 2h gt= gives 246 m.h = (b) The above method assumed that 0t > when the square root was taken. The negative root (with 0)tΔ = gives an answer of 2.51 m, clearly not a “cliff”. This would correspond to an object that was initially near the bottom of this “cliff” being thrown upward and taking 1.30 s to rise to the top and fall to the bottom. Although physically possible, the conditions of the problem preclude this answer. EVALUATE: For the first two-thirds of the distance, 0 164 my y− = , 0 0yv = , and 29.80 m/sya = . 02 ( ) 56.7 m/sy yv a y y= − = . Then for the last third of the distance, 0 82.0 my y− = , 0 56.7 m/syv = and 29.80 m/sya = . 21 0 0 2y yy y v t a t− = + gives 2 2(4.90 m/s ) (56.7 m/s) 82.0 m 0t t+ − = . ( )21 56.7 (56.7) 4(4.9)(82.0) s 1.30 s 9.8 t = − + + = , as required. 3-1 MOTION IN TWO OR THREE DIMENSIONS 3.1. IDENTIFY and SET UP: Use Eq.(3.2), in component form. EXECUTE: ( ) 2 1 av 2 1 5.3 m 1.1 m 1.4 m/s 3.0 s 0x x x xv t t t Δ − − = = = = Δ − − ( ) 2 1 av 2 1 0.5 m 3.4 m 1.3 m/s 3.0 s 0y y y yv t t t Δ − − − = = = = − Δ − − EVALUATE: Our calculation gives that avv ! is in the 4th quadrant. This corresponds to increasing x and decreasing y. 3.2. IDENTIFY: Use Eq.(3.2), written in component form. The distance from the origin is the magnitude of r! . SET UP: At time 1t , 1 1 0x y= = . EXECUTE: (a) av-( )Δ ( 3.8m s)(12.0 s) 45.6 mxx v t= = − = − and av-( )Δ (4.9m s)(12.0 s) 58.8 myy v t= = = . (b) 2 2 2 2( 45.6 m) (58.8 m) 74.4 m.r x y= + = − + = EVALUATE: Δr! is in the direction of avv ! . Therefore, xΔ is negative since av-xv is negative and yΔ is positive since av-yv is positive. 3.3. (a) IDENTIFY and SET UP: From r! we can calculate x and y for any t. Then use Eq.(3.2), in component form. EXECUTE: ( ) ( )2 2 ˆ ˆ4.0 cm 2.5 cm/s 5.0 cm/st t⎡ ⎤= + +⎣ ⎦r i j! At 0,t = ( ) ˆ4.0 cm .=r i! At 2.0 s,t = ( ) ( )ˆ ˆ14.0 cm 10.0 cm .= +r i j! ( )av 10.0 cm 5.0 cm/s. 2.0 sx xv t Δ = = = Δ ( )av 10.0 cm 5.0 cm/s. 2.0 sy yv t Δ = = = Δ ( ) ( ) av av 1.3 m/stan 0.9286 1.4 m/s y x v v α − = = = − 360 42.9 317α = ° − ° = ° ( ) ( )2 2 av av avx y v v v= + 2 2 av (1.4 m/s) ( 1.3 m/s) 1.9 m/sv = + − = Figure 3.1 3 3-2 Chapter 3 ( ) ( )2 2 av av av 7.1 cm/s x y v v v= + = ( ) ( ) av av tan 1.00y x v v α = = 45 .θ = ° Figure 3.3a EVALUATE: Both x and y increase, so avv ! is in the 1st quadrant. (b) IDENTIFY and SET UP: Calculate r! by taking the time derivative of ( ).tr! EXECUTE: ( ) ( )2 ˆ ˆ5.0 cm/s 5.0 cm/sd t dt ⎡ ⎤= = +⎣ ⎦ rv i j !! 0 :t = 0,xv = 5.0 cm/s;yv = 5.0 cm/sv = and 90θ = ° 1.0 s:t = 5.0 cm/s,xv = 5.0 cm/s;yv = 7.1 cm/sv = and 45θ = ° 2.0 s:t = 10.0 cm/s,xv = 5.0 cm/s;yv = 11 cm/sv = and 27θ = ° (c) The trajectory is a graph of y versus x. 2 24.0 cm (2.5 cm/s ) ,x t= + (5.0 cm/s)y t= For values of t between 0 and 2.0 s, calculate x and y and plot y versus x. Figure 3.3b EVALUATE: The sketch shows that the instantaneous velocity at any t is tangent to the trajectory. 3.4. IDENTIFY: d dtv = r/! ! . This vector will make a 45° -angle with both axes when its x- and y-components are equal. SET UP: 1( )n nd t nt dt −= . EXECUTE: 2ˆ ˆ2 3bt ctv = i + j! . x yv v= gives 2 3t b c= . EVALUATE: Both components of v! change with t. 3.5. IDENTIFY and SET UP: Use Eq.(3.8) in component form to calculate ( )av x a and ( )av .ya Motion in Two or Three Dimensions 3-5 v! is always tangent to the path; v! at 2.0 st = shown in part (c) is tangent to the path at this t, conforming to this general rule. a! is constant and in the -direction;y− the direction of v! is turning toward the -direction.y− 3.8. IDENTIFY: The component ⊥a ! of a! perpendicular to the path is related to the change in direction of v! and the component a" ! of a! parallel to the path is related to the change in the magnitude of v! . SET UP: When the speed is increasing, a" ! is in the direction of v! and when the speed is decreasing, a" ! is opposite to the direction of v! . When v is constant, a" is zero and when the path is a straight line, a⊥ is zero. EXECUTE: The acceleration vectors in each case are sketched in Figure 3.8a-c. EVALUATE: ⊥a ! is toward the center of curvature of the path. Figure 3.8a-c 3.9. IDENTIFY: The book moves in projectile motion once it leaves the table top. Its initial velocity is horizontal. SET UP: Take the positive y-direction to be upward. Take the origin of coordinates at the initial position of the book, at the point where it leaves the table top. x-component: 0,xa = 0 1.10 m/s,xv = 0.350 st = y-component: 29.80 m/s ,ya = − 0 0,yv = 0.350 st = Figure 3.9a Use constant acceleration equations for the x and y components of the motion, with 0xa = and .ya g= − EXECUTE: (a) 0 ?y y− = 2 2 21 1 0 0 2 20 ( 9.80 m/s )(0.350 s) 0.600 m.y yy y v t a t− = + = + − = − The table top is 0.600 m above the floor. (b) 0 ?x x− = 21 0 0 2 (1.10 m/s)(0.350 s) 0 0.358 m.x xx x v t a t− = + = + = (c) 0 1.10 m/sx x xv v a t= + = (The x-component of the velocity is constant, since 0.)xa = 2 0 0 ( 9.80 m/s )(0.350 s) 3.43 m/sy y yv v a t= + = + − = − 2 2 3.60 m/sx yv v v= + = 3.43 m/stan 3.118 1.10 m/s y x v v α − = = = − 72.2α = − ° Direction of v! is 72.2° below the horizontal Figure 3.9b 3-6 Chapter 3 (d) The graphs are given in Figure 3.9c Figure 3.9c EVALUATE: In the x-direction, 0xa = and xv is constant. In the y-direction, 29.80 m/sya = − and yv is downward and increasing in magnitude since ya and yv are in the same directions. The x and y motions occur independently, connected only by the time. The time it takes the book to fall 0.600 m is the time it travels horizontally. 3.10. IDENTIFY: The bomb moves in projectile motion. Treat the horizontal and vertical components of the motion separately. The vertical motion determines the time in the air. SET UP: The initial velocity of the bomb is the same as that of the helicopter. Take y+ downward, so 0xa = , 29.80 m/sya = + , 0 60.0 m/sxv = and 0 0yv = . EXECUTE: (a) 21 0 0 2y yy y v t a t− = + with 0 300 my y− = gives 0 2 2( ) 2(300 m) 7.82 s 9.80 m/sy y yt a − = = = . (b) The bomb travels a horizontal distance 21 0 0 2 (60.0 m/s)(7.82 s) 470 mx xx x v t a t− = + = = . (c) 0 60.0 m/sx xv v= = . 2 0 (9.80 m/s )(7.82 s) 76.6 m/sy y yv v a t= + = = . (d) The graphs are given in Figure 3.10. (e) Because the airplane and the bomb always have the same x-component of velocity and position, the plane will be 300 m directly above the bomb at impact. EVALUATE: The initial horizontal velocity of the bomb doesn’t affect its vertical motion. Figure 3.10 3.11. IDENTIFY: Each object moves in projectile motion. SET UP: Take y+ to be downward. For each cricket, 0xa = and 29.80 m/sya = + . For Chirpy, 0 0 0x yv v= = . For Milada, 0 0.950 m/sxv = , 0 0yv = EXECUTE: Milada's horizontal component of velocity has no effect on her vertical motion. She also reaches the ground in 3.50 s. 21 0 0 2 (0.950 m/s)(3.50 s) 3.32 mx xx x v t a t− = + = = EVALUATE: The x and y components of motion are totally separate and are connected only by the fact that the time is the same for both. 3.12. IDENTIFY: The person moves in projectile motion. She must travel 1.75 m horizontally during the time she falls 9.00 m vertically. SET UP: Take y+ downward. 0xa = , 29.80 m/sya = + . 0 0xv v= , 0 0yv = . EXECUTE: Time to fall 9.00 m: 21 0 0 2y yy y v t a t− = + gives 0 2 2( ) 2(9.00 m) 1.36 s 9.80 m/sy y yt a − = = = . Speed needed to travel 1.75 m horizontally during this time: 21 0 0 2x xx x v t a t− = + gives 0 0 0 1.75 m 1.29 m/s 1.36 sx x xv v t − = = = = . EVALUATE: If she increases her initial speed she still takes 1.36 s to reach the level of the ledge, but has traveled horizontally farther than 1.75 m. 3.13. IDENTIFY: The car moves in projectile motion. The car travels 21.3 m 1.80 m 19.5 m− = downward during the time it travels 61.0 m horizontally. SET UP: Take y+ to be downward. 0xa = , 29.80 m/sya = + . 0 0xv v= , 0 0yv = . Motion in Two or Three Dimensions 3-7 EXECUTE: Use the vertical motion to find the time in the air: 21 0 0 2y yy y v t a t− = + gives 0 2 2( ) 2(19.5 m) 1.995 s 9.80 m/sy y yt a − = = = Then 21 0 0 2x xx x v t a t− = + gives 0 0 0 61.0 m 30.6 m/s 1.995 sx x xv v t − = = = = . (b) 30.6m sxv = since 0xa = . 0 19.6m sy y yv v a t= + = − . 2 2 36.3m sx yv v v= + = . EVALUATE: We calculate the final velocity by calculating its x and y components. 3.14. IDENTIFY: The marble moves with projectile motion, with initial velocity that is horizontal and has magnitude 0v . Treat the horizontal and vertical motions separately. If 0v is too small the marble will land to the left of the hole and if 0v is too large the marble will land to the right of the hole. SET UP: Let x+ be horizontal to the right and let y+ be upward. 0 0xv v= , 0 0yv = , 0xa = , 29.80 m/sya = − EXECUTE: Use the vertical motion to find the time it takes the marble to reach the height of the level ground; 0 2.75 my y− = − . 21 0 0 2y yy y v t a t− = + gives 0 2 2( ) 2( 2.75 m) 0.749 s 9.80 m/sy y yt a − − = = = − . The time does not depend on 0v . Minimum 0 :v 0 2.00 mx x− = , 0.749 st = . 21 0 0 2x xx x v t a t− = + gives 0 0 2.00 m 2.67 m/s 0.749 s x xv t − = = = . Maximum 0v : 0 3.50 mx x− = and 0 3.50 m 4.67 m/s 0.749 s v = = . EVALUATE: The horizontal and vertical motions are independent and are treated separately. Their only connection is that the time is the same for both. 3.15. IDENTIFY: The ball moves with projectile motion with an initial velocity that is horizontal and has magnitude 0v . The height h of the table and 0v are the same; the acceleration due to gravity changes from 2 E 9.80 m/sg = on earth to Xg on planet X. SET UP: Let x+ be horizontal and in the direction of the initial velocity of the marble and let y+ be upward. 0 0xv v= , 0 0yv = , 0xa = , ya g= − , where g is either Eg or Xg . EXECUTE: Use the vertical motion to find the time in the air: 0y y h− = − . 21 0 0 2y yy y v t a t− = + gives 2ht g = . Then 21 0 0 2x xx x v t a t− = + gives 0 0 0 2 x hx x v t v g − = = . 0x x D− = on earth and 2.76D on Planet X. 0 0( ) 2x x g v h− = , which is constant, so E X2.76D g D g= . 2E X E2 0.131 1.28 m/s (2.76) gg g= = = . EVALUATE: On Planet X the acceleration due to gravity is less, it takes the ball longer to reach the floor, and it travels farther horizontally. 3.16. IDENTIFY: The football moves in projectile motion. SET UP: Let y+ be upward. 0xa = , ya g= − . At the highest point in the trajectory, 0yv = . EXECUTE: (a) 0y y yv v a t= + . The time t is 0 2 16.0m s 1.63 s 9.80m s yv g = = . (b) Different constant acceleration equations give different expressions but the same numerical result: 2 021 1 02 2 13.1 m 2 y y v gt v t g = = = . (c) Regardless of how the algebra is done, the time will be twice that found in part (a), or 3.27 s (d) 0xa = , so 0 0 (20.0 m s)(3.27 s) 65 3 mxx x v t .− = = = . (e) The graphs are sketched in Figure 3.16. 3-10 Chapter 3 EVALUATE: When the shot returns to its initial height, 9.32 m/syv = − . The shot continues to accelerate downward as it travels downward 1.81 m to the ground and the magnitude of yv at the ground is larger than 9.32 m/s. Figure 3.20 3.21. IDENTIFY: Take the origin of coordinates at the point where the quarter leaves your hand and take positive y to be upward. The quarter moves in projectile motion, with 0,xa = and .ya g= − It travels vertically for the time it takes it to travel horizontally 2.1 m. Figure 3.21 (a) SET UP: Use the horizontal (x-component) of motion to solve for t, the time the quarter travels through the air: ?,t = 0 2.1 m,x x− = 0 3.2 m/s,xv = 0xa = 21 0 0 02 ,x x xx x v t a t v t− = + = since 0xa = EXECUTE: 0 0 2.1 m 0.656 s 3.2 m/sx x xt v − = = = SET UP: Now find the vertical displacement of the quarter after this time: 0 ?,y y− = 29.80 m/s ,ya = − 0 5.54 m/s,yv = + 0.656 st = 21 0 0 2y yy y v t a t− + + EXECUTE: 2 21 0 2(5.54 m/s)(0.656 s) ( 9.80 m/s )(0.656 s) 3.63 m 2.11 m 1.5 m.y y− = + − = − = (b) SET UP: ?,yv = 0.656 s,t = 29.80 m/s ,ya = − 0 5.54 m/syv = + 0y y yv v a t= + EXECUTE: 25.54 m/s ( 9.80 m/s )(0.656 s) 0.89 m/s.yv = + − = − EVALUATE: The minus sign for yv indicates that the y-component of v! is downward. At this point the quarter has passed through the highest point in its path and is on its way down. The horizontal range if it returned to its original height (it doesn’t!) would be 3.6 m. It reaches its maximum height after traveling horizontally 1.8 m, so at 0 2.1 mx x− = it is on its way down. 3.22. IDENTIFY: Use the analysis of Example 3.10. SET UP: From Example 3.10, 0 0cos dt v α = and 21 dart 0 0 2( sin )y v t gtα= − . EXECUTE: Substituting for t in terms of d in the expression for darty gives dart 0 2 2 0 0 tan . 2 cos gdy d v α α ⎛ ⎞ = −⎜ ⎟ ⎝ ⎠ Using the given values for d and 0α to express this as a function of 0v , 2 2 2 0 26.62 m s(3.00 m) 0.90 .y v ⎛ ⎞ = −⎜ ⎟ ⎝ ⎠ (a) 0 12.0 m/sv = gives 2.14 my = . (b) 0 8.0 m/sv = gives 1.45 my = . 0 0 0cos (6.4 m/s)cos60xv v α= = ° 0 3.20 m/sxv = 0 0 0sin (6.4 m/s)sin 60yv v α= = ° 0 5.54 m/syv = Motion in Two or Three Dimensions 3-11 (c) 0 4.0 m/sv = gives 2.29 my = − . In this case, the dart was fired with so slow a speed that it hit the ground before traveling the 3-meter horizontal distance. EVALUATE: For (a) and (d) the trajectory of the dart has the shape shown in Figure 3.26 in the textbook. For (c) the dart moves in a parabola and returns to the ground before it reaches the x-coordinate of the monkey. 3.23. IDENTIFY: Take the origin of coordinates at the roof and let the -directiony+ be upward. The rock moves in projectile motion, with 0xa = and .ya g= − Apply constant acceleration equations for the x and y components of the motion. SET UP: 0 0 0cos 25.2 m/sxv v α= = 0 0 0sin 16.3 m/syv v α= = Figure 3.23a (a) At the maximum height 0.yv = 29.80 m/s ,ya = − 0,yv = 0 16.3 m/s,yv = + 0 ?y y− = 2 2 0 02 ( )y y yv v a y y= + − EXECUTE: 2 2 2 0 0 2 0 (16.3 m/s) 13.6 m 2 2( 9.80 m/s ) y y y v v y y a − − − = = = + − (b) SET UP: Find the velocity by solving for its x and y components. 0 25.2 m/sx xv v= = (since 0xa = ) ?,yv = 29.80 m/s ,ya = − 0 15.0 my y− = − (negative because at the ground the rock is below its initial position), 0 16.3 m/syv = 2 2 0 02 ( )y y yv v a y y= + − 2 0 02 ( )y y yv v a y y= − + − ( yv is negative because at the ground the rock is traveling downward.) EXECUTE: 2 2(16.3 m/s) 2( 9.80 m/s )( 15.0 m) 23.7 m/syv = − + − − = − Then 2 2 2 2(25.2 m/s) ( 23.7 m/s) 34.6 m/s.x yv v v= + = + − = (c) SET UP: Use the vertical motion (y-component) to find the time the rock is in the air: ?,t = 23.7 m/syv = − (from part (b)), 29.80 m/s ,ya = − 0 16.3 m/syv = + EXECUTE: 0 2 23.7 m/s 16.3 m/s 4.08 s 9.80 m/s y y y v v t a − − − = = = + − SET UP: Can use this t to calculate the horizontal range: 4.08 s,t = 0 25.2 m/s,xv = 0,xa = 0 ?x x− = EXECUTE: 21 0 0 2 (25.2 m/s)(4.08 s) 0 103 mx xx x v t a t− = + = + = (d) Graphs of x versus t, y versus t, xv versus t, and yv versus t: Figure 3.23b 3-12 Chapter 3 EVALUATE: The time it takes the rock to travel vertically to the ground is the time it has to travel horizontally. With 0 16.3 m/syv = + the time it takes the rock to return to the level of the roof ( 0)y = is 02 / 3.33 s.yt v g= = The time in the air is greater than this because the rock travels an additional 15.0 m to the ground. 3.24. IDENTIFY: Consider the horizontal and vertical components of the projectile motion. The water travels 45.0 m horizontally in 3.00 s. SET UP: Let y+ be upward. 0xa = , 29.80 m/sya = − . 0 0 0cosxv v θ= , 0 0 0sinyv v θ= . EXECUTE: (a) 21 0 0 2x xx x v t a t− = + gives 0 0 0(cos )x x v tθ− = and 0 45.0 mcos 0.600 (25.0 m/s)(3.00 s) θ = = ; 0 53.1θ = ° (b) At the highest point 0 (25.0 m/s)cos53.1 15.0 m/sx xv v= = ° = , 0yv = and 2 2 15.0 m/sx yv v v= + = . At all points in the motion, 29.80 m/sa = downward. (c) Find 0y y− when 3 00st = . : 2 2 21 1 0 0 2 2(25.0 m/s)(sin53.1 )(3.00 s) ( 9.80 m/s )(3.00 s) 15.9 my yy y v t a t− = + = ° + − = 0 15.0 m/sx xv v= = , 2 0 (25.0 m/s)(sin53.1 ) (9.80m/s )(3.00 s) 9.41 m/sy y yv v a t= + = ° − = − , and 2 2 2 2(15.0 m/s) ( 9.41 m/s) 17.7 m/sx yv v v= + = + − = EVALUATE: The acceleration is the same at all points of the motion. It takes the water 0 2 20.0 m/s 2.04 s 9.80 m/s y y v t a = − = − = − to reach its maximum height. When the water reaches the building it has passed its maximum height and its vertical component of velocity is downward. 3.25. IDENTIFY and SET UP: The stone moves in projectile motion. Its initial velocity is the same as that of the balloon. Use constant acceleration equations for the x and y components of its motion. Take y+ to be upward. EXECUTE: (a) Use the vertical motion of the rock to find the initial height. 6.00 s,t = 0 20.0 m/s,yv = + 29.80 m/s ,ya = + 0 ?y y− = 21 0 0 2y yy y v t a t− = + gives 0 296 my y− = (b) In 6.00 s the balloon travels downward a distance 0 (20.0 s)(6.00 s) 120 m.y y− = = So, its height above ground when the rock hits is 296 m 120 m 176 m.− = (c) The horizontal distance the rock travels in 6.00 s is 90.0 m. The vertical component of the distance between the rock and the basket is 176 m, so the rock is 2 2(176 m) (90 m) 198 m+ = from the basket when it hits the ground. (d) (i) The basket has no horizontal velocity, so the rock has horizontal velocity 15.0 m/s relative to the basket. Just before the rock hits the ground, its vertical component of velocity is 0y y yv v a t= + = 220.0 m/s (9.80 m/s )(6.00 s) 78.8 m/s,+ = downward, relative to the ground. The basket is moving downward at 20.0 m/s, so relative to the basket the rock has downward component of velocity 58.8 m/s. (e) horizontal: 15.0 m/s; vertical: 78.8 m/s EVALUATE: The rock has a constant horizontal velocity and accelerates downward 3.26. IDENTIFY: The shell moves as a projectile. To just clear the top of the cliff, the shell must have 0 25.0 my y− = when it has 0 60.0 mx x− = . SET UP: Let y+ be upward. 0xa = , ya g= − . 0 0 cos43xv v= ° , 0 0 sin 43yv v= ° . EXECUTE: (a) horizontal motion: 0 0 0 60.0 m so ( cos43 )xx x v t t v t − = = ° . vertical motion: 2 2 21 1 0 0 02 2 gives 25.0m ( sin 43.0 ) ( 9.80m/s )y yy y v t a t v t t− = + = ° + − . Solving these two simultaneous equations for 0v and t gives 0 3.26 m/sv = and 2.51 st = . (b) yv when shell reaches cliff: 2 0 (32.6 m/s) sin 43.0 (9.80 m/s )(2.51 s) 2.4 m/s y y yv v a t= + = ° − = − The shell is traveling downward when it reaches the cliff, so it lands right at the edge of the cliff. EVALUATE: The shell reaches its maximum height at 0 2.27 sy y v t a = − = , which confirms that at 2.51 st = it has passed its maximum height and is on its way down when it strikes the edge of the cliff. 3.27. IDENTIFY: The suitcase moves in projectile motion. The initial velocity of the suitcase equals the velocity of the airplane. Motion in Two or Three Dimensions 3-15 (a) 5.0 m/s to the right (b) 16.0 m/s to the left (c) 13.0 m/s to the left. EVALUATE: The scooter has the largest speed relative to the ground when it is moving to the right relative to the flatcar, since in that case the two velocities S/Fv ! and F/Gv ! are in the same direction and their magnitudes add. 3.37. IDENTIFY: Relative velocity problem. The time to walk the length of the moving sidewalk is the length divided by the velocity of the woman relative to the ground. SET UP: Let W stand for the woman, G for the ground, and S for the sidewalk. Take the positive direction to be the direction in which the sidewalk is moving. The velocities are W/Gv (woman relative to the ground), W/Sv (woman relative to the sidewalk), and S/Gv (sidewalk relative to the ground). Eq.(3.33) becomes W/G W/S S/G.v v v= + The time to reach the other end is given by W/G distance traveled relative to groundt v = EXECUTE: (a) S/G 1.0 m/sv = W/S 1.5 m/sv = + W/G W/S S/G 1.5 m/s 1.0 m/s 2.5 m/s.v v v= + = + = W/G 35.0 m 35.0 m 14 s. 2.5 m/s t v = = = (b) S/G 1.0 m/sv = W/S 1.5 m/sv = − W/G W/S S/G 1.5 m/s 1.0 m/s 0.5 m/s.v v v= + = − + = − (Since W/Gv now is negative, she must get on the moving sidewalk at the opposite end from in part (a).) W/G 35.0 m 35.0 m 70 s. 0.5 m/s t v − − = = = − EVALUATE: Her speed relative to the ground is much greater in part (a) when she walks with the motion of the sidewalk. 3.38. IDENTIFY: Calculate the rower’s speed relative to the shore for each segment of the round trip. SET UP: The boat’s speed relative to the shore is 6.8 km/h downstream and 1.2 km/h upstream. EXECUTE: The walker moves a total distance of 3.0 km at a speed of 4.0 km/h, and takes a time of three fourths of an hour (45.0 min). The total time the rower takes is 1.5 km 1.5 km 1.47 h 88.2 min. 6.8 km/h 1.2 km/h + = = EVALUATE: It takes the rower longer, even though for half the distance his speed is greater than 4.0 km/h. The rower spends more time at the slower speed. 3.39. IDENTIFY: Apply the relative velocity relation. SET UP: The relative velocities are C/Ev ! , the canoe relative to the earth, R/Ev! , the velocity of the river relative to the earth and C/Rv! , the velocity of the canoe relative to the river. EXECUTE: C/E C/R R/Ev = v + v! ! ! and therefore C/R C/E R/E−v = v v! ! ! . The velocity components of C/Rv! are 0.50 m/s (0.40 m/s)/ 2, east and (0.40 m/s)/ 2, south,− + for a velocity relative to the river of 0.36 m/s, at 52.5° south of west. EVALUATE: The velocity of the canoe relative to the river has a smaller magnitude than the velocity of the canoe relative to the earth. 3.40. IDENTIFY: Use the relation that relates the relative velocities. SET UP: The relative velocities are the velocity of the plane relative to the ground, P/Gv ! , the velocity of the plane relative to the air, P/Av ! , and the velocity of the air relative to the ground, A/Gv! . P/Gv ! must due west and A/Gv! must be south. A/G 80 km/hv = and P/A 320 km/hv = . P/G P/A A/Gv = v + v! ! ! . The relative velocity addition diagram is given in Figure 3.40. EXECUTE: (a) A/G P/A 80 km/hsin 320 km/h v v θ = = and 14θ = ° , north of west. (b) 2 2 2 2 P/G P/A A/G (320 km/h) (80.0 km/h) 310 km/hv v v= − = − = . 3-16 Chapter 3 EVALUATE: To travel due west the velocity of the plane relative to the air must have a westward component and also a component that is northward, opposite to the wind direction. Figure 3.40 3.41. IDENTIFY: Relative velocity problem in two dimensions. His motion relative to the earth (time displacement) depends on his velocity relative to the earth so we must solve for this velocity. (a) SET UP: View the motion from above. The velocity vectors in the problem are: M/E ,v! the velocity of the man relative to the earth W/E ,v! the velocity of the water relative to the earth M/W ,v! the velocity of the man relative to the water The rule for adding these velocities is M/E M/W W/Ev = v + v! ! ! Figure 3.41a The problem tells us that W/Ev! has magnitude 2.0 m/s and direction due south. It also tells us that M/Wv! has magnitude 4.2 m/s and direction due east. The vector addition diagram is then as shown in Figure 3.41b This diagram shows the vector addition M/E M/W W/Ev = v + v! ! ! and also has M/Wv! and W/Ev! in their specified directions. Note that the vector diagram forms a right triangle. Figure 3.41b The Pythagorean theorem applied to the vector addition diagram gives 2 2 2 M/E M/W W/E.v v v= + EXECUTE: 2 2 2 2 M/E M/W W/E (4.2 m/s) (2.0 m/s) 4.7 m/sv v v= + = + = M/W W/E 4.2 m/stan 2.10; 2.0 m/s v v θ = = = 65 ;θ = ° or 90 25 .φ θ= ° − = ° The velocity of the man relative to the earth has magnitude 4.7 m/s and direction 25 S° of E. (b) This requires careful thought. To cross the river the man must travel 800 m due east relative to the earth. The man’s velocity relative to the earth is M/E.v! But, from the vector addition diagram the eastward component of M/Ev equals M/W 4.2 m/s.v = Thus 0 800 m 190 s. 4.2 m/sx x xt v − = = = (c) The southward component of M/Ev! equals W/E 2.0 m/s.v = Therefore, in the 190 s it takes him to cross the river the distance south the man travels relative to the earth is 0 (2.0 m/s)(190 s) 380 m.yy y v t− = = = EVALUATE: If there were no current he would cross in the same time, (800 m) /(4.2 m/s) 190 s.= The current carries him downstream but doesn’t affect his motion in the perpendicular direction, from bank to bank. 3.42. IDENTIFY: Use the relation that relates the relative velocities. SET UP: The relative velocities are the water relative to the earth, W/Ev! , the boat relative to the water, B/Wv! , and the boat relative to the earth, B/Ev ! . B/Ev ! is due east, W/Ev! is due south and has magnitude 2.0 m/s. B/W 4.2 m/sv = . B/E B/W W/E= +v v v! ! ! . The velocity addition diagram is given in Figure 3.42. Motion in Two or Three Dimensions 3-17 EXECUTE: (a) Find the direction of B/Wv! . W/E B/W 2.0 m/ssin 4.2 m/s v v θ = = . 28.4θ = ° , north of east. (b) 2 2 2 2 B/E B/W W/E (4.2 m/s) (2.0 m/s) 3.7 m/sv v v= − = − = (c) B/E 800 m 800 m 216 s 3.7 m/s t v = = = . EVALUATE: It takes longer to cross the river in this problem than it did in Problem 3.41. In the direction straight across the river (east) the component of his velocity relative to the earth is lass than 4.2 m/s. Figure 3.42 3.43. IDENTIFY: Relative velocity problem in two dimensions. (a) SET UP: P/Av ! is the velocity of the plane relative to the air. The problem states that P/Av ! has magnitude 35 m/s and direction south. A/Ev! is the velocity of the air relative to the earth. The problem states that A/Ev! is to the southwest ( 45 S° of W) and has magnitude 10 m/s. The relative velocity equation is P/E P/A A/E.= +v v v! ! ! Figure 3.43a EXECUTE: (b) P/A( ) 0,xv = P/A( ) 35 m/syv = − A/E( ) (10 m/s)cos45 7.07 m/s,xv = − ° = − A/E( ) (10 m/s)sin 45 7.07 m/syv = − ° = − P/E P/A A/E( ) ( ) ( ) 0 7.07 m/s 7.1 m/sx x xv v v= + = − = − P/E P/A A/E( ) ( ) ( ) 35 m/s 7.07 m/s 42 m/sy y yv v v= + = − − = − (c) 2 2 P/E P/E P/E( ) ( )x yv v v= + 2 2 P/E ( 7.1 m/s) ( 42 m/s) 43 m/sv = − + − = P/E P/E ( ) 7.1tan 0.169 ( ) 42 x y v v φ − = = = − 9.6 ;φ = ° ( 9.6° west of south) Figure 3.43b EVALUATE: The relative velocity addition diagram does not form a right triangle so the vector addition must be done using components. The wind adds both southward and westward components to the velocity of the plane relative to the ground. 3-20 Chapter 3 EXECUTE: (a) The bird’s tangential velocity can be found from circumference 2 (8.00 m) 50.27 m 10.05 m/s time of rotation 5.00 s 5.00 sxv π = = = = Thus its velocity consists of the components 10.05 m/sxv = and 3.00 m/syv = . The speed relative to the ground is then 2 2 10.5 m/sx yv v v= + = . (b) The bird’s speed is constant, so its acceleration is strictly centripetal–entirely in the horizontal direction, toward the center of its spiral path–and has magnitude 2 2 2 rad (10.05 m/s) 12.6 m/s 8.00 m xva r = = = . (c) Using the vertical and horizontal velocity components 1 3.00 m/stan 16.6 10.05 m/s θ −= = ° . EVALUATE: The angle between the bird’s velocity and the horizontal remains constant as the bird rises. 3.51. IDENTIFY: Take y+ to be downward. Both objects have the same vertical motion, with 0 yv and .ya g= + Use constant acceleration equations for the x and y components of the motion. SET UP: Use the vertical motion to find the time in the air: 0 0,yv = 9.80 m/s ,ya 2= 0 25 m,y y− = ?t = EXECUTE: 21 0 0 2y yy y v t a t− = + gives 2.259 st = During this time the dart must travel 90 m, so the horizontal component of its velocity must be 0 0 90 m 40 m/s 2.25 sx x xv t − = = = EVALUATE: Both objects hit the ground at the same time. The dart hits the monkey for any muzzle velocity greater than 40 m/s. 3.52. IDENTIFY: The person moves in projectile motion. Her vertical motion determines her time in the air. SET UP: Take y+ upward. 0 15.0 m/sxv = , 0 10.0 m/syv = + , 0xa = , 29.80 m/sya = − . EXECUTE: (a) Use the vertical motion to find the time in the air: 21 0 0 2y yy y v t a t− = + with 0 30.0 my y− = − gives 2 230.0 m (10.0 m/s) (4.90 m/s )t t− = − . The quadratic formula gives ( )21 10.0 ( 10.0) 4(4.9)( 30) s 2(4.9) t = + ± − − − . The positive solution is 3.70 st = . During this time she travels a horizontal distance 21 0 0 2 (15.0 m/s)(3.70 s) 55.5 mx xx x v t a t− = + = = . She will land 55.5 m south of the point where she drops from the helicopter and this is where the mats should have been placed. (b) The x-t, y-t, xv -t and yv -t graphs are sketched in Figure 3.52. EVALUATE: If she had dropped from rest at a height of 30.0 m it would have taken her 2 2(30.0 m) 2.47 s 9.80 m/s t = = . She is in the air longer than this because she has an initial vertical component of velocity that is upward. Figure 3.52 3.53. IDENTIFY: The cannister moves in projectile motion. Its initial velocity is horizontal. Apply constant acceleration equations for the x and y components of motion. Motion in Two or Three Dimensions 3-21 SET UP: Take the origin of coordinates at the point where the canister is released. Take y+ to be upward. The initial velocity of the canister is the velocity of the plane, 64.0 m/s in the -direction.x+ Figure 3.53 Use the vertical motion to find the time of fall: ?,t = 0 0,yv = 29.80 m/s ,ya = − 0 90.0 my y− = − (When the canister reaches the ground it is 90.0 m below the origin.) 21 0 0 2y yy y v t a t− = + EXECUTE: Since 0 0,yv = 0 2 2( ) 2( 90.0 m) 4.286 s. 9.80 m/sy y yt a − − = = = − SET UP: Then use the horizontal component of the motion to calculate how far the canister falls in this time: 0 ?,x x− = 0,xa − 0 64.0 m/s,xv = EXECUTE: 21 0 0 2 (64.0 m/s)(4.286 s) 0 274 m.x x v t at− = + = + = EVALUATE: The time it takes the cannister to fall 90.0 m, starting from rest, is the time it travels horizontally at constant speed. 3.54. IDENTIFY: The equipment moves in projectile motion. The distance D is the horizontal range of the equipment plus the distance the ship moves while the equipment is in the air. SET UP: For the motion of the equipment take x+ to be to the right and y+ to be upwards. Then 0xa = , 29.80 m/sya = − , 0 0 0cos 7.50 m/sxv v α= = and 0 0 0sin 13.0 m/syv v α= = . When the equipment lands in the front of the ship, 0 8.75 my y− = − . EXECUTE: Use the vertical motion of the equipment to find its time in the air: 21 0 0 2y yy y v t a t− = + gives ( )21 13.0 ( 13.0) 4(4.90)(8.75) s 9.80 t = ± − + . The positive root is 3.21 st = . The horizontal range of the equipment is 21 0 0 2 (7.50 m/s)(3.21 s) 24.1 mx xx x v t a t− = + = = . In 3.21 s the ship moves a horizontal distance (0.450 m/s)(3.21 s) 1.44 m= , so 24.1 m 1.44 m 25.5 mD = + = . EVALUATE: The equation 2 0 0sin 2vR g α = from Example 3.8 can't be used because the starting and ending points of the projectile motion are at different heights. 3.55. IDENTIFY: Projectile motion problem. Take the origin of coordinates at the point where the ball leaves the bat, and take y+ to be upward. 0 0 0cosxv v α= 0 0 0sin ,yv v α= but we don’t know 0.v Figure 3.55 Write down the equation for the horizontal displacement when the ball hits the ground and the corresponding equation for the vertical displacement. The time t is the same for both components, so this will give us two equations in two unknowns ( 0v and t). 3-22 Chapter 3 (a) SET UP: y-component: 29.80 m/s ,ya = − 0 0.9 m,y y− = − 0 0 sin 45yv v= ° 21 0 0 2y yy y v t a t− = + EXECUTE: 2 21 0 20.9 m ( sin 45 ) ( 9.80 m/s )v t t− = ° + − SET UP: x-component: 0,xa = 0 188 m,x x− = 0 0 cos45xv v= ° 21 0 0 2x xx x v t a t− = + EXECUTE: 0 0 0 188 m cos45x x xt v v − = = ° Put the expression for t from the x-component motion into the y-component equation and solve for 0.v (Note that sin 45 cos45 .° = ° ) 2 2 0 0 0 188 m 188 m0.9 m ( sin 45 ) (4.90 m/s ) cos45 cos45 v v v ⎛ ⎞ ⎛ ⎞ − = ° −⎜ ⎟ ⎜ ⎟° °⎝ ⎠ ⎝ ⎠ 2 2 0 188 m4.90 m/s 188 m 0.9 m 188.9 m cos45v ⎛ ⎞ = + =⎜ ⎟°⎝ ⎠ 2 2 0 cos45 4.90 m/s , 188 m 188.9 m v °⎛ ⎞ =⎜ ⎟ ⎝ ⎠ 2 0 188 m 4.90 m/s 42.8 m/s cos45 188.9 m v ⎛ ⎞= =⎜ ⎟°⎝ ⎠ (b) Use the horizontal motion to find the time it takes the ball to reach the fence: SET UP: x-component: 0 116 m,x x− = 0,xa = 0 0 cos45 (42.8 m/s)cos45 30.3 m/s,xv v= ° = ° = ?t = 21 0 0 2x xx x v t a t− = + EXECUTE: 0 0 116 m 3.83 s 30.3 m/sx x xt v − = = = SET UP: Find the vertical displacement of the ball at this t: y-component: 0 ?,y y− = 29.80 m/s ,ya = − 0 0 sin 45 30.3 m/s,yv v= ° = 3.83 st = 21 0 0 2y yy y v t a t− = + EXECUTE: 21 0 2(30.3 s)(3.83 s) ( 9.80 m/s )(3.83 s)y y 2− = + − 0 116.0 m 71.9 m 44.1 m,y y− = − = + above the point where the ball was hit. The height of the ball above the ground is 44.1 m 0.90 m 45.0 m.+ = It’s height then above the top of the fence is 45.0 m 3.0 m 42.0 m.− = EVALUATE: With 0 42.8 m/s,v = 0 30.3 m/syv = and it takes the ball 6.18 s to return to the height where it was hit and only slightly longer to reach a point 0.9 m below this height. 0(188 m) /( cos45 )t v= ° gives 6.21 s,t = which agrees with this estimate. The ball reaches its maximum height approximately (188 m) / 2 94 m= from home plate, so at the fence the ball is not far past its maximum height of 47.6 m, so a height of 45.0 m at the fence is reasonable. 3.56. IDENTIFY: The water moves in projectile motion. SET UP: Let 0 0 0x y= = and take y+ to be positive. 0xa = , ya g= − . EXECUTE: The equations of motions are 21 0 2( sin )y v α t gt= − and 0( cos )x v α t= . When the water goes in the tank for the minimum velocity, 2y D= and 6x D= . When the water goes in the tank for the maximum velocity, 2y D= and 7x D= . In both cases, sin cos 2 / 2.α α= = To reach the minimum distance: 0 26 2 D v t= , and 21 0 2 22 2 D v t gt= − . Solving the first equation for t gives 0 6 2Dt v = . Substituting this into the second equation gives 2 1 2 0 6 22 6 DD D g v ⎛ ⎞ = − ⎜ ⎟⎜ ⎟ ⎝ ⎠ . Solving this for 0v gives 0 3v gD= . Motion in Two or Three Dimensions 3-25 Call the range 1R when the angle is 0α and 2R when the angle is 90 .α° − 0 0 1 0 0 2 sin( cos ) vR v g αα ⎛ ⎞ = ⎜ ⎟ ⎝ ⎠ 0 0 2 0 0 2 sin(90 )( cos(90 )) vR v g αα ⎛ ⎞° − = ° − ⎜ ⎟ ⎝ ⎠ The problem asks us to show that 1 2.R R= EXECUTE: We can use the trig identities in Appendix B to show: 0 0 0cos(90 ) cos( 90 ) sinα α α° − = − ° = 0 0 0 0sin(90 ) sin( 90 ) ( cos ) cosα α α α° − = − − ° = − − = + Thus 0 0 0 0 2 0 0 0 0 1 2 cos 2 sin( sin ) ( cos ) .v vR v v R g g α αα α ⎛ ⎞ ⎛ ⎞ = = =⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ (b) 2 0 0sin 2vR g α = so 2 0 2 2 0 (0.25 m)(9.80 m/s )sin 2 . (2.2 m/s) Rg v α = = This gives 15α = ° or 75 .° EVALUATE: 2 0 0( sin 2 ) / ,R v gα= so the result in part (a) requires that 2 2 0 0sin (2 ) sin (180 2 ),α α= ° − which is true. (Try some values of 0α and see!) 3.62. IDENTIFY: Mary Belle moves in projectile motion. SET UP: Let y+ be upward. 0xa = , ya g= − . EXECUTE: (a) Eq.(3.27) with 8.2 mx = , 6.1 my = and 0 53α = ° gives 0 13 8 m/sv .= . (b) When she reached Joe Bob, 0 8.2 m 0.9874 s cos53 t v = = ° . 0 8.31 m/sx xv v= = and 0 1.34 m/sy y yv v a t= + = + . 8 4 m/sv .= , at an angle of 9.16° . (c) The graph of ( )xv t is a horizontal line. The other graphs are sketched in Figure 3.62. (d) Use Eq. (3.27), which becomes 1 2(1.327) (0.071115 m )y x x−= − . Setting 8.6 my = − gives 23.8 mx = as the positive solution. Figure 3.62 3.63. (a) IDENTIFY: Projectile motion. Take the origin of coordinates at the top of the ramp and take y+ to be upward. The problem specifies that the object is displaced 40.0 m to the right when it is 15.0 m below the origin. Figure 3.63 We don’t know t, the time in the air, and we don’t know 0.v Write down the equations for the horizontal and vertical displacements. Combine these two equations to eliminate one unknown. SET UP: y-component: 0 15.0 m,y y− = − 29.80 m/s ,ya = − 0 0 sin53.0yv v= ° 21 0 0 2y yy y v t a t− = + EXECUTE: 2 2 015.0 m ( sin53.0 ) (4.90 m/s )v t t− = ° − 3-26 Chapter 3 SET UP: x-component: 0 40.0 m,x x− = 0,xa = 0 0 cos53.0xv v= ° 21 0 0 2x xx x v t a t− = + EXECUTE: 040.0 m ( )cos53.0v t= ° The second equation says 0 40.0 m 66.47 m. cos53.0 v t = = ° Use this to replace 0v t in the first equation: 2 215.0 m (66.47 m)sin53 (4.90 m/s )t− = ° − 2 2 (66.46 m)sin53 15.0 m 68.08 m 3.727 s. 4.90 m/s 4.90 m/s t ° + = = = Now that we have t we can use the x-component equation to solve for 0:v 0 40.0 m 40.0 m 17.8 m/s. cos53.0 (3.727 s)cos53.0 v t = = = ° ° EVALUATE: Using these values of 0v and t in the 1 0 0 2y yy y v a t 2= = + equation verifies that 0 15.0 m.y y− = − (b) IDENTIFY: 0 (17.8 m/s) / 2 8.9 m/sv = = This is less than the speed required to make it to the other side, so he lands in the river. Use the vertical motion to find the time it takes him to reach the water: SET UP: 0 100 m;y y− = − 0 0 sin53.0 7.11 m/s;yv v= + ° = 29.80 m/sya = − 21 0 0 2y yy y v t a t− = + gives 100 7.11 4.90t t 2− = − EXECUTE: 24.90 7.11 100 0t t− − = and ( )21 9.80 7.11 (7.11) 4(4.90)( 100)t = ± − − 0.726 s 4.57 st = ± so 5.30 s.t = The horizontal distance he travels in this time is 0 0 0( cos53.0 ) (5.36 m/s)(5.30 s) 28.4 m.xx x v t v t− = = ° = = He lands in the river a horizontal distance of 28.4 m from his launch point. EVALUATE: He has half the minimum speed and makes it only about halfway across. 3.64. IDENTIFY: The rock moves in projectile motion. SET UP: Let y+ be upward. 0xa = , ya g= − . Eqs.(3.22) and (3.23) give xv and yv . EXECUTE: Combining equations 3.25, 3.22 and 3.23 gives 2 2 2 2 2 2 2 2 0 0 0 0 0 0 0 0 0cos ( sin ) (sin cos ) 2 sin ( )v v v gt v v gt gtα α α α α= + − = + − + . 2 2 2 2 0 0 0 0 12 ( sin ) 2 2 v v g v t gt v gyα= − − = − , where Eq.(3.21) has been used to eliminate t in favor of y. For the case of a rock thrown from the roof of a building of height h, the speed at the ground is found by substituting y h= − into the above expression, yielding 2 0 2v v gh= + , which is independent of 0α . EVALUATE: This result, as will be seen in the chapter dealing with conservation of energy (Chapter 7), is valid for any y, positive, negative or zero, as long as 2 0 2 0v gy− > . 3.65. IDENTIFY and SET UP: Take y+ to be upward. The rocket moves with projectile motion, with 0 40.0 m/syv = + and 0 30.0 m/sxv = relative to the ground. The vertical motion of the rocket is unaffected by its horizontal velocity. EXECUTE: (a) 0yv = (at maximum height), 0 40.0 m/s,yv = + 29.80 m/s ,ya = − 0 ?y y− = 2 2 0 02 ( )y y yv v a y y= + − gives 0 81.6 my y− = (b) Both the cart and the rocket have the same constant horizontal velocity, so both travel the same horizontal distance while the rocket is in the air and the rocket lands in the cart. (c) Use the vertical motion of the rocket to find the time it is in the air. 0 40 m/s,yv = 29.80 m/s ,ya = − 40 m/s,yv = − ?t = 0y y yv v a t= + gives 8.164 st = Then 0 0 (30.0 m/s)(8.164 s) 245 m.xx x v t− = = = Motion in Two or Three Dimensions 3-27 (d) Relative to the ground the rocket has initial velocity components 0 30.0 m/sxv = and 0 40.0 m/s,yv = so it is traveling at 53.1° above the horizontal. (e) (i) Figure 3.65a Relative to the cart, the rocket travels straight up and then straight down (ii) Figure 3.65b Relative to the ground the rocket travels in a parabola. EVALUATE: Both the cart and rocket have the same constant horizontal velocity. The rocket lands in the cart. 3.66. IDENTIFY: The ball moves in projectile motion. SET UP: The woman and ball travel for the same time and must travel the same horizontal distance, so for the ball 0 6.00 m/sxv = . EXECUTE: (a) 0 0 0cosxv v θ= . 0 0 0 6.00 m/scos 20.0 m/s xv v θ = = and 0 72.5θ = ° . (b) Relative to the ground the ball moves in a parabola. The ball and the runner have the same horizontal component of velocity, so relative to the runner the ball has only vertical motion. The trajectories as seen by each observer are sketched in Figure 3.66. EVALUATE: The ball could be thrown with a different speed, so long as the angle at which it was thrown was adjusted to keep 0 6.00 m/sxv = . Figure 3.66 3.67. IDENTIFY: The boulder moves in projectile motion. SET UP: Take y+ downward. 0 0xv v= , 0 0yv = . 0xa = , 29.80 m/sya = + . EXECUTE: (a) Use the vertical motion to find the time for the boulder to reach the level of the lake: 21 0 0 2y yy y v t a t− = + with 0 20 my y− = + gives 0 2 2( ) 2(20 m) 2.02 s 9.80 m/sy y yt a − = = = . The rock must travel horizontally 100 m during this time. 21 0 0 2x xx x v t a t− = + gives 0 0 0 100 m 49.5 m/s 2.02 sx x xv v t − = = = = (b) In going from the edge of the cliff to the plain, the boulder travels downward a distance of 0 45 my y− = . 0 2 2( ) 2(45 m) 3.03 s 9.80 m/sy y yt a − = = = and 0 0 (49.5 m/s)(3.03 s) 150 mxx x v t− = = = . The rock lands 150 m 100 m 50 m− = beyond the foot of the dam. EVALUATE: The boulder passes over the dam 2.02 s after it leaves the cliff and then travels an additional 1.01 s before landing on the plain. If the boulder has an initial speed that is less than 49 m/s, then it lands in the lake. 3.68. IDENTIFY: The bagels move in projectile motion. Find Henrietta’s location when the bagels reach the ground, and require the bagels to have this horizontal range.