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Motion in Two Dimensions
CHAPTER OUTLINE 4.1 The Position, Velocity, and Acceleration Vectors 4.2 Two-Dimensional Motion with Constant Acceleration 4.3 Projectile Motion 4.4 Uniform Circular Motion 4.5 Tangential and Radial Acceleration 4.6 Relative Velocity and Relative Acceleration
ANSWERS TO QUESTIONS
- Q4.1 The car’s acceleration must have an inward component and a forward component: answer (f). Another argument: Draw a final velocity vector of two units west. Add to it a vector of one unit south. This represents subtracting the initial velocity from the final velocity, on the way to finding the acceleration. The direction of the resultant is that of vector (f).
Q4.2 No, you cannot determine the instantaneous velocity. Yes, you can determine the average velocity. The points could be widely separated. In this case, you can only determine the average velocity, which is
v^ ^ avg^ = t
x
65
Q4.3 (a) (b)
*Q4.4 (i) The 45° angle means that at point A the horizontal and vertical velocity components are equal. The horizontal velocity component is the same at A, B , and C. The vertical velocity component is zero at B and negative at C. The assembled answer is a = b = c = e > d = 0 > f (ii) The x -component of acceleration is everywhere zero and the y -component is everywhere
- 9.8 ms^2. Then we have a = c = e = 0 > b = d = f.
Q4.5 A parabola results, because the originally forward velocity component stays constant and the rocket motor gives the spacecraft constant acceleration in a perpendicular direction.
Q4.6 (a) yes (b) no: the escaping jet exhaust exerts an extra force on the plane. (c) no (d) yes (e) no: the stone is only a few times more dense than water, so friction is a significant force on the stone. The answer is (a) and (d).
Q4.7 The projectile is in free fall. Its vertical component of acceleration is the downward acceleration of gravity. Its horizontal component of acceleration is zero.
Q4.8 (a) no (b) yes (c) yes (d) no. Answer: (b) and (c)
*Q4.9 The projectile on the moon is in flight for a time interval six times larger, with the same range of vertical speeds and with the same constant horizontal speed as on Earth. Then (i) its range is (d) six times larger and (ii) its maximum altitude is (d) six times larger. Apollo astronauts performed the experiment with golf balls.
Q4.10 (a) no. Its velocity is constant in magnitude and direction. (b) yes. The particle is continuously changing the direction of its velocity vector.
Q4.11 (a) straight ahead (b) either in a circle or straight ahead. The acceleration magnitude can be constant either with a nonzero or with a zero value.
*Q4.12 (i) a = v^2 r becomes 3^2 3 = 3 times larger: answer (b). (ii) T = 2 π r v changes by a factor of 3 3 = 1. The answer is (a).
Q4.
The skater starts at the center of the eight, goes clockwise around the left circle and then counter- clockwise around the right circle.
*Q4.14 With radius half as large, speed should be smaller by a factor of 1 2 , so that a = v^2 r can be the same. The answer is (d).
*Q4.15 The wrench will hit (b) at the base of the mast. If air resistance is a factor, it will hit slightly leeward of the base of the mast, displaced in the direction in which air is moving relative to the deck. If the boat is scudding before the wind, for example, the wrench’s impact point can be in front of the mast.
*Q4.16 Let the positive x direction be that of the girl’s motion. The x component of the velocity of the ball relative to the ground is +5 – 12 m s = –7 ms. The x -velocity of the ball relative to the girl is –7 – 8 m s = –15 m s. The relative speed of the ball is +15 m s, answer (d).
66 Chapter 4
SOLUTIONS TO PROBLEMS
Section 4.1 The Position, Velocity, and Acceleration Vectors
P4.1 x m ( )
m 2 330m
y m
(a) Net displacement^ =^ +^ at tan^1 (^ )
x^2 2 y x 4 87
y − R. km at 28 6. °S of W
(b) Average speed s s m s = ( 20 0. m )( 180 ) + ( 25 0. ) 120 s m s s s s s
( ) + ( )( )
3 3 3. m s
(c) Average velocity m 360 s = m s alo
×
. nng
R
FIG. P4.
68 Chapter 4
(b)^ θ =^
⎠ −^ =^ +
tan −. .
= ° °from x aaxis
(c) At t = 25.0 s its position is specified by its coordinates and the direction of its motion is specified by the direction angle of its velocity:
x f = x i + vxi t + a tx = + ( ) + ( )
2
2
m
y (^) f y (^) i vyi t a ty 5 5 0
(. ) + (− 0 300. )( 25 0. )^2 = −
72.7 m
v xf vxi a tx == + ( ) =
m s v (^) yf vyi a ty 5 ms
θ =
tan −^ = tan− ⎛⎝−. ⎞⎠ = − .
v v
y x
P4.6 (a) ^
v =^ d r^ = ⎛⎝ ⎞⎠ ( i − j (^) ) = − j dt
d dt
3 00. ˆ^ 6 00. t^2 ˆ^ 12 0. t ˆ mm s
a = v^ = ⎛⎝ ⎞⎠ (− j (^) ) = − j d dt dt
t d 12 0. ˆ^ 12 0. ˆ m s 2
(b) by substitution, r ^ = (^) ( 3 00. i ˆ^ −6 00. ˆ j (^) )^ m; v = −12 0. j ˆm s
*P4.7 (a) From a^ = d v^ dt (^) ∫ (^) id v^ =∫ a dt
f i
f / , we have
Then v − 5 i = (^) ∫ 6 j = 6 j = 3 2
3 2 (^0 )
ˆ (^) m /s ˆ^3 /
/ t dt t t
t t // (^2) ˆ ˆ 3 2/ ˆ j so v = 5 i + 4 t j
(b) From v ^ = r^ r^ = v d dt (^) ∫ (^) id (^) ∫ dt
f i
f / , we have
Then
r − = (^) ( i j (^) ) = i + j
/ t dt t t ⎜⎜
⎠⎟^
∫ 0 =^ + 0
t 5 1 6 5 2 t t ˆ i^. t / ˆ j
P4.8 a = 3 00. ˆ j m s 2 ;
v (^) i = 5 00. i ˆ m s; r i (^) = 0 i ˆ^ + 0 ˆ j
(a) r^ (^) f = r i^ (^) + v ^ (^) i t + a t = ⎡ t i + t j ⎣⎢^
(^2). ˆ (^) 3 00. 2 ˆ⎥⎥ m
v (^) v (^) a i j f =^ i +^ t^ =^ (^ 5 00.^ ˆ^ +3 00.^ t ˆ^ )m s
(b) t = 2 00. s, r f = 5 00 2 00( ) i + 1 ( )( ) j = i +
.. ˆ^ 3 00. 2 00. 2 ˆ^ (10 0. ˆ^6 .. 00 ˆ j ) m
so x (^) f = 10 0. m , y (^) f = 6 00. m
v i j i j f =^ 5 00.^ ˆ^ +^ 3 00 2 00.^ (^.^ )^ ˆ^ =^ (^ 5 00.^ ˆ^ +6 00.^ ˆ)^ m s
v ff = v^ f = vxf^2 + vyf^2 = ( 5 00. ) + ( 2 6 00. ) 2 = 7 81. m s
Motion in Two Dimensions 69
Section 4.3 Projectile Motion
P4.9 (a) The mug leaves the counter horizontally with a velocity vxi (say). If time t elapses before it hits the ground, then since there is no horizontal acceleration, x (^) f = vxi t , i.e.,
t
x (^) f xi xi
v v
1 40. m
In the same time it falls a distance of 0.860 m with acceleration downward of (^) 9 80. m s 2. Then
y (^) f = yi + vyit +^1 a ty 2
9 0 1 40^
2 = + (^) (− )
. m. 8 m s 2. m vxi
Thus,
vxi = ( )( ) =
m s m 0.860 m
m s
2 2
(b) The vertical velocity component with which it hits the floor is
v (^) yf = vyi + a ty = + −( )
⎝⎜^
m s m 3.34 m s
2 ⎟⎟ = −4 11.^ m s Hence, the angle θ at which the mug strikes the floor is given by
θ =
⎠ =^ −
tan− tan−. .
v v
yf xf
P4.10 The mug is a projectile from just after leaving the counter until just before it reaches the floor. Taking the origin at the point where the mug leaves the bar, the coordinates of the mug at any time are
x (^) f = vxi t + a tx = vxit +
(^2 0) and y (^) f = vyi t + a ty = − g t
2 2
When the mug reaches the floor, y (^) f = − h so
− h = − g t
2
which gives the time of impact as
t h g
(a) Since x (^) f = d when the mug reaches the floor, x (^) f = vxi t becomes d h xi g = v
giving the initial velocity as
vxi d g h
FIG. P4.
continued on next page
Motion in Two Dimensions 71
P4.12 (a) To identify the maximum height we let i be the launch point and f be the highest point:
v v v
yf yi y f i i i
a y y g y
2 2 2 2
= sin θ + (^) (− ) (( (^) max− )
y = g max^ v^ i^ sin^ i.
2 2 2
θ
To identify the range we let i be the launch and f be the impact point; where t is not zero:
y t a t
t g t
t
f i yi y
i i i
= + + ( − )
=
y v
v v
2
sin θ^2 ssin
cos sin
θ
θ θ
i
f i xi x
i i i^ i
g
x x t a t
d
v
v v
2
gg
For this rock, d = y max vi i vi i i
i i
i
g g
2 2 2 2
sin 2 sin cos
sin cos tan
θ θ θ
θ θ θ
θ i 76 0. °
(b) Since g divides out, the answer is the same on every planet.
(c) The maximum range is attained for θ i = 45° : d d
g g
i i i i
max cos^ sin cos sin
v v v v
So d d max =^
P4.13 h g
= vi^ i
2 2 2
sin θ (^) ; R g
= vi^ (^ i )
(^2) sin 2 θ ; 3 h = R
so 3 2
v^2 (^) i^2 i vi^2^2 i g g
sin θ = (sin θ)
or 2 3 2 2
2 = sin = sin
θ tan θ
i θ i
i
thus θ i = tan −^1 ⎛⎝^4 ⎞⎠ =. 3
72 Chapter 4
P4.14 The horizontal component of displacement is x (^) f = vxi t = ( vi cos θ (^) i ) t. Therefore, the time required
to reach the building a distance d away is t d i i
v cos θ
. At this time, the altitude of the water is
y t a t d g d f yi y i i i i i
v v − v v
(^2) sin cos c
θ θ oos θ i
2
Therefore the water strikes the building at a height h above ground level of
h y (^) f d (^) i gd i i
= = tan − cos
θ θ
2 2 v^2
P4.15 (a) x f = vxi t = 8 00. cos 20 0. ° ( 3 00. ) = 22 6. m
(b) Taking y positive downwards,
y t g t
y
f yi
f
v
2
. sin. °.. ( 3 3 00. )^2 = 52 3. m
(c) (^) 10 0 8 00 20 0 1 2
. =. ( sin. ° ) + t ( 9 80. ) t^2
2 2
t t
t
− ± (^) ( ) + = 1 1 18. s
*P4.16 The time of fl ight of a water drop is given by y (^) f = y (^) i + vyi t +^1 a ty 2
=. m + − (^) ( 9 8. m s (^2) )t 12
For t 1 > 0 , the root is t 1 2 2 35 9 8
= (^ 0 693
m. m s 2 s. (a) The horizontal range of the font is x (^) f 1 x (^) i xit^1 a tx^2 2 0 1 70 0 693 0 1 1
v
. m s. s. 8 8 m
This is about the width of a town sidewalk, so there is space for a walkway behind the waterfall. Unless the lip of the channel is well designed, water may drip on the visitors. A tall or inattentive person may get his head wet.
(b) Now the flight time t 2 is given by 0 0
= y + − gt^2.
t y g
y g
y g
t 2
= = From the same equation as in part (a) for
horizontal range, x (^) 2 = v 2 2 t.
x 1 t 12 =^ v^^2121 v^
v 2 1 1
1 12 12
=^ x = = = 0 491 t
. m s (^). m s
The rule that the scale factor for speed is the square root of the scale factor for distance is Froude’s law, published in 1870.
FIG. P4.
1.70 m /s
2.35 m
74 Chapter 4
(b) As it passes over the wall, the ball is above the street by y (^) f = y (^) i + vyi t + a ty
2
y (^) f = 0 + ( 18 1 (^) )( 53 )( 2 2 ) + (^1) (− ) 2
. m s sin °. s 9 8. m s 2 ( 2 .. 2 s )^2 =8 13. m
So it clears the parapet by 8 13. m − 7 m = 1 13. m.
(c) Note that the highest point of the ball’s trajectory is not directly above the wall. For the whole flight, we have from the trajectory equation
y x g f i f x i i
= ( ) − f
tan cos
θ 2 2 2 θ
2 v or 6 53
m m s m s
2
( (^) )
tan
. cos
x (^) f
⎠⎟^
x (^) f^2
Solving,
(^ 0 041 2.^ m^ −^1 ) x^^2 f −^ 1 33.^ xf +^6 m=^0
and
x f = ±^ −^ (^
( −)
2 1
. m This yields two results: x (^) f = 26 8. m or 5.44 m The ball passes twice through the level of the roof. It hits the roof at distance from the wall
26 8. m − 24 m = 2 79. m
P4.20 From the instant he leaves the floor until just before he lands, the basketball star is a projectile.
His vertical velocity and vertical displacement are related by the equation v yf^2 = vyi^2 + 2 ay ( y f − yi ).
Applying this to the upward part of his flight gives 0 = v^2 yi + (^2) ( −9 80. m s (^2) ) (1 85. −1 02. )m. From this, vyi = 4 03. m s. [Note that this is the answer to part (c) of this problem.] For the downward part of the flight, the equation gives vyf^2 = 0 + (^2) ( −9 80. m s (^2) ) ( 0 900. −1 85. )m. Thus the vertical velocity just before he lands is vyf = −4 32. m s
(a) His hang time may then be found from v (^) yf = vyi + a ty :
− 4 32. m s = 4 03. m s + −( 9 80. m s (^2) )t
or t = 0 852. s.
(b) Looking at the total horizontal displacement during the leap, x = vxi t becomes
2 80. m = vxi ( 0 852. s)
which yields vxi = 3 29. m s.
(c) vyi = 4.03 m s. See above for proof.
continued on next page
Motion in Two Dimensions 75
(d) The takeoff angle is: θ =
⎝⎜^
tan− 1 v tan− 1 4 03. ⎞ v
yi xi
m s 3.29 m s⎠⎠⎟
(e) Similarly for the deer, the upward part of the flight gives
v yf^2 = vyi^2 + 2 ay ( y f − yi ):
0 = v^2 yi + (^2) ( −9 80. m s (^2) ) ( 2 50. −1 20. )m
so vyi = 5 04. m s.
For the downward part, v yf^2 = vyi^2 + 2 a y ( y f − yi ) yields
vyf^2 = 0 + (^2) ( −9 80. m s (^2) ) ( 0 700. −2 50. )mand vyf = −5 94. m s
The hang time is then found as v (^) yf = vyi + a ty : − 5 94. m s = 5 04. m s + −( 9 80. m s (^2) )t and
t =1 12. s
P4.21 The horizontal kick gives zero vertical velocity to the rock. Then its time of flight follows from
y y t a t
t
f =^ i +^ yi + y
− = + + (^) (− )
v^1 2 40 0 0 0 1 2
2
. m. m s 2 22 t = 2 86. s. The extra time 3 00. s − 2 86. s = 0 143. sis the time required for the sound she hears to travel straight back to the player. It covers distance ( 343 m s ) 0 143. s = 49 0. m = x^2 + ( 40 0. m)^2 where x represents the horizontal distance the rock travels. x (^) xit t
xi
.^2
m m s
m s
v
v
*P4.22 We match the given equations
x (^) f = + ( ) t
= +
. cos. ...
m s
m m
( m s ) sin 18 5. − (^) (. m s (^2) )
° t 9 80 t^2
to the equations for the coordinates of the final position of a projectile x x t
y y t gt
f i xi
f i yi
v
v^1 2
2
For the equations to represent the same functions of time, all coefficients must agree: x (^) i = 0 , yi = 0 840. m, vxi = (11 2. m s) cos18 5 .°, vyi = ( 11 2. m s) sin 18 5.° and g = 9 80. m s 2.
(a) Then the original position of the athlete’s center of mass is the point with coordinates (^ x^ i ,^ yi ) =^ (^0 ,^ 0 840 m.^ ). That is, his original position has position vector r = 0 ˆ i (^) +0 840. m j ˆ.
continued on next page
Motion in Two Dimensions 77
P4.26 a c (^) r
v^2
v = a rc = 3 9 8(. m s (^2) )( 9 45. m ) =16 7. m s
Each revolution carries the astronaut over a distance of 2 π r = 2 π( 9 45. m ) =59 4. m. Then the
rotation rate is
16 7
. m s 0 281. rev 59.4 m rev s
P4.27 (a) v = r ω
At 8.00 rev s, v = ( 0 600. m (^) )( 8 00. rev s (^) )( 2 π rad rev (^) ) = 30 2. m s= 9 .. 60 π m s.
At 6.00 rev s, v = ( 0 900. m (^) )( 6 00. rev s)( 2 π rad rev (^) ) = 33 9. m s= 10 0 8. π m s. 6 00. rev s gives the larger linear speed.
(b) Acceleration =^ = v^2 ( 9 60 ) (^2) = × 3 0 600
r
π m s. m
m s (^2).
(c) At 6.00 rev s, acceleration (^) = (^ )^ = ×
2
. 3 .
π m s m m s 2. So 8 rev/s gives the higher acceleration.
Section 4.5 Tangential and Radial Acceleration
*P4.28 The particle’s centripetal acceleration is v^2 r = (3 ms)^2 2 m = 4.50 ms^2. The total acceleration magnitude can be larger than or equal to this, but not smaller.
(a) Yes. The particle can be either speeding up or slowing down, with a tangential component of acceleration of magnitude 6 2 − 4 5. 2 =3 97. m /s 2.
(b) No. The magnitude of the acceleration cannot be less than v^2 r = 4.5 ms^2.
P4.29 We assume the train is still slowing down at the instant in question.
a r
a t
c
t
(− )( )
v
v
2
3
m s
km h m km
2
hh s s
m s 2
a = a (^) c + at = ( mm s 2 ) + −( m s^2 ) 2 2 0 741.
at an angle of tan −^ tan−
a a
t c
a^ = 1 48. m s 2 inward and 29.9° backward
FIG. P4.
78 Chapter 4
P4.30 (a) See fi gure to the right.
(b) The components of the 20.2 and the 22 5. m s 2 along the rope together constitute the centripetal acceleration: ac = ( 22 5. m s (^2) ) cos ( 90 0. ° −36 9. °) + ( 20 2. m s^2 )cos 3 36 9. ° = 29 7. m s 2
(c) a c (^) r
v^2
so v = a rc = 29 7. m s 2 ( 1 50. m ) =6 67. m s tangent to ciircle
v = 6 67. m s at 36.9° above the horizontal
P4.31 r = 2 50. m, a = 15 0. m s 2
(a) ac = a cos 30 0. ° = ( 15 0. m s (^2) )( cos 30 °) = 13 0. m s^2
(b) (^) a c (^) r
v^2
so v^2 = rac = 2 50. m (^) (13 0. m s (^2) ) =32 5. m 2 s^2
v = 32 5. m s = 5 70. m s
(c) a^2 = at^2^ + ar^2
so a (^) t = a^2 − ar^2 = (^) ( 15 0. m s (^2) ) − (^2 13 0. m s (^2) ) = 7 50. m s^2
P4.32 Let i be the starting point and f be one revolution later. The curvilinear motion with constant tangential acceleration is described by
and v (^) xf = vxi + a tx , v (^) f a tt r t
= 0 + =^4 π^. The magnitude of the radial acceleration is a r
r r t r =^ vf =
(^2 2 ) 2
16 π (^).
Then tan θ π π π
a a
r t t r
t r
2 2 2 θ =^ 4 55.^ °
Section 4.6 Relative Velocity and Relative Acceleration
P4.33 v ce = the velocity of the car relative to the earth. v wc =^ the velocity of the water relative to the car. v we =^ the velocity of the water relative to the earth. These velocities are related as shown in the diagram at the right. (a) Since v we is vertical, v (^) wc sin 60 0. ° = vce = 50 0. km h or v wc =^ 57 7.^ km h at^ 60 0.°^ west of vertical. (b) Since
v ce has zero vertical component, v (^) we = vwc cos 60 0. ° = ( 57 7. km h) cos 60 0. °= 28 9. km h downward
∆ x t a t
r a t
a r t
xi x
t
t
v
2
2
2
π π
FIG. P4.
FIG. P4.
FIG. P4.
a
at ar
v ce
60º
v we (^) v wc
v (^) we = v ce (^) + v wc FIG. P4.
80 Chapter 4
P4.37 To guess the answer, think of v just a little less than the speed c of the river. Then poor Alan will spend most of his time paddling upstream making very little progress. His time-averaged speed will be low and Beth will win the race. Now we calculate: For Alan, his speed downstream is c + v , while his speed upstream is c − v. Therefore, the total time for Alan is
t L c
L c
L c (^1 2) c 2
v v − v
For Beth, her cross-stream speed (both ways) is
c 2 − v^2
Thus, the total time for Beth is t L c
L c c (^2 2 2 2 )
v − v
Since 1 1
2 − 2 < v c
,^ t^1 >^ t 2 , or Beth, who swims cross-stream, returns first.
*P4.38 We can find the time of fl ight of the can by considering its horizontal motion:
16 m = (9.5 ms) t + 0 t = 1.68 s
(a) For the boy to catch the can at the same location on the truck bed, he must throw it straight up, at 0° to the vertical.
(b) For the free fall of the can, yf = yi + vyit + (12) a (^) yt^2 : 0 = 0 + vyi (1.68 s) − (12)(9.8 ms 2 )(1.68 s)^2 vyi = 8.25 ms
(c) The boy sees the can always over his head, traversing a straight line segment upward and then downward.
(d) The ground observer sees the can move as a projectile on a symmetric section of a parabola opening downward. Its initial velocity is (9.5^2 + 8.25^2 )^1 ^2 ms = 12.6 m/s north at tan−^1 (8.25/9.5) = 41.0°^ above the horizontal
P4.39 Identify the student as the S ′ observer and the professor as the S observer. For the initial motion in S ′, we have ′ ′
v v
y x
tan 60 0. ° 3
Let u represent the speed of S ′ relative to S. Then because there is no x -motion in S , we can write v (^) x = vx ′ + u = 0 so that v ′ = − x u = −10 0. m s. Hence the ball is thrown backwards in S ′. Then,
v (^) y = v (^) y ′ = 3 vx ′ =10 0. 3 m s
Using vy^2^ = 2 gh we find
h =
( )
2 . .
m s m s 2 m
The motion of the ball as seen by the student in S ′ is shown in diagram (b). The view of the professor in S is shown in diagram (c).
FIG. P4.
Motion in Two Dimensions 81
*P4.40 (a) To an observer at rest in the train car, the bolt accelerates downward and toward the rear of the train. a = ( ) + ( ) =
=
tan
m s m s m s m
2
θ ss m s
to the south from
2 9 80 2 0 255 14 3
θ = ° tthe vertical
To this observer, the bolt moves as if it were in a gravitational field of 9.80 ms^2 down + 2.50 ms^2 south.
(b) a = 9 80. m s vertically downward^2
(c) If it is at rest relative to the ceiling at release, the bolt moves on a straight line downward and southward at 14.3 degrees from the vertical.
(d) The bolt moves on a parabola with a vertical axis.
P4.41 Choose the x axis along the 20-km distance. The y components of the displacements of the ship and the speedboat must agree: 26 40 15 50 11
( km h ) ( − ) = ( km h)
= −
t sin t sin
sin
° ° α
α 1 1 0 50
The speedboat should head
15 ° + 12 7. ° = 27 7. °east of north
Additional Problems
P4.42 (a) The speed at the top is v (^) x = vi cos θ i = ( 143 m s ) cos 45 ° = 101 m s.
(b) In free fall the plane reaches altitude given by v (^) yf^2 vyi^2 ay y (^) f yi 2
= ( m s sin ° ) + −. m s ft
ft m f
y
y
f
f
31 000 522 3 28.^ tt m
ft 1
⎛⎝ ⎞⎠ = 3 27. × 10 4.
(c) For the whole free fall motion v (^) yf = vyi + a ty − = + − ( ) =
m s m s m s s
.^2
t t (d) a c (^) r
v^2
v = a rc = 0 8 9 8. (. m s 2 ) 4 130, m = 180 m s
FIG. P4.
15º
N
E
40º 25º
α
x
y
Motion in Two Dimensions 83
P4.45 Refer to the sketch. We find it convenient to solve part (b) f irst.
(b) ∆ x = vxi t ; substitution yields 130 = ( vi cos 35 0. °) t.
∆ y = v (^) yit + at
(^2) ; substitution yields
. = ( vi sin. °) + t ( −9 80. ) t^2
Solving the above by substituting vi t = 159 gives 20 = 91 − 4.9 t^2 so t = 3 81. s.
(a) substituting back gives vi = 41 7. m s
(c) v (^) yf = vi sin θ (^) i − gt , v (^) x = vi cos θ i
At t = 3 81. s, vyf = 41 7. sin 35 0. ° − ( 9 80. )( 3 81. ) = −13 4. m s
v v v v
x
f x yf
. cos.. .
° m s m ss.
P4.46 At any time t , the two drops have identical y -coordinates. The distance between the two drops is then just twice the magnitude of the horizontal displacement either drop has undergone. Therefore,
d = 2 x t ( ) = 2 ( v (^) xit ) = 2 ( vi cos θ i ) t = 2 vit cosθ i
P4.47 (a) (^) a c (^) r
( )
v^2 5 00^2 1 00
m s m m s 2
a (^) t = g = 9 80. m s 2
(b) See figure to the right.
(c) (^) a = a (^) c^2^ + at^2 = ( 25 0. m s 2 ) + (^2 9 80. m s 2 ) 2 = 26 8. m s^2
φ =
tan −^ = tan − =
a a
t c
m s m s
2 2 4 4°
P4.48 (a) The moon’s gravitational acceleration is the probe’s centripetal acceleration: (For the moon’s radius, see end papers of text.)
a r
( ) =^ × = ×
v
v
v
2
2 6 6
m s m m
2
(^2) ss 2 = 1 69. km s
(b) v = 2 π r T
T r = =
×
×
= ×
π π^63 v
m) 1.69 10 m s 3 s==^ 1 80.^ h
FIG. P4.
FIG. P4.
84 Chapter 4
*P4.49 (a) We find the x coordinate from x = 12 t. We find the y coordinate from 49 t − 4.9 t^2. Then we fi nd the projectile’s distance from the origin as ( x^2 + y^2 ) 1/2^ , with these results:
t (s) 0 1 2 3 4 5 6 7 8 9 10 r (m) 0 45.7 82.0 109 127 136 138 133 124 117 120
(b) From the table, it looks like the magnitude of r is largest at a bit less than 6 s. The vector v tells how r is changing. If v at a particular point has a component along r , then r will be increasing in magnitude (if v is at an angle less than 90° from r ) or decreasing (if the angle between v and r is more than 90°). To be at a maximum, the distance from the origin must be momentarily staying constant, and the only way this can happen is for the angle between velocity and displacement to be a right angle. Then r will be changing in direction at that point, but not in magnitude.
(c) The requirement for perpendicularity can be defined as equality between the tangent of the angle between v and the x direction and the tangent of the angle between r and the y direc- tion. In symbols this is (9.8 t − 49) 12 = 12 t (49 t − 4.9 t^2 ), which has the solution t = 5.70 s, giving in turn r = 138 m. Alternatively, we can require dr^2 dt = 0 = ( d dt )[(12 t )^2 + (49 t − 4.9 t^2 )^2 ], which results in the same equation with the same solution.
*P4.50 (a) The time of fl ight must be positive. It is determined by yf = yi + vyit − (12) a (^) yt^2
0 = 1.2 + v 0 sin 35° t – 4.9 t^2 from the quadratic formula as (^) t = 0 574^ +^ 0 329^ +23 52 9 8
0 0
v v
Then the range follows from x = vxit + 0 = v 0 t as x (^) ( v (^) 0 ) = v (^) 0 0 164 3. + 0 002 299. v (^) 02 +0 047 94. v 02 where x is in meters and v 0 is in meters per second.
(b) Substituting v 0 = 0.1 gives x ( v 0 ) = 0.0410 m (c) Substituting v 0 = 100 gives x ( v 0 ) = (^) 961 m (d) When v 0 is small, v 02 becomes negligible. The expression x ( v 0 ) simplifies to v (^) 0 0 164 3. + 0 + 0 = 0 405. v 0 Note that this gives nearly the answer to part (b). (e) When v 0 is large, v 0 is negligible in comparison to v 02. Then x ( v 0 ) simplifies to x (^) ( v (^) 0 ) ≈ v (^) 0 0 + 0 002 299. v (^) 02 + 0 047 94. v 02 =0 0959. v 02 This nearly gives the answer to part (c).
(f) The graph of x versus v 0 starts from the origin as a straight line with slope 0.405 s. Then it curves upward above this tangent line, getting closer and closer to the parabola x = (0.095 9 s^2 m) v 02
