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PH100-5T Mid-Term February 24, 2002 Dr. Shealy Name Instructions. Total exam with worth 100 points. The number of points for each question are given. There is a 20 point extra credit problem at the end of exam. Partial credit will be given based on the work you show on your paper. If you need extra space, write on back of exam pages and give directions to where your work is located for each question. Circle your answer.
1. Consider a particle whose position as a function of time is given by
x t ) 1 2 t 3 t^2 m. # a. (5 pts) What is the average velocity (speed) between the times t ) 0 and t ) 2 x (^) av ) x ^2 s^ 2 " sx ^0 ) 8 ms b. (5 pts) What is the instantaneous velocity (speed) at t ) 1 s? v t ) lim (^) t v 0 x t ^ t^ t " x t ) 2 6 t ms v 1 s ) 8 ms c. (5 pts) What is the average acceleration between the times t ) 0 and t ) 2 s? a (^) av ) v ^2 s^ 2 " s v ^0 ) 6 ms 2 d. (5 pts) What is the instantaneous acceleration at t ) 1 s? a t ) lim (^) t v 0 v t ^ t^ t " v t ) (^6) sm 2
2. (10 pts) The brakes on your automobile are capable of creating a deceleration of 17 (^) sft 2. If you are going 85 mph and suddenly see a state tropper, what is the minimum time in which you can get your car under the 55 mph speed limit? (Hint: 1 mph ) 1. 467 fts .) v ) v 0 at a ) (^417) sft 2 ; v 0 ) 85 mph ; v ) 55 mph
t ) v " av^0 )
55 " 85 mph '1.467 (^) mphft / s " (^17) sft 2 )^ 2. 59 s
3. A ball is thrown vertically upwards from the ground with a speed of 25. 2 ms. a. (5pts) How long does it take to reach its highest point? v ) v 0 4 gt ) 0 at the highest point t ) v g^0 ) 25.^
ms 9.8 (^) sm 2 )^ 2. 57 s b. (5 pts) How high does it rise? y ) v 0 t 4 g 2 t^2
y 2. 57 s ) 25. 2 ms 9 2. 57 s 4
9.8 (^) sm 2 2 2. 57 s
(^2) ) 32. 4 m
c. (10 pts) At what times will it be 27. 0 m above the ground and what are the speeds of the object at these times? Assume y t ) 27. 0 m ) v 0 t 4 g 2 t^2 Rearranging equation gives g 2 t
(^2 4) v 0 t 27 m ) 0
Find t by solving the above quadratic equation for t
t )
v 0 o v 02 " 27 m ' 2 ' g g )^
25.2 ms o10.2879 ms 9.8 mss^ )^ 3. 612 and^ 1. 5216 s
v 1. 5216 s ) v 0 4 g 1. 5216 s ) 10. 28 ms v 3. 612 s ) v 0 4 g 3. 612 s ) 4 10. 28 ms
4. (15 points) Suppose a particle starts at the origin with velocity v 0 ) 4 3. 5 i 4. 7 j , in m/s. Its acceleration is 2. 1 i 1. 1 j , in m/s 2. What are the coordinates and velocity 5. 0 s later? What is the speed of the particle at t ) 5. 0 s? For motion with constant acceleration, the displacement of object as a funtion of time is given by r ) r 0 v 0 t a t
2 2 Evaluating the displacement t ) 5. 0 s gives r 5 s ) 5 s 4 3. 5 i 4. 7 j ms ^5 s^
2 2 2. 1 i^ ^ 1. 1 j ^
m s^2
r 5 s ) 8. 75 i 37. 25 j m
For motion with constant acceleration, the velocity of a particle as a function of time is given by v t ) v 0 a t ) 4 3. 5 i 4. 7 j ms 2. 1 i 1. 1 j (^) sm 2 t Evaluating the velocity of particle at t ) 5. 0 s gives v 5 s ) 7 i 10. 2 j ms The speed is the magnitude of the velocity which is for this case speed ) v ) v (^) x^2 v (^) y^2 ) 49 104. 04 ms ) 12. 3 ms
5. (15 points) 2. A projectile is fired with an initial velocity of 55 m / s , 25( above the horizontal. How long does it take to get to the highest point on its trajectory? How far above the launch point is the highest point? How far down range is the highest point? The x and y components of initial velocity are given by v (^) y 0 ) v 0 sin 25( ) 23. 244 ms and v (^) x 0 ) v 0 cos 25( ) 49. 845 ms The time for projectile to reach the top of its path is given when the vertical component of its velocity is equal to zero v (^) y ) 0 ) v (^) yo 4 gt Solving for t gives t (^) top ) v^0 sin 25
( g )^
9.8 s^ )^ 2. 37 s The highest point is found be evaluating y t (^) top y t (^) top ) v (^) y 0 t (^) top 4 g 2 t (^) top^2 ) 27. 57 m The distance down range of highest point is found by evaluating x t (^) top x t (^) top ) v (^) x 0 t (^) top ) 118. 1 m
6. (10 points) 3. An outfielder can throw a baseball at a speed of 35 m / s. At what angle above the horizontal should he throw it if he wants it to be caught by an infielder 95 m away? Assume the throwing and catching heights are thesame. We are given v 0 ) 35 ms What is the initial projection angle P 0. We know from the range formula of a projectile R )
v 02 sin (^220) g )^95 m Solving for P 0 P 0 ) 12 sin"^1 gRv 02
). 5 sin"^1 9.8 35 ' 295 ) 24. 7(
7. (10pts)2. A rock is thrown horizontally from a 115 m high cliff and strikes the ground 92. 5 m from the base of the cliff. At what speed was the rock initially thrown? x ) v (^) x 0 t ) 92. 5 m
