Physics 103
Exam # 1
1. A little mouse runs in a straight line, which we shall call the x axis. As an eager scientist, you record its velocity as a function of time and use the information you obtain to construct the graph plotted in the figure below, which shows the mouse’s velocity v as a function of time t. You have adopted the usual sign convention according to which quantities to the right are positive and those to the left are negative. Use the information on the graph to answer the following questions:
(a)What is the magnitude of the mouse's initial velocity?
The initial velocity of the mouse is +20 m/s.
(b)Find the mouse's greatest speed.
The mouse’s greatest speed is 40 m/s.
(c)When is the mouse moving to the right?
The mouse will be moving to the right as long as the velocity is positive, from t = 0s to t = 6s and from t = 9s to t = 10s.
(d)When does the mouse change direction?
The mouse changes direction at t=6s and at t=9s when the sigh of its velocity changes.
(e)When is the mouse speeding up?
The mouse is speeding up from t = 0s to 3s, t = 6s to 7s, and from t = 9s to 10s because that is when the absolute value of the velocity is increasing.
(f)When is the mouse not accelerating?
The mouse is not accelerating from t = 3s to 5s because its velocity is constant during this time interval.
(g)Do the kinematics expressions v = vo + at, and x = xo + vot + ½ at2 apply to this mouse’s motion throughout the 10s? Explain.
These expressions do not apply throughout the mouse’s motion because these expressions apply only when the acceleration is constant. The acceleration will be constant if the v vs. t curve is linear. The mouse’s motion is constant acceleration during the time intervals of t = 0s to 3s (a =6.67 m/s2), t = 3s to 5s (a =0 m/s2) and t = 5s to 6s (a =-40 m/s2). After t = 6s the v vs t curve is not linear so the acceleration is not constant.
2. As soon as a traffic light turns green, a car speeds up from rest with a constant acceleration of 3 m/s2. In the adjoining bike lane, a cyclist moves at constant velocity of 12 m/s. When the car starts to accelerate, the cyclist is 50 m ahead of the car. The car maintains a constant velocity after reaching and passing the bicycle.
(a) Draw a velocity versus time graph for the two objects for the first 15 s of motion.
(b) Draw a position versus time graph for the two objects for the first 15 s of motion.
(c) For how long is the bicycle ahead of the car?
Find the time at which the position of the car and bicycle are equal.
The position of the car is given by , while the position of the bicycle is given by. Setting these two equations equal to each other yields the following quadratic equation.
This quadratic equation (a = 1.5, b = -12 and c = -50) has two roots for Δt; 11.02s and -3.02s. Take the positive root for this problem. The car has the same position as the bicycle at 11.02 s so the bicycle is ahead of the car for 11.02 s.
(d) What is the speed of the car when it passes the bicycle?
Use the constant acceleration formula .
(e) At what moment is the speed of the car the same as the speed of the bicycle?
Again, use the constant acceleration formula but solve for Δt.
(f) By how much is the cyclist ahead of the car at this moment?
As in part (c), the position of the car is given by , while the position of the bicycle is given by. The distance the cyclist is ahead of the car is . So…
3.An ice sled powered by a rocket engine starts from rest on a large frozen lake and accelerates at 2.5 m/s2. After a time t1 = 12 s, the rocket engine is shut down and the sled moves with constant velocity v for a time t2, until the sled travels a total distance d = 980 m.
(a) Find the velocity v.
Use the constant acceleration equation , where vi = 0, a = 2.5 m/s2 and Δt = 12 s.
The sled’s maximum velocity is 30 m/s.
(b) Find the time t2.
The distance travelled by the sled in the constant acceleration phase from 0 s to 12 s is given by
So, the sled has to travel another (980 m – 180 m) = 800 m while at a constant velocity of 30 m/s (from part (a)). The sled will take to travel the additional 800 m reaching a total distance of 980 m. So the total time from start is 12 s + 26.7 s = 38.7 s. The time t2 this is 38.7 s.
At the 980 m mark, the sled begins to accelerate at –2 m/s2.
(c) What is the final position of the sled when it comes to rest?
Use to determine how far the sled takes to come to rest after initiating the deceleration.
The final position of the sled will be 980 m + 225 m = 1, 205 m from its starting point.
(d) What is the duration of the entire trip?
There are three phases to the trip. Find the times for each phase and then add them up.
Phase / How to determine Δt / ΔtAccelerating / Given as 12 s / 12 s
Coasting / Determined in part (b) / 26.7 s
Decelerating / Use where vf is 0 m/s, vi is 30 m/s and a is -2 m/s2. / 15 s
Overall / 53.7 s
The duration of the entire trip is 53.7 seconds.
(e) Draw the graph acceleration versus time for the motion.
(f) Draw the graph velocity versus time for the motion.
(g) Draw the graph position versus time for the motion.
4.A rocket is fired vertically with an upward acceleration of 20 m/s2. After 25 s, the engine shuts off and the rocket continues as a free particle.
(a) Draw the graph acceleration versus time for the motion.
(b) Draw the graph velocity versus time for the motion.
(c) Draw the graph position versus time for the motion.
(d) On each graph mark by the letter A the point where the engine is shut off.
(e) On each graph marl by letter B the point where the rocket reaches the maximum height.
(f) Find the highest point the rocket reaches.
This is a two part problem: one part for each period of constant acceleration.
1. When a = 20 m/s2, find the velocity the rocket has at the end of this first 25 s using
2. When a = -9.8 m/s2, find the height the rocket achieves using
Therefore the rocket reaches a maximum height of 6,250 m + 12,755 m = 19,005 m
Note the time it takes to reach maximum height after engine cutoff is given by
(g)Find the total time the rocket is in the air.
The rocket is in free fall from its maximum height of 19,005 m, so use
The rocket will free fall for 62.3 seconds. Thus the total flight time from launch to crash is 25s + 51s+ 62.3s = 138.3s.
(h) What is the speed of the rocket just before it hits the ground?
The rocket is in free fall from its maximum height of 19,005 m, so use
Note the rocket is traveling downward at the time of the crash so vf = -610 m/s.
5. A fireman 44.0 m away from a burning building directs a stream of water from a ground-level fire hose at an angle of 36.0° above the horizontal. If the speed of the stream as it leaves the hose is 40.0 m/s, at what height will the stream of water strike the building?
This is a projectile problem where the motion can be decomposed into two independent motions; constant velocity in the horizontal direction and constant acceleration in the vertical direction. As with 99.99% of projectile problems you must find the “time of flight” first. The time of flight in this case is the time it takes the water to traverse 44 meters horizontally.
Having found the time of flight we can now find the final height of the water stream, yf.
The stream of water will strike the building 22.9 m above the hose level.
6.A student decides to measure the muzzle velocity of a pellet shot from his gun. He points the gun horizontally. He places a target on a vertical wall a distance x = 3.6 m away from the gun. The pellet hits the target a vertical distance y = 0.18 m below the gun. What is the initial speed of the pellet?
This is a projectile problem where there are two independent types of motion
- The horizontal motion is constant velocity (if we ignore air resistance)
- The vertical motion is constant acceleration (free fall)
- The combined motion is
- Solve the two motion equations for vx and Δt
Solving for Δt gives 0.192 s and then solving for vx gives 18.8 m/s.
The initial speed of the pellet is 18.9 m/s.