Practice Problems: Kinematics Solutions

1. (easy) How fast will an object (in motion along the x-axis) be moving at t = 10 s if it had a speed of  2 m/s at t = 0 and a constant acceleration of 2 m/s²?
v = v_o + at
v = 2 + 2(10)
v = 22 m/s

2. (easy) A car is rolling toward a cliff with an initial speed of 15 m/s. The maximum negative acceleration that the brakes can provide is -0.3 m/s². If the cliff is 350 m from the initial position of the car, will the car go over the cliff?
In order to stop, the final speed must be zero before the car moves 350 m.
v² = v_o² + 2a(x - x_o)
0 = 15² + 2 (-0.3)(x - 0)
x = 375 m
Thus, the car cannot stop in time and does go over the cliff.

3. (moderate) Which would have the greatest effect on the displacement of an object that is accelerating uniformly in one-dimensional motion: doubling the initial velocity or doubling the time of the acceleration? Additionally, does the magnitude of the acceleration play any role in the difference of effect between these two parameters?
Analysis of the second kinematic equation, x - x_o = v_ot + ½at², indicates that doubling either the time or the initial velocity would increase the displacement, but since the time is evident in both parts of the solution, it will have the greatest effect. The acceleration magnitude does mitigate the effect that time contributes. A larger acceleration indicates that the time effect would larger.

4. (moderate) Cart A moves with a uniform speed past point 1 on a straight track at 0.3 m/s. Cart B moves past point 1 at 0.1 m/s but is uniformly accelerating at 0.1 m/s². Point 2 is 1.0 m past point 1. Which cart gets to point 2 first?
Find the time for each cart to reach point 2:
Cart A:
x - x_o = v_ot + ½at²
1.0 - 0 = 0.3t + 0
t = 3.3 s
Cart B:
x - x_o = v_ot + ½at²
1.0 - 0 = 0.1t + ½(0.1)t²
Use the quadratic equation to deterime t
t = 3.6 s
Cart A arrives at point 2 first.

5. (easy) A small ball is released from a window at t = 0. Assuming free-fall conditions, how far does it travel in 2.8 seconds? If the ball had more mass would it fall a greater distance? 
Assume y_o = 0
y - y_o = v_ot + ½gt²
y - 0 = 0 +-4.9(2.8)²
y = -38.4 m

This means that the ball has fallen 38.4 m below its initial position.
Free-fall acceleration is independent of mass, so the distance fallen by a more massive object would be the same.

6. (moderate) Three spheres are held at at various positions above a table. Sphere 1 is closest to the table, sphere 3 is furthest from the table. Assume all collisions with the table are perfectly elastic. That is, no energy is lost upon impact and the sphere returns to its initial height before it falls again as if it was dropped from rest. Consider motion up away from the table to be positive motion.

For questions A and B assume the following:

Sphere 1 initial height = 2.0 m

Sphere 2 initial height = 3.0 m

Sphere 3 initial height = 4.0 m

Sphere 1 is released from rest.

A. At what speed must sphere 2 be initially traveling upward if it first hits the table at the same time that sphere 1 hits the table for a second time?
Time for sphere 1 to hit the first time:
x - x_o = v_ot + ½at² 
-2 = 0 - 4.9t²
t = 0.64s
It takes 3(0.64s)=1.92s to hit the second time (down, up, down)
Now, sphere 2 must take the same time to hit the first time:
x - x_o = -3 = v_ot + ½gt² 
-3 = v_o(1.92) - 4.9(1.92)²
v_o = 7.8 m/s

B. At what speed must sphere 3 be initially traveling downward if it first hits the table at the same time that sphere 1 hits the table for the first time.
x - x_o = -4 = v_ot + ½gt²= v_o(0.64) - 4.9(0.64)²
v_o = -3.1 m/s
This means that the sphere has a speed of 3.1 m/s.

C. Now assume that the initial upward speed of sphere 2 is 1 m/s and the initial upward speed of sphere 3 is 2 m/s, determine the difference in initial heights of spheres 2 and 3 if they hit the table for the first time when sphere 1 hits the table for the sixth time. Sphere 1 is released from rest.
Time for sphere 1 to hit the sixth time = 11(0.64) = 7.0 s
For sphere 2:
x - x_o = v_ot + ½gt²= 1(7) - 4.9(7)²
x - x_o = -233 m (meaning that it has to fall that far to hit the table)
For sphere 3:
x - x_o = v_ot + ½gt²= 2(7) -4.9(7)²
x - x_o = -226 m (meaning that it has to fall that far to hit the table)
The difference in the two initial heights is 7 m.

7. (easy) A cart is at x=5m at time t=0. The cart accelerates at 4 m/s². If the speed of the cart at t=0 is 3 m/s, find the position of the cart at t=2 s and also determine where the cart is when it reaches a speed of 5 m/s.
x - x_o = v_ot + ½at²
x - 5 = 3(2) + ½(4)(2)²
x = 19 m
v² = v_o² + 2a(x - x_o)
5² = 3² + (2)(4)(x - 5)
x = 7 m

8. (moderate) A car moving at 20 m/s passes a street corner. The car maintains this speed even though the speed limit is 10 m/s. The police car that was sitting at the corner begins to chase the car by accelerating at 2 m/s². How long will it take for the police car to catch the speeder? How far from the corner is the catch-up point? How fast will the police car be traveling at that time?
For the car: x = 20t
For the police car: x = ½(2)t²
At catch up point: 20t = ½(2)t²
Therefore, t = 20sec
position at catch up: x_{catch up} = 20t = 20(20) = 400 m
Speed at catch-up: v = v_o+ at = 0 + (2)(20) = 40 m/s

9. (hard) Two spheres are rolling toward each other. At t = 0, sphere 1 is at x = 0 and has a speed of 10 m/s to the right while sphere 2  is moving 2 m/s to the left and has an initial position of x = 1000 m. Observations of the spheres show the following data: -After 2 s, sphere 2 has picked up speed and is moving at 10 m/s to the left. This acceleration is maintained until the spheres collide.
-Sphere 1 is seen to have an acceleration of 2 m/s² to the right. How fast will each sphere be traveling when they collide?

Find the acceleration of sphere 2:
v = v_o + at = -10 = -2 + a(2)
a = -4 m/s²
Find the position at time of impact (t):
For sphere 1: x₁ = v_ot + ½at² 
x₁ = 10t + ½(2)t² = 10t + t²
For sphere 2: x₂ - x_o = v_ot + ½at² 
x₂ - 1000 = -2t - ½(-4)t² 
x₂ = 1000 - 2t - 2t² 
Now find the time to impact by setting x₁ = x₂
10t + t² = 1000 - 2t - 2t²
0 = -3t² -12t +1000
Using the quadratic formula: t = 16.4 s
Find the velocity (relative to the Earth) for each sphere at time of impact:
v₁ = = v_o + at = 10 + 2(16.4) = 43 m/s
v₂ = = v_o + at = -2 - 4(16.4) = -68m/s
Their speed are the magnitudes of these solutions.

10. (moderate) This problem is a followup to one you saw in the previous presentation about cars at a red light:
Cars are lined up (with 5.0 m of distance between each car) at a red light. Assume that each car is 4.6 m in length. When the light turns green, all cars accelerate at 1.22 m/s² for 10.0 seconds, and then proceed at a constant speed. If the light stays green for 90.0 seconds, how many cars make it to or through the intersection?
x₁₀ = distance each car moves in first 10 seconds
x₁₀ = v_ot + ½at² = 0 + ½ (1.22)10² = 61 m
v₁₀ = speed at 10 seconds = v_o + at
v₁₀ = 0 + (1.22)(10) = 12.2 m/s
x₈₀ = distance each car moves in final 80 seconds
x₈₀ = v₁₀(80s) = 976 m
Total distance the cars move in 90 s = 61 + 976 = 1037
Distance from front of one car to front of another car = 4.6 + 5 = 9.6 m
#cars through intersection = 1037/9.6 = 108
The final car makes just to edge of the intersection.
Thus, the final answer is that 109 cars make it to or through the interesection.

11. (moderate) Determine the distance between two steel spheres (after 1.4 s) dropped from a tower if the second sphere was dropped 0.5 seconds after the first. Assume free-fall and that the spheres are dropped from rest. 
(Use y_o = 0 and + is up)
Sphere 1 after 1.4 s:
y₁ = y_o + v_ot + ½gt²
y₁ = 0 + 0 - 4.9(1.4)² 
y₁ = -9.6 m
Sphere 2 after 0.9 s:
y₂ = y_o + v_ot + ½gt²
y₂ = 0 + 0 - 4.9(0.9)² 
y₁ = -4.0 m
The spheres will be 5.6 m apart at that time.

12. (hard) If a ball is tossed up (free-fall conditions) with an initial speed of 2.0 m/s does it spend more time in the top 0.1 m of the toss or the bottom 0.1 m of the toss?
(Use y_o = 0 and + is up)
First find the max height (where v = 0):
v² = v_o² + 2gΔy
0 = 2² - 19.6Δy
Δy = max height = 0.2 m
Thus, the top motion is from y = 0.1 m (on the way up) to y = 0.1 (on the way down):
We need to find the speed at y = 0.1:
v² = v_o² + 2gΔy
v² = 2² - 19.6(0.1)
v = 1.4 m/s
(by symmetry the velocity at y = 0.1 on the way down is opposite)
Now find time for the top motion:
v = v_o + gt
-1.4 = 1.4 - 9.8t
t = 0.29s
For the bottom motion on the way up:
y - y_o = ½(v + v_o)t
0.1 - 0 = ½(1.4 + 2)t 
t = 0.06 s
By symmetry, for the bottom motion on the way down, t = 0.06 s.
Thus, the top of the motion lasts 0.29 s while the bottom of the motion lasts 0.12 s. This helps to explain why a volleyball player with a big vertical leap can appear to hover in the air.

13. (moderate) A model rocket enthusiast launches a rocket with a motion sensor in the launchpad. Assume y = 0 at the launchpad, that positive is up, and that the fuel mass is very small compared to the rocket body mass. Create qualitative y-t, v-t, and a-t graphs for an experiment that starts at lift off and ends when the rocket hits the Earth on the way back down. Assume free fall after the rocket fuel is used up.