Section Test:Translational Motion

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Topics Covered

This Section Test deals with the basic one-dimensional and two-dimensional kinematics problems. Questions regarding objects in free fall, vectors, and projectiles are covered.


Section Test

1. A bowling ball is dropped from the top of a 200m high building. How long does it take the bowling ball to reach a speed of 98m/s (ignoring air resistance)?

0.1s
10s
20s
40s
→ Using the formula v_{f} = v_{i} + at, we can solve this question. In our situation, vi is zero, vf is 98m/s, acceleration is just gravity, or 9.8m/s*s and t is what we are solving for.

2. Starting from rest, a marble rolls freely down an incline that is 10 m long in 2.0 s. The acceleration of the marble is?

2.5 m/s2
5.0 m/s2
10 m/s2
20 m/s2
→ While this problem seems overly complicated in that one could potentially try to solve it using components of x and y we could not in fact solve it this way as we do not know the angle of the incline. The correct method is to recognize that we have initial velocity, 0m/s, distance, 10m, time, 2 seconds and the formula d = v_{i}t + \frac{1}{2}at^2 will do if we rearrange and solve for a. In doing this we get, a = \frac{2(d - v_{i}t)}{t^2} = \frac{2(10)}{2^2} = 5.

3. A driver of a race car has their foot to the accelerator but then begins to release it half way, the car's ______________.

speed will increase while its acceleration will decrease.
speed will increase while its acceleration will be negative.
speed will decrease while its acceleration will increase.
speed will decrease while its acceleration will be negative.
→ Consider what is happening, the car is moving forwards with a high acceleration (since the foot is to the floor) but when the foot is lifted a bit, the acceleration is now smaller (but still positive! no brakes were applied!) therefore, the speed will still increase (but not as fast) and the acceleration will decrease (but still be positive).

4. A student on a flat field throws a baseball 8√2 m/s at 45 degrees above the horizontal. How long does the ball remain in flight before it hits the ground (ignoring air resistance)?

0.4 seconds
0.8 seconds
1.6 seconds
3.2 seconds
→ Firstly, to answer this question one must determine how much velocity the ball is given in the vertical direction (since the horizontal direction will play no rule in deciding how long the ball is in the air). Drawing a triangle, with 45 degrees in it, we [get the sides to be 1, 1, and √2 (for the hyp). sin(45) = opp/hyp = 1/√2. In our problem the hyp is 8√2 thus, we have sin(45) = verticalcomp/8√2. Or (8√2)(1/√2) = verticalcomp = 8m/s. Secondly, to solve for how long the ball is in the air, we can simplify by solving for how long it takes for the ball to get to the peak where the velocity is zero, and then just double that time. We have initial velocity (8m/s), final velocity (at peak), acceleration (gravity), and we want time. We can solve this problem easily with v_{f} = v_{i} + at. Solving for t we get t = (vf -vi)/a = (0 - 8)/(-9.8) ~ 0.8seconds. This time however is the time it takes the ball to reach the peak height, thus we must double it to get the total time because it will fall back in exactly the same time (everything in reverse).

5. A car traveling north at 10 m/s shifts into reverse and experiences a uniform acceleration of 4 m/s2 south. What is the final velocity of the car after 5 seconds?

5 m/s north
5 m/s south
10 m/s north
10 m/s south
→ We have the cars initial velocity (10m/s north), the acceleration ( 4m/s2 south, or -4 north), the time (5s), and we want to solve for the final velocity. The equation v_{f} = v_{i} + at can be used directly without rearranging. Therefore, vf = 10 + (-4)(5) = 10 - 20 = -10. The negative implies the opposite of what we chose to be positive, which was north. Therefore our answer is 10 m/s south.

6. A ball is thrown horizontally off a cliff at 4m/s and after 6 seconds hits the ground below. How far will the ball have traveled in the horizontal direction after 3 seconds? (Gravity is 9. m/s2)

6 m
12 m
22.4 m
32.6 m
→ This question is a trap. A lot of useless information is provided to trick one into thinking they should be trying to involve acceleration of gravity some how. The fact of the matter is however, that this question is asking about the horizontal direction only, how far along the x direction has the ball gone. We are given an initial velocity, which is also in the horizontal direction, and a time of 3 seconds, none of the other information matters. The ball's x component will simply start at 4m/s and continue in the direction without changing until the ball finally hits the ground. Thus, the distance the ball travels in the x direction is just d = vt, or (4m/s)(3s)=12m.

7. A student standing on a bridge 32 m above the water drops a stone. After 5 seconds, the student throws a second stone straight down. What initial velocity must the second stone be given to ensure both stones hit the water below at the same time?

5 m/s
25 m/s
75 m/s
150 m/s
→ This problem must be solved in multiple parts. Firstly we will need to find out how long it takes the first stone to reach the ground. Using this time, we can then subtract the delay between the drop and the throw to determine how long the second stone is in the air. (The time of the second stone in the air plus the delay must equal the first stones time!) Using this time, we will be able to solve for the initial velocity of the second stone. The first stone is dropped from 32 meters, thus it follows that gravity will accelerate it down and we can use the basic formula d_{f}=d_{i} + v_{i}t + \frac{1}{2}at^{2}, which becomes \frac{2 \times 32}{\sqrt[2]{g} }, or 20.4. Using this time minus the 5 seconds, we can then solve for the initial velocity of the second stone, since we know the other terms (a = gravity, df = 30 m). Infact, we can use the same formula, d_{f}=d_{i} + v_{i}t + \frac{1}{2}at^{2}, but this time solve for vi and use 20.4 - 5 seconds. Solving for v we get v_{i} = \frac{d - \frac{1}{2}gt^{2}}{t} = \frac{32 - (\frac{1}{2}){(-9.8)}(15.4)^{2}}{15.4} = 77. (Note that in this case gravity was negative because we chose the distance down to be positive.)

8. A cannon ball is fired off a ship at 35 degrees above the horizontal, ignoring air resistance, what happens to the magnitude of the velocity for the duration of the flight?

The horizontal component remains constant while the vertical increases.
The horizontal component decreases while the vertical component increases but then decreases.
The horizontal component remains constant while the vertical component decreases but then increases.
The horizontal component decreases and the vertical component decreases.
→ From the horizontal point of view, the cannon ball is given an initial velocity and remains at that velocity because there are no forces (in the horizontal, there is no gravity, and from the question there is no air resistance) acting on the ball. In the vertical direction, the cannon ball is given an initial velocity, but as time passes the cannon balls velocity decreases because gravity is pulling down. Once the cannon ball reaches the peak of its height, the velocity becomes zero and then begins to increase in the opposite direction (down) and will continue to increase due to gravity until the ball hits the floor.

9. How long does it take a marble to hit the ground if it is dropped from the top of a building 200m high?

2.2s
3.8s
4.6s
6.4s
→ For this problem we know the distance, d=200m, and we know the acceleration, a=9.8m/ss. Using the formula, d = di+vit + (1/2)at2, we can solve this easily as di and vi are zero. Thus we get, d = (1/2)at2. Solving for t we get, t = sqrroot( 2d/a ) = sqrroot( 2 * 200 / 9.8 ) ~= 6.4s.

10. A sail boat at the center of a 4km wide river with a current of 5km/h south is pushed by the wind east at 10km/h. How far south, down the river from the bank, does the boat sail?

0.5 km
1 km
1.5 km
2 km
→ This problems deals with vectors, but it can be solved in parts. Firstly, we can determine how long it takes the boat to be pushed to the shore from the center of the river. The river is 4km wide, thus the boat is 2km from the shore, and the wind is pushing the boat at 10km/h towards the shore (east). Thus it would take, t = d/v = (2km)/(10km/h) = 0.2h to reach the shore. Now, using this time, we can determine how far the river current would push the boat south. d = vt = (5km/h south)(0.2h) = 1km.

11. A soccer ball is kicked off a cliff that is 50m, at 45 degrees above horizon at a velocity of 5√2 m/s. What is the distance from the base of the cliff to the place the soccer ball lands?

20 m
40 m
60 m
80 m
→ We must break this problem into components. The easiest way to solve this problem is to first determine how long the ball is in the air for, and then using this time of flight and knowing that the horizontal component of velocity will stay the same, solve for the distance with d=vt. Start with finding the time it takes for the ball to hit the ground going in the up/down (vertical) direction. Velocity in the vertical direction is 5√2m/s*sin (45 deg) = 5√2/√2 = 5m/s. Then, find the time it takes for the ball to hit the ground at 50 m below the cliff. This portion of the problem however is not so simple. If we try to find the total time using the formula d_{f}=d_{i} + v_{i}t + \frac{1}{2}at^{2} and setting df to -50m.
it will require solving a quadratic, which without a calculator is quite difficult, therefore we will break it up into two parts.
→ If we solve for the final velocity using the formula v_{f}^{2} = v_{i}^{2} + 2 a \Delta d we can get the final velcoity, which when combined with the formula v_{f} = v_{i} + at can allow us to solve for time. v_{f}^{2} = v_{i}^{2} + 2 a \Delta d = 5^{2} + 2(-9.8)(-100) which if we assume ~10 for accerlation it becomes 2025, the square root of this is the 45m/s. Using the second formula now, we can solve for the total time,t = \frac{v_{f} + v_{i}}{a} = \frac{-45 + 5}{9.8}, which equals about 4.
→ Using this time now, we can determine how far in the x direction the ball travels by multiplying the time by the x component of velcocity. 5√2m/s*cos (45 deg) = 5√2/√2 = 5m/s. So distance = 5m/s * 4s = 20m.

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