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MCAT Physical : Motion in One Dimension

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Example Question #11 : Motion In One Dimension

In five seconds a runner moves from the 50m mark to the 75m mark along a straight path. What is the runner's average velocity?

Possible Answers:

$5\frac{m}{s}$

$12\frac{m}{s}$

$3\frac{m}{s}$

$8\frac{m}{s}$

$10\frac{m}{s}$

Correct answer:

$5\frac{m}{s}$

Explanation:

To find the average velocity, the displacement needs to be measured. The displacement the difference betweent eh final position and the initial position.

75m-50m=25m

Using the equation for velocity, $velocity=\frac{displacement}{time}$ , we can find the average velocity.

$v=\frac{25m}{5s}=5\frac{m}{s}$

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Example Question #12 : Motion In One Dimension

A child throws a ball straight up into the air. He throws the ball with an initial velocity of $\small 8\frac{m}{s}$ . Assume there is no air resistance.

How long will it take for the ball to reach a velocity of $\small 0\frac{m}{s}$ , while in the air?

Possible Answers:

1.6s

1.2 s

0.8s

The baseball will never have a velocity of $\small 0\frac{m}{s}$ .

Correct answer:

0.8s

Explanation:

In order to solve for the time at which the ball has a velocity of $\small 0\frac{m}{s}$ , we need to use an equation which incorporates all of the known variables. We know the acceleration due to gravity, the initial velocity, and the final velocity. As a result, the best equation to use is the one that allows us to simply solve for the unknown variable, time.

$\small v = v_{0} + at$

$\small 0\frac{m}{s} = 8\frac{m}{s} + (-10\frac{m}{s^{2}})(t)$

$\small t = 0.8 s$

Keep in mind that 1.6s is also a time at which the velocity is $\small 0\frac{m}{s}$ , but this time refers to the point when the boy catches the ball. The question asks for the specific time while the ball is still in the air, so the correct answer, 0.8s, refers to the point where the ball is at its peak height.

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Example Question #1 : Motion In One Dimension

If a 15kg ball takes five seconds to strike the ground when released from rest, at what height was the ball dropped?

Possible Answers:

50m

75m

125m

250m

100m

Correct answer:

125m

Explanation:

Using the equation d=v_ot + (1/2)a(t^2) we can find the distance at which the ball was dropped. Notice that the mass of the ball does not matter in this problem. We are told that the ball is dropped from rest making, v_o =0 , thus we have d=(1/2)a(t^2) . When we plug in our values, and assuming that acceleration is equal to gravity (10m/s²) we find that d=(1/2)(10)(5^2) = 125m.

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Example Question #2 : Motion In One Dimension

How far will an object travel after ten seconds if it is dropped into a bottomless pit?

Possible Answers:

250m

500m

25m

50m

300m

Correct answer:

500m

Explanation:

Since the object is dropped, the inital velocity is zero. Gravity is the only acceleration, the time is ten seconds, and the distance at which the object travels is unknown.

The equation d=v_ot +(1/2)at^2 can be used to find the distance traveled.

d=0+(1/2)(10m/s^2)(10s)^2=500m

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Example Question #1 : Motion In One Dimension

How long does it take an object to travel a distance of 30m from rest at a constant acceleration of 2m/s²?

Possible Answers:

2.7s

12.6s

10.9s

7.3s

5.5s

Correct answer:

5.5s

Explanation:

Using the equation d=v_ot + (1/2)at^2 , we can solve for time.

Since the object started at rest, v_0=0 . Now we are left with the equation d=(1/2)at^2 .

$30m=\frac{1}{2}(2\frac{m}{s^2})t^2$

30s^2=t^2

t=5.5s

Plugging in the remaining values we can find that t = 5.5s.

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Example Question #13 : Motion In One Dimension

A car is moving with a constant velocity of $10\frac{m}{s}$ when it abruptly stops, applying a constant breaking acceleration of $-5\frac{m}{s^2}$ . Over what distance does the car come to a stop?

Possible Answers:

10m

15m

50m

20m

Correct answer:

10m

Explanation:

This problem can be easily solved using the formula, $v{_{f}}^{2} = v{_{0}}^{2} + 2ax$ , and solving for . Our initial velocity is $10\frac{m}{s}$ , acceleration is $-5\frac{m}{s^2}$ , and final velocity is $0\frac{m}{s}$ .

$(0\frac{m}{s})^2=(10\frac{m}{s})^2+2(-5\frac{m}{s^2})x$

$0=100\frac{m^2}{s^2}-10\frac{m}{s^2}x$

$-100\frac{m^2}{s^2}=-10\frac{m}{s^2}x$

$\frac{-100\frac{m^2}{s^2}}{-10\frac{m}{s^2}}=10m=x$

The car comes to a stop after applying the breaks over 10m.

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Example Question #14 : Motion In One Dimension

A sprinter running a 100m race accelerates constantly at $0.5\frac{m}{s^2}$ from rest. What is his approximate final velocity as he crosses the finish line?

Possible Answers:

$7\frac{m}{s}$

$14\frac{m}{s}$

$50\frac{m}{s}$

$10\frac{m}{s}$

Correct answer:

$10\frac{m}{s}$

Explanation:

To answer this question, we must have a solid understanding of the kinematics equations. For this question, we must use an equation relating final velocity, distance, and acceleration.

The best fit for this is $\nu_{f}^{2} - \nu_{o}^{2} = 2ax$ .

Since we are solving for final velocity, and we started from rest, we can simplify the equation.

$\nu_o=0\frac{m}{s}, a=0.5\frac{m}{s^2}, x=100m$

$\nu_{f}^{2} - (0\frac{m}{s})^{2} = 2(0.5\frac{m}{s^2})(100m)$

$\nu_{f} = \sqrt{2(0.5\frac{m}{s^2})(100m)} = \sqrt{100\frac{m^2}{s^2}}=10\frac{m}{s}$

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Example Question #15 : Motion In One Dimension

An car accelerates from rest and travels 25m in 0.7s . What is the approximate acceleration of the car?

Possible Answers:

Correct answer:

Explanation:

The kinematics equation relating distance, time, and acceleration is $x = \upsilon_{o}t + \frac{1}{2}at^{2}$ .

Since we know that the car was initially at rest, we can rewrite the equation with zero initial velocity.

$x=\frac{1}{2}at^2$

Now we can plug in our values from the question, and solve for the acceleration.

$25m=\frac{1}{2}a(0.7s)^2$

$25m=\frac{1}{2}a(0.49s^2)=(0.245s^2)a$

$a=\frac{25m}{0.245s^2}=102\frac{m}{s^2}\approx 100\frac{m}{s^2}$

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Example Question #16 : Motion In One Dimension

If an ball is dropped from a cliff $\small 80m$ high, how long will it take the ball to strike the ground?

Possible Answers:

5.05s

16.3s

4.04s

12.7s

Correct answer:

4.04s

Explanation:

Since acceleration is constant, we can use the appropriate kinematics equation to solve:

$d=v_0t + \frac{1}{2}at^2$

We are given the height of the cliff, which will be equal to the distance traveled. The initial velocity is zero since the ball starts from rest and the acceleration will be equal to the acceleration due to gravity. Use these values to calculate the time.

$80m=(0\frac{m}{s})t+\frac{1}{2}(9.8\frac{m}{s^2})t^2$

$t^2=\frac{80m}{\frac{1}{2}(9.8\frac{m}{s^2})}=16.3s^2$

$t=\sqrt{16.3s^2}=4.04s$

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Example Question #17 : Motion In One Dimension

A 2kg mass is suspended on a rope that wraps around a frictionless pulley. The pulley is attached to the ceiling and has a mass of 0.01kg and a radius of 0.25m. The other end of the rope is attached to a massless suspended platform, upon which 0.5kg weights may be placed. While the system is initially at equilibrium, the rope is later cut above the weight, and the platform subsequently raised by pulling on the rope.

Screen_shot_2013-10-09_at_10.32.21_pm

What is the velocity of the platform two seconds after the rope is cut?

Possible Answers:

$3.4\frac{m}{s}$

$19.6\frac{m}{s}$

$26.4\frac{m}{s}$

$9.8\frac{m}{s}$

$5.1\frac{m}{s}$

Correct answer:

$19.6\frac{m}{s}$

Explanation:

Think back to the 3 main kinematics equations that we know.

v_f² = v_i² + 2aΔx

v_f = v_i + at

Δx = v_it + ½at²

We need to determine which formula will allow us to find final velocity after a given amount of time.

v_f = v_i + at = (0m/s) + (9.8m/s²)(2s) = 19.6m/s

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MCAT Physical : Motion in One Dimension

Study concepts, example questions & explanations for MCAT Physical

All MCAT Physical Resources

Example Questions

Example Question #11 : Motion In One Dimension

Example Question #12 : Motion In One Dimension

Example Question #1 : Motion In One Dimension

Example Question #2 : Motion In One Dimension

Example Question #1 : Motion In One Dimension

Example Question #13 : Motion In One Dimension

Example Question #14 : Motion In One Dimension

Example Question #15 : Motion In One Dimension

Example Question #16 : Motion In One Dimension

Example Question #17 : Motion In One Dimension

All MCAT Physical Resources

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