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High School Physics : Understanding Motion with Constant Acceleration

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Example Questions

Example Question #1 : Understanding Motion With Constant Acceleration

A ball rolls to a stop after 4.2s $\displaystyle 4.2s$. If it had a starting velocity of $8.1\frac{m}{s}$ $\displaystyle 8.1\frac{m}{s}$, what is the deceleration on the ball due to friction?

Possible Answers:

$-1.93\frac{m}{s^2}$ $\displaystyle -1.93\frac{m}{s^2}$

$-0.526\frac{m}{s^2}$ $\displaystyle -0.526\frac{m}{s^2}$

$6.93\frac{m}{s^2}$ $\displaystyle 6.93\frac{m}{s^2}$

$-4.2\frac{m}{s^2}$ $\displaystyle -4.2\frac{m}{s^2}$

$3.9\frac{m}{s^2}$ $\displaystyle 3.9\frac{m}{s^2}$

Correct answer:

$-1.93\frac{m}{s^2}$ $\displaystyle -1.93\frac{m}{s^2}$

Explanation:

We are given the initial velocity, time, and final velocity (zero because the ball stops). Using these values and the appropriate motion equation, we can solve for the acceleration.

Acceleration is given by the change in velocity over time:

$a=\frac{v_2-v_1}{t}$ $\displaystyle a=\frac{v_2-v_1}{t}$

We can use our values to solve for the acceleration.

$a=\frac{0\frac{m}{s}-8.1\frac{m}{s}}{4.2s}$ $\displaystyle a=\frac{0\frac{m}{s}-8.1\frac{m}{s}}{4.2s}$

$a=\frac{-8.1\frac{m}{s}}{4.2s}$ $\displaystyle a=\frac{-8.1\frac{m}{s}}{4.2s}$

$a=-1.93\frac{m}{s^2}$ $\displaystyle a=-1.93\frac{m}{s^2}$

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Example Question #1 : Understanding Motion With Constant Acceleration

An object starts moving with a velocity, $\displaystyle v$. If it accelerates in such a way so that the velocity doubles every second, what will be the velocity after $\displaystyle 3s$?

Possible Answers:

$\displaystyle 8v$

$\displaystyle 4v$

$\displaystyle v$

$\displaystyle 6v$

$\displaystyle 2v$

Correct answer:

$\displaystyle 8v$

Explanation:

If the velocity doubles every second, and it starts with a velocity of $\displaystyle v$, then after $\displaystyle 1s$ it would have a velocity of:

2*v=2v $\displaystyle 2*v=2v$

After $\displaystyle 2s$ it would have a velocity of:

2*2v=4v $\displaystyle 2*2v=4v$

After $\displaystyle 3s$ it would have a velocity of:

2*4v=8v $\displaystyle 2*4v=8v$

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Example Question #1 : Understanding Motion With Constant Acceleration

A ball is dropped off of a cliff with an initial velocity of $0\frac{m}{s}$ $\displaystyle 0\frac{m}{s}$. If it is in the air for $\displaystyle s$ seconds, what will be its final velocity as it hits the ground?

Possible Answers:

$\frac{g}{s}$ $\displaystyle \frac{g}{s}$

$\displaystyle (g)(s)^2$

We need to know the height of the cliff in order to solve

(g)(s) $\displaystyle (g)(s)$

We need to know the mass of the ball in order to solve

Correct answer:

(g)(s) $\displaystyle (g)(s)$

Explanation:

The basic formula for final velocity is:

$\displaystyle v_f=v_i+at$

Since the ball starts at rest, we can simplify the equation by removing the initial velocity.

v_f=at $\displaystyle v_f=at$

Since the acceleration on the ball will be the acceleration due to gravity, $\displaystyle g$, and the time will be $\displaystyle s$, as per the problem, we can insert these variables into the equation to get our answer.

$\displaystyle v_f=(g)(s)$

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Example Question #1 : Understanding Motion With Constant Acceleration

A box slides along the floor. If it starts with a velocity of $2\frac{m}{s}$ $\displaystyle 2\frac{m}{s}$ , has a final velocity of $6\frac{m}{s}$ $\displaystyle 6\frac{m}{s}$, and has been sliding for $\displaystyle 4s$, what is the average acceleration of the box?

Possible Answers:

$16\frac{m}{s^2}$ $\displaystyle 16\frac{m}{s^2}$

$0\frac{m}{s^2}$ $\displaystyle 0\frac{m}{s^2}$

$8\frac{m}{s^2}$ $\displaystyle 8\frac{m}{s^2}$

$1\frac{m}{s^2}$ $\displaystyle 1\frac{m}{s^2}$

$4\frac{m}{s^2}$ $\displaystyle 4\frac{m}{s^2}$

Correct answer:

$1\frac{m}{s^2}$ $\displaystyle 1\frac{m}{s^2}$

Explanation:

Acceleration is the change in velocity over the change in time.

$a=\frac{\Delta v}{\Delta t}=\frac{v_f-v_i}{\Delta t}$ $\displaystyle a=\frac{\Delta v}{\Delta t}=\frac{v_f-v_i}{\Delta t}$

We are given the initial and final velocities, as well as the change in time. We can use these values to calculate the acceleration.

$a=\frac{6\frac{m}{s}-2\frac{m}{s}}{4s}$ $\displaystyle a=\frac{6\frac{m}{s}-2\frac{m}{s}}{4s}$

$a=\frac{4\frac{m}{s}}{4s}$ $\displaystyle a=\frac{4\frac{m}{s}}{4s}$

$a=1\frac{m}{s^2}$ $\displaystyle a=1\frac{m}{s^2}$

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Example Question #1 : Understanding Motion With Constant Acceleration

A baseball leaves a pitcher's hand at $41\frac{m}{s}$ $\displaystyle 41\frac{m}{s}$. When it crosses home plate 0.24s $\displaystyle 0.24s$ later, it is moving at $40.22\frac{m}{s}$ $\displaystyle 40.22\frac{m}{s}$. What is the deceleration due to air resistance?

Possible Answers:

$-0.78\frac{m}{s^2}$ $\displaystyle -0.78\frac{m}{s^2}$

$-0.54\frac{m}{s^2}$ $\displaystyle -0.54\frac{m}{s^2}$

$-3.25\frac{m}{s^2}$ $\displaystyle -3.25\frac{m}{s^2}$

$-0.187\frac{m}{s^2}$ $\displaystyle -0.187\frac{m}{s^2}$

$-0.308\frac{m}{s^2}$ $\displaystyle -0.308\frac{m}{s^2}$

Correct answer:

$-3.25\frac{m}{s^2}$ $\displaystyle -3.25\frac{m}{s^2}$

Explanation:

For this question, we will need to use the acceleration formula:

$a=\frac{\Delta v}{\Delta t}=\frac{v_2-v_1}{\Delta t}$ $\displaystyle a=\frac{\Delta v}{\Delta t}=\frac{v_2-v_1}{\Delta t}$

We are given the initial velocity, final velocity, and time. Using these values in the equation, we can solve for the acceleration.

$a=\frac{40.22\frac{m}{s}-41\frac{m}{s}}{0.24s}$ $\displaystyle a=\frac{40.22\frac{m}{s}-41\frac{m}{s}}{0.24s}$

$a=\frac{-0.78\frac{m}{s}}{0.24s}$ $\displaystyle a=\frac{-0.78\frac{m}{s}}{0.24s}$

$a=-3.25\frac{m}{s^2}$ $\displaystyle a=-3.25\frac{m}{s^2}$

Note that the acceleration is negative, showing that the ball is slowing down, or decelerating.

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Example Question #4 : Understanding Motion With Constant Acceleration

What is the final velocity of a car that starts from rest and accelerates at $3.1\frac{m}{s^2}$ $\displaystyle 3.1\frac{m}{s^2}$ for 12s $\displaystyle 12s$?

Possible Answers:

$37.2\frac{m}{s}$ $\displaystyle 37.2\frac{m}{s}$

$1.25\frac{m}{s}$ $\displaystyle 1.25\frac{m}{s}$

$3.87\frac{m}{s}$ $\displaystyle 3.87\frac{m}{s}$

$115.32\frac{m}{s}$ $\displaystyle 115.32\frac{m}{s}$

$15.1\frac{m}{s}$ $\displaystyle 15.1\frac{m}{s}$

Correct answer:

$37.2\frac{m}{s}$ $\displaystyle 37.2\frac{m}{s}$

Explanation:

To solve this problem use the equation:

$v_f=v_i+a\Delta t$ $\displaystyle v_f=v_i+a\Delta t$

We are given the acceleration and time, and we can assume the initial velocity is zero since the car starts from rest. Use these values to solve for the final velocity.

$v_f=0\frac{m}{s}+(3.1\frac{m}{s^2})(12s)$ $\displaystyle v_f=0\frac{m}{s}+(3.1\frac{m}{s^2})(12s)$

$v_f=37.2\frac{m}{s}$ $\displaystyle v_f=37.2\frac{m}{s}$

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Example Question #5 : Understanding Motion With Constant Acceleration

A car starts at rest and accelerates at a constant rate of $0.88\frac{m}{s^2}$ $\displaystyle 0.88\frac{m}{s^2}$. What is its velocity after 31s $\displaystyle 31s$?

Possible Answers:

$27.28\frac{m}{s}$ $\displaystyle 27.28\frac{m}{s}$

$31.88\frac{m}{s}$ $\displaystyle 31.88\frac{m}{s}$

$35.23\frac{m}{s}$ $\displaystyle 35.23\frac{m}{s}$

$62.88\frac{m}{s}$ $\displaystyle 62.88\frac{m}{s}$

$272\frac{m}{s}$ $\displaystyle 272\frac{m}{s}$

Correct answer:

$27.28\frac{m}{s}$ $\displaystyle 27.28\frac{m}{s}$

Explanation:

Acceleration is equal to a change in velocity divided by a change in time.

$a=\frac{\Delta v}{\Delta t}$ $\displaystyle a=\frac{\Delta v}{\Delta t}$.

Since the car starts at rest, the initial velocity is zero. We know the acceleration and the change in time. Using these values, we can solve for the final velocity.

We can plug in our given values and solve:

$a=\frac{\Delta v}{\Delta t}=\frac{v_f-v_i}{\Delta t}$ $\displaystyle a=\frac{\Delta v}{\Delta t}=\frac{v_f-v_i}{\Delta t}$

$0.88\frac{m}{s^2}=\frac{v_f-0\frac{m}{s}}{31s}$ $\displaystyle 0.88\frac{m}{s^2}=\frac{v_f-0\frac{m}{s}}{31s}$

$(31s)(0.88\frac{m}{s^2})=v_f-0\frac{m}s{}$ $\displaystyle (31s)(0.88\frac{m}{s^2})=v_f-0\frac{m}s{}$

$27.28\frac{m}{s}=v_f$ $\displaystyle 27.28\frac{m}{s}=v_f$

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Example Question #1 : Understanding Motion With Constant Acceleration

A car starts from rest, speeds up with constant acceleration and travels 100m in 5s. What is the final velocity and the acceleration of the car?

Possible Answers:

$V_{f}=32\frac{m}{s};a=6.7\frac{m}{s^2}$ $\displaystyle V_{f}=32\frac{m}{s};a=6.7\frac{m}{s^2}$

$V_{f}=24\frac{m}{s};a=8\frac{m}{s^2}$ $\displaystyle V_{f}=24\frac{m}{s};a=8\frac{m}{s^2}$

$V_{f}=50\frac{m}{s};a=10\frac{m}{s^2}$ $\displaystyle V_{f}=50\frac{m}{s};a=10\frac{m}{s^2}$

$V_{f}=40\frac{m}{s};a=8\frac{m}{s^2}$ $\displaystyle V_{f}=40\frac{m}{s};a=8\frac{m}{s^2}$

$V_{f}=25\frac{m}{s};a=15\frac{m}{s^2}$ $\displaystyle V_{f}=25\frac{m}{s};a=15\frac{m}{s^2}$

Correct answer:

$V_{f}=40\frac{m}{s};a=8\frac{m}{s^2}$ $\displaystyle V_{f}=40\frac{m}{s};a=8\frac{m}{s^2}$

Explanation:

The easiest way to approach this problem is find the average velocity, multiply by 2 because the car started from rest, and then divide the final velocity by time to get the acceleration.

Average velocity:

$V_{avg}=\frac{d}{t}=\frac{100m}{5s}=20\frac{m}{s}$ $\displaystyle V_{avg}=\frac{d}{t}=\frac{100m}{5s}=20\frac{m}{s}$

Final velocity:

The final velocity is $40\frac{m}{s}$ $\displaystyle 40\frac{m}{s}$ because the car started from rest. This is evident from the equation for average velocity if we solve for final velocity:

$V_{avg}=\frac{V_{0}+V_{f}}{2}$ $\displaystyle V_{avg}=\frac{V_{0}+V_{f}}{2}$

$V_0=2V_{avg}-V_f$ $\displaystyle V_0=2V_{avg}-V_f$

$V_0=2*40-40=40\frac{m}{s}$ $\displaystyle V_0=2*40-40=40\frac{m}{s}$

Acceleration:

$a=\frac{V_0-V_f}{t}=\frac{40\frac{m}{s}}{5s}=8\frac{m}{s^2}$ $\displaystyle a=\frac{V_0-V_f}{t}=\frac{40\frac{m}{s}}{5s}=8\frac{m}{s^2}$

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High School Physics : Understanding Motion with Constant Acceleration

Study concepts, example questions & explanations for High School Physics

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Example Questions

Example Question #1 : Understanding Motion With Constant Acceleration

Example Question #1 : Understanding Motion With Constant Acceleration

Example Question #1 : Understanding Motion With Constant Acceleration

Example Question #1 : Understanding Motion With Constant Acceleration

Example Question #1 : Understanding Motion With Constant Acceleration

Example Question #4 : Understanding Motion With Constant Acceleration

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Example Question #1 : Understanding Motion With Constant Acceleration

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