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High School Physics : High School Physics

Study concepts, example questions & explanations for High School Physics

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6 Diagnostic Tests 233 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #132 : Linear Motion

During a storm, you can usually see the lightning before you hear the thunder, unless you are very close to the lightning strike. What causes this discrepancy?

Possible Answers:

We need to know the voltage of the lightning in order to determine the answer

The speed of sound is much faster than the speed of light

There is no definitive scientific reason for this phenomenon

The speed of light is much faster than the speed of sound

We need to know the current of the lightning in order to determine the answer

Correct answer:

The speed of light is much faster than the speed of sound

Explanation:

Assuming you stand in one place, the distance between you and the lightning strike does not change.

The formula for velocity is:

$v = \frac{\Delta x}{\Delta t}$

In this scenario, the distance travelled, $\Delta x$ , does not change. The time taken to travel this distance, $\Delta t$ , does change. That means that the velocity must also be changing.

This is an indirect relationship. As $\Delta t$ increases, will decrease; thus, the object with a greater time of travel (sound) will have a slower velocity.

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Example Question #1 : Understanding Motion With Constant Acceleration

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

Possible Answers:

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

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

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

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

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

Correct answer:

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

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:

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

We can use our values to solve for the acceleration.

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

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

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

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Example Question #1 : Understanding Motion With Constant Acceleration

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

Possible Answers:

Correct answer:

Explanation:

If the velocity doubles every second, and it starts with a velocity of , then after it would have a velocity of:

2*v=2v

After it would have a velocity of:

2*2v=4v

After it would have a velocity of:

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 $0\frac{m}{s}$ . If it is in the air for seconds, what will be its final velocity as it hits the ground?

Possible Answers:

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

(g)(s)^2

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

$\frac{g}{s}$

(g)(s)

Correct answer:

(g)(s)

Explanation:

The basic formula for final velocity is:

v_f=v_i+at

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

v_f=at

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

v_f=(g)(s)

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Example Question #3 : Understanding Motion With Constant Acceleration

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

Possible Answers:

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

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

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

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

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

Correct answer:

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

Explanation:

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

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

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

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

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

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

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Example Question #202 : High School Physics

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

Possible Answers:

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

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

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

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

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

Correct answer:

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

Explanation:

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

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

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

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

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

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

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 $3.1\frac{m}{s^2}$ for 12s ?

Possible Answers:

$3.87\frac{m}{s}$

$115.32\frac{m}{s}$

$1.25\frac{m}{s}$

$15.1\frac{m}{s}$

$37.2\frac{m}{s}$

Correct answer:

$37.2\frac{m}{s}$

Explanation:

To solve this problem use the equation:

$v_f=v_i+a\Delta t$

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.

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

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

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Example Question #5 : Understanding Motion With Constant Acceleration

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

Possible Answers:

$27.28\frac{m}{s}$

$31.88\frac{m}{s}$

$35.23\frac{m}{s}$

$62.88\frac{m}{s}$

$272\frac{m}{s}$

Correct answer:

$27.28\frac{m}{s}$

Explanation:

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

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

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:

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

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

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

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

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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:

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

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

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

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

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

Correct answer:

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

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:

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

Final velocity:

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

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

$V_0=2V_{avg}-V_f$

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

Acceleration:

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

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Example Question #1 : Understanding Circular Motion

Ellen is swinging a 0.01kg yo-yo in a circular path perpendicular to the ground. The yo-yo moves in a clockwise direction with a constant speed of $2\frac{m}{s}$ .

What is the velocity of the yo-yo at the bottom of the circle?

Possible Answers:

$2\frac{m}{s}\text{ down}$

$0\frac{m}{s}$

$2\frac{m}{s}\text{ to the left}$

$2\frac{m}{s}\text{ up}$

$2\frac{m}{s}\text{ to the right}$

Correct answer:

$2\frac{m}{s}\text{ to the left}$

Explanation:

Remember, when working with circular motion, the velocity is ALWAYS tangential to the circle. This means that, even though the speed is constant, the direction is always tangent to the edge of the circle. If the circle below represents the path of the yo-yo, and it moves in a clockwise direction, then the velocity at the bottom of the path will be to the left.

Screen_shot_2013-12-16_at_12.54.09_pm

The magnitude of the velocity is constant, so the final answer will be $2\frac{m}{s}\text{ to the left}$ .

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High School Physics : High School Physics

Study concepts, example questions & explanations for High School Physics

All High School Physics Resources

Example Questions

Example Question #132 : Linear Motion

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 #3 : Understanding Motion With Constant Acceleration

Example Question #202 : High School Physics

Example Question #4 : Understanding Motion With Constant Acceleration

Example Question #5 : Understanding Motion With Constant Acceleration

Example Question #1 : Understanding Motion With Constant Acceleration

Example Question #1 : Understanding Circular Motion

All High School Physics Resources

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