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High School Physics : Understanding Distance, Velocity, and Acceleration

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Example Questions

Example Question #92 : Linear Motion

Leslie rolls a ball out of a window from 10 meters above the ground, such that the initial y-velocity is zero. How long will it be before the ball hits the ground?

$\small g=-9.8\frac{m}{s^2}$

Possible Answers:

0.73 s

1.43 s

1.01 s

2.04 s

1.86 s

Correct answer:

1.43 s

Explanation:

We are given the initial velocity, acceleration, and distance traveled. Using the equation below, we can solve for the time.

$\Delta y=V_i+\frac{1}{2}at^2$

$(y_f-y_i)=V_i+\frac{1}{2}at^2$

$(0m-10m)=0\frac{m}{s}+\frac{1}{2}(-9.8\frac{m}{s^2})t^2$

$(-10m)=(-4.9\frac{m}{s^2})t^2$

The distance is negative, which makes since because the ball is traveling downward.

$\frac{-10m}{-4.9\frac{m}{s^2}}=t^2$

$2.04s^2\approx t^2$

$\sqrt{2.04s^2}\approx \sqrt{t^2}$

$1.43s\approx t$

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Example Question #1 : Understanding Distance, Velocity, And Acceleration

Derek rolls a ball along a flat surface with an initial velocity of $\small 2.2\frac{m}{s}$ . If it stops after 12 seconds, what was the total distance it travelled?

Possible Answers:

26.4 m

13.2 m

14.4 m

144 m

52.8 m

Correct answer:

13.2 m

Explanation:

We can solve this question by multiplying the average velocity by the time.

$\Delta x=\frac{V_i+V_f}{2}(t)$

We are given the initial and final velocities and the time travelled.

$\Delta x=\frac{2.2\frac{m}{s}+0\frac{m}{s}}{2}(12s)$

$\Delta x=(1.1\frac{m}{s})(12s)$

$\Delta x=13.2m$

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Example Question #94 : Linear Motion

Peter starts from rest and runs down a hallway 31 seconds. If his final velocity is $\small 13\frac{m}{s}$ , how far did he run?

Possible Answers:

222 m

13.3 m

0.387 m

201.5 m

100.75 m

Correct answer:

201.5 m

Explanation:

Using the given values for the initial velocity, final velocity, and time, we can solve for the distance. The distance will be the product of the average velocity and the time.

$\Delta x=\frac{V_f+V_i}{2}(t)$

$\Delta x=\frac{13\frac{m}{s}+0\frac{m}{s}}{2}(31s)$

$\Delta x=(6.5\frac{m}{s})(31s)$

$\Delta x=201.5m$

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Example Question #95 : Linear Motion

Walter throws a disc from 1.5 meters above the ground with purely horizontal motion. If he throws it with an initial velocity of $\small 3.3\frac{m}{s}$ and it stays in the air for 0.553 second, how far will it travel?

Possible Answers:

1.83m

1.5 m

2.12 m

2.2 m

1.08 m

Correct answer:

1.83m

Explanation:

The horizontal velocity will determine how far the disc travels; gravity and vertical velocity will not affect the horizontal distance. We can solve by using the distance formula.

$\Delta x=V_xt$

$\Delta x=3.3\frac{m}{s}(0.553s)$

$\Delta x=1.83m$

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Example Question #2 : Understanding Distance, Velocity, And Acceleration

Julia throws a 2kg rock at $80^{\circ}$ above the horizontal, with an initial velocity of $1.4\frac{m}{s}$ .

If she were throwing a 4kg rock instead, how would this affect the total horizontal displacement?

Possible Answers:

It will travel half as far

It will travel 9.8 times as far

It will travel twice as far

It will travel 4.9 times as far

It will travel the same distance

Correct answer:

It will travel the same distance

Explanation:

The equation for horizontal distance travelled during parabolic (projectile) motion is:

$\Delta x =v_xt$

The distance is related to the velocity and the flight time only. The mass of the projectile will not affect any of these variables, and therefore will not affect the distance travelled.

We can also analyze this formula in terms of the units and dimensional analysis.

$m=\frac{m}{s}\cdot s$

There is no units for mass or anywhere in the calculation, so mass will not be involved in determining the final answer.

A change in the mass will not change the distance travelled; the projectile will travel the same distance.

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Example Question #3 : Understanding Distance, Velocity, And Acceleration

A 1.55kg box starts at rest and reaches a velocity of $3.45\frac{m}{s}$ after traveling a distance of 12m . What was the acceleration on the box?

Possible Answers:

$0.992\frac{m}{s^2}$

$2.02\frac{m}{s^2}$

$12.1\frac{m}{s^2}$

$0.496\frac{m}{s^2}$

$35.9\frac{m}{s^2}$

Correct answer:

$0.496\frac{m}{s^2}$

Explanation:

The best formula for this will be $v_f^2=v_i^2+2a\Delta x$ .

We are given the initial velocity (zero because the box starts at rest), final velocity, and distance. Using these values, we can solve for the acceleration. We are also given the mass, but this is extraneous information.

$v_f^2=v_i^2+2a\Delta x$

$(3.45\frac{m}{s})^2=(0\frac{m}{s})^2+2a(12m)$

$11.9\frac{m^2}{s^2}=a(24m)$

Divide both sides by $\small 24m$ .

$\frac{11.9\frac{m^2}{s^2}}{24m}=\frac{a( 24m)}{24m}{}$

$0.496\frac{m}{s^2}=a$

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Example Question #1 : Understanding Distance, Velocity, And Acceleration

A crate slides across a floor. After it has a velocity of $3.4\frac{m}{s}$ . After it has a velocity of $3.4\frac{m}{s}$ . What is the acceleration?

Possible Answers:

$3.4\frac{m}{s^2}$

$0.57\frac{m}{s^2}$

$1.76\frac{m}{s^2}$

$0.68\frac{m}{s^2}$

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

Correct answer:

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

Explanation:

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

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

In this case, even though we look at the crate after different times, the velocity hasn't changed at all.

$a=\frac{v_2-v_1}{\Delta t}=\frac{3.4\frac{m}{s}-3.4\frac{m}{s}}{\Delta t}$

$a=\frac{0\frac{m}{s}}{\Delta t}$

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

If there is no change in the velocity, then the acceleration must be equal to zero.

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Example Question #4 : Understanding Distance, Velocity, And Acceleration

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:

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

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

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

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

There is no definitive scientific reason for this phenomenon

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 #5 : Understanding Distance, Velocity, And Acceleration

A boy jogs down a street in . What was his average velocity?

Possible Answers:

$3\frac{m}{s}$

$8\frac{m}{s}$

$0\frac{m}{s}$

$12\frac{m}{s}$

$6\frac{m}{s}$

Correct answer:

$3\frac{m}{s}$

Explanation:

Velocity is change in displacement over change in time.

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

We are given the distance traveled and the time period. Using these values, we can find the velocity.

$v=\frac{6m}{2s}$

$v=3\frac{m}{s}$

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Example Question #1 : Understanding Distance, Velocity, And Acceleration

Which of these is the correct relationship between velocity, distance, and time?

Possible Answers:

$\Delta t=\Delta x *v$

$v=\Delta x + \Delta t$

$v=\frac{\Delta x *\Delta t}{2}$

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

$v=\Delta x * \Delta t$

Correct answer:

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

Explanation:

Velocity is defined by a change in distance during a period of time. It is a rate of movement.

The magnitude of velocity is equal to the quotient of the change in distance and the change in time.

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

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High School Physics : Understanding Distance, Velocity, and Acceleration

Study concepts, example questions & explanations for High School Physics

All High School Physics Resources

Example Questions

Example Question #92 : Linear Motion

Example Question #1 : Understanding Distance, Velocity, And Acceleration

Example Question #94 : Linear Motion

Example Question #95 : Linear Motion

Example Question #2 : Understanding Distance, Velocity, And Acceleration

Example Question #3 : Understanding Distance, Velocity, And Acceleration

Example Question #1 : Understanding Distance, Velocity, And Acceleration

Example Question #4 : Understanding Distance, Velocity, And Acceleration

Example Question #5 : Understanding Distance, Velocity, And Acceleration

Example Question #1 : Understanding Distance, Velocity, And Acceleration

All High School Physics Resources

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