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High School Physics : Linear Motion

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Example Questions

Example Question #11 : Understanding Motion In One Dimension

A ball is thrown vertically with an initial velocity of $3.3\frac{m}{s}$ . What is the maximum height it will reach?

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

Possible Answers:

0.67m

0.98

0.17m

1.68m

0.34m

Correct answer:

0.17m

Explanation:

To solve this problem, remember that when something is thrown vertically, its velocity will be at the highest point. Using this principle, we know the initial velocity, final velocity (zero), and the acceleration. With the appropriate motion equation, we can solve for the distance traveled.

The best equation for this problem is:

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

Use the given values to solve for the distance.

$0\frac{m}{s}^2=3.3\frac{m}{s}+2(-9.8\frac{m}{s^2})\Delta y$

$-3.3\frac{m}{s}=(-19.6\frac{m}{s^2})\Delta y$

$\frac{-3.3\frac{m}{s}}{-19.6\frac{m}{s^2}}=\Delta y$

$0.17m=\Delta y$

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Example Question #11 : Linear Motion

Two balls, one with mass 2kg and one with mass 4kg , are dropped from 30m above the ground. Which ball hits the ground first?

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

Possible Answers:

They hit the ground at the same time

The 2kg mass hits first

The 4kg mass hits first

We must know the forces to draw a conclusion

We must know the final velocities to draw a conclusion

Correct answer:

They hit the ground at the same time

Explanation:

The mass of an object is completely unrelated to its free-fall motion. The equation for the vertical motion for an object in freefall is:

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

Notice, there is no mention of mass anywhere in this equation. The only thing that affects the time an object takes to hit the ground is the acceleration due to gravity and the distance travelled. Since these objects travel the same distance and are affected by the same gravitational force, they will fall for the same amount of time and hit the ground together.

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Example Question #12 : Linear Motion

Two balls, one with mass 2kg and one with mass 4kg , are dropped from 30m above the ground. How long does it take the 4kg ball to hit the ground?

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

Possible Answers:

2.47s

17.15s

31.23s

1.27s

6.12s

Correct answer:

2.47s

Explanation:

The equation for the vertical motion for an object in freefall is:

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

Notice that there is no place for mass anywhere in this equation. This means that the two balls will be in the air for the same amount of time. We simply need to use the distance and acceleration to solve for the time.

Remember that even though the height is 30m , the DISPLACEMENT will be -30m . Displacement is a vector; since the direction of the distance is downward, the displacement will be negative.

$-30m=\frac{1}{2}(-9.8\frac{m}{s^2 })t^2$

$\frac{-30m}{-9.8\frac{m}{s^2}}=\frac{1}{2}t^2$

$3.06s^2=\frac{1}{2}t^2$

2*3.06s^2=t^2

6.12s^2=t^2

$\sqrt{6.12s^2}=t^2$

2.47s=t

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Example Question #14 : Motion And Mechanics

Two balls, one with mass 2kg and one with mass 4kg , are dropped from 30m above the ground. How long does it take the 2kg ball to hit the ground?

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

Possible Answers:

2.47s

6.12s

9.8s

1.27s

31.23s

Correct answer:

2.47s

Explanation:

The equation for the vertical motion for an object in freefall is:

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

Remember that even though the height is 30m , the DISPLACEMENT will be -30m . Displacement is a vector; since the direction of the distance is downward, the displacement will be negative.

$-30m=\frac{1}{2}(-9.8\frac{m}{s^2 })t^2$

$\frac{-30m}{-9.8\frac{m}{s^2}}=\frac{1}{2}t^2$

$3.06s^2=\frac{1}{2}t^2$

2*3.06s^2=t^2

6.12s^2=t^2

$\sqrt{6.12s^2}=t^2$

2.47s=t

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Example Question #11 : Understanding Motion In One Dimension

Two balls, one with mass 2kg and one with mass 4kg , are dropped from rest. It takes 18s for them to hit the ground. What is the final velocity of the 2kg ball right before it hits the ground?

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

Possible Answers:

$-4.9\frac{m}{s}$

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

$-9.8\frac{m}{s}$

$-19.6\frac{m}{s}$

$-176.4\frac{m}{s}$

Correct answer:

$-176.4\frac{m}{s}$

Explanation:

When dealing with simple vertical motion, the final velocity can be found from the initial velocity, acceleration, and time.

v_f=v_i+at

Remember that because the ball is dropped from rest, the initial velocity will be zero.

Notice that mass is not a variable in this calculation; the final velocity will be the same for any mass dropped from rest that is in the air for 18s .

Use the values given in the question to solve for the final velocity.

v_f=v_i+at

$v_f=0\frac{m}{s}+(-9.8\frac{m}{s^2})(18s)$

$v_f=-176.4\frac{m}{s}$

Note that the velocity is negative because the object is traveling downward.

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Example Question #16 : Motion And Mechanics

Two balls, one with mass 2kg and one with mass 4kg , are dropped from rest. It takes 18s for them to hit the ground. What is the final velocity of the 4kg ball right before it hits the ground?

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

Possible Answers:

$-176.4\frac{m}{s}$

$-352.8\frac{m}{s}$

$-39.2\frac{m}{s}$

$-88.2\frac{m}{s}$

$-9.8\frac{m}{s}$

Correct answer:

$-176.4\frac{m}{s}$

Explanation:

When dealing with simple vertical motion, the final velocity can be found from the initial velocity, acceleration, and time.

v_f=v_i+at

Remember that because the ball is dropped from rest, the initial velocity will be zero.

Notice that mass is not a variable in this calculation; the final velocity will be the same for any mass dropped from rest that is in the air for 18s .

Use the values given in the question to solve for the final velocity.

v_f=v_i+at

$v_f=0\frac{m}{s}+(-9.8\frac{m}{s^2})(18s)$

$v_f=-176.4\frac{m}{s}$

Note that the velocity is negative because the object is traveling downward.

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Example Question #11 : Linear Motion

A 7kg crate slides along a frictionless surface. If it maintains a constant velocity of $3\frac{m}{s}$ , what is the net acceleration on the crate?

Possible Answers:

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

$6.67\frac{m}{s^2}$

$60\frac{m}{s^2}$

$3\frac{m}{s^2}$

$20\frac{m}{s^2}$

Correct answer:

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

Explanation:

Acceleration is the change in velocity per unit time.

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

Since the velocity does not change from one moment to the next, then there must be no net acceleration on the object.

$a=\frac{3\frac{m}{s}-3\frac{m}{s}}{t}=\frac{0\frac{m}{s}}{t}$

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

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Example Question #12 : Linear Motion

A crate has an initial velocity of $9\frac{m}{s}$ . If it accelerates at a constant rate of $0.5\frac{m}{s^2}$ for 1.5s , what is the final velocity?

Possible Answers:

$1.5\frac{m}{s}$

$9.75\frac{m}{s}$

$8.25\frac{m}{s}$

$0.75\frac{m}{s}$

$5.25\frac{m}{s}$

Correct answer:

$9.75\frac{m}{s}$

Explanation:

For this problem, we use the formula v_f=v_i+at .

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

$v_f=9\frac{m}{s}+(0.5\frac{m}{s^2}*1.5s)$

$v_f=9\frac{m}{s}+0.75\frac{m}{s}$

$v_f=9.75\frac{m}{s}$

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Example Question #81 : High School Physics

A box moves at $0.11\frac{m}{s}$ across a frictionless surface. Assuming no other forces, how far will it travel in ?

Possible Answers:

0.11m

0.037m

3.3m

33m

0.33m

Correct answer:

0.33m

Explanation:

The relationship between distance, velocity, and time is given by the formula for velocity:

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

We can rearrange this equation to solve for the distance.

$\Delta x=v*\Delta t$

Using the values given for the velocity and the time, we can find the distance traveled.

$\Delta x=0.11\frac{m}{s}*3s$

$\Delta x =0.33m$

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Example Question #20 : Motion And Mechanics

A ball rolls at $2.22\frac{m}{s}$ . Assuming no outside forces, how long will it take for the ball to roll 12m ?

Possible Answers:

26.64s

0.185s

9.91s

5.41s

16.64s

Correct answer:

5.41s

Explanation:

The relationship between distance, velocity, and time is given by the equation for velocity:

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

We can rearrange this formula for solve for the time.

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

Using the given values for the velocity and distance traveled, we can solve for the time.

$\Delta t =\frac{12m}{2.22\frac{m}{s}}$

$\Delta t=5.41s$

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High School Physics : Linear Motion

Study concepts, example questions & explanations for High School Physics

All High School Physics Resources

Example Questions

Example Question #11 : Understanding Motion In One Dimension

Example Question #11 : Linear Motion

Example Question #12 : Linear Motion

Example Question #14 : Motion And Mechanics

Example Question #11 : Understanding Motion In One Dimension

Example Question #16 : Motion And Mechanics

Example Question #11 : Linear Motion

Example Question #12 : Linear Motion

Example Question #81 : High School Physics

Example Question #20 : Motion And Mechanics

All High School Physics Resources

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