Understanding Motion in One Dimension - Physics

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Question

A picture hanging above the ground falls off the wall and hits the ground. What is its final velocity?

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

Answer

The problem gives us the distance, the acceleration due to gravity, and implies that the initial velocity of the picture is zero, as it starts at rest.

We can find the final velocity using the appropriate motion equation:

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

We can use our values to solve for the final velocity. Keep in mind that the displacement will be negative because the ball is traveling in the downward direction!

$v_f^2=(0\frac{m}{s})^2+2(-9.8\frac{m}{s^2})(-1.5m)$

$v_f^2=29.4\frac{m^2}{s^2}$

$\sqrt{v_f^2}=\sqrt{29.4\frac{m^2}{s^2}}$

$v_f=5.42\frac{m}{s}$

The square root of a term can be either positive or negative, depending on the direction. Since velocity is a vector, and the painting is falling downward, our final answer will be negative: $-5.42\frac{m}{s}$ .

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Question

Marcus throws a ball directly up in the air with an initial velocity of $\small 35\frac{m}{s}$ . How high will the ball go?

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

Answer

Remember that the final velocity at an object's highest point is equal to zero. We can use the following equation to solve for our height ( $\Delta{y}}$ ), using the initial velocity, final velocity, and acceleration due to gravity.

$V_{f}^{2}=V_{i}^{2}+2a\Delta{y}}$

$(0\frac{m}{s})^2=({35\frac{m}{s}})^{2}+2(-9.8\frac{m}{s^2})\Delta{y}}$

$0=1225\frac{m^2}{s^2}+(-19.6\frac{m}{s^2})\Delta{y}}$

$-1225\frac{m^2}{s^2}=(-19.6\frac{m}{s^2})\Delta{y}}$

$\frac{-1225\frac{m^2}{s^2}}{-19.6\frac{m}{s^2}} =\Delta{y}}$

$62.5m=\Delta{y}$

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Question

Jenny throws a ball directly up in the air. She notices that it changes direction after approximately 3 seconds. What was the initial velocity of the ball?

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

Answer

The ball will travel upwards to the highest point, then change direction and travel downwards. Remember that the velocity in at the highest point is equal to zero. We can use the following equation and our known data to solve for the initial velocity.

$V_{f}=V_{i}+at$

$0\frac{m}{s}=V_{i}+(-9.8\frac{m}{s^2})(3s)$

$0\frac{m}{s}=V_{i}-29.4\frac{m}{s}$

$29.4\frac{m}{s}=V_{i}$

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Question

Laura throws a ball vertically. She notices it reaches a maximum height of 10 meters. What was the initial velocity of the ball?

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

Answer

Remember that at the highest point, the velocity in the y-direction is equal to zero. Using the given values and the equation below, we can solve for the initial velocity.

$V_f^2=V_i^2+2a\Delta y$

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

$0\frac{m^2}{s^2}=V_i^2-196\frac{m^2}{s^2}$

$196\frac{m^2}{s^2}=V_i^2$

$\sqrt{196\frac{m^2}{s^2}}=\sqrt{V_i^2}$

$14\frac{m}{s}=V_i$

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Question

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

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

Answer

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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Question

A ball rolls along a table with a constant velocity of $0.8\frac{m}{s}$ . If it rolls for , how long was it rolling?

Answer

The relationship between constant velocity, distance, and time can best be illustrated as $\Delta x =vt$ .

We are given the velocity and the distance, allowing us to solve for the time that the ball was in motion.

$\Delta x =vt$

$10m =0.8\frac{m}{s}*t$

Isolate the variable for time.

$\frac{10m}{0.8\frac{m}{s}} =t$

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Question

A picture hanging above the ground falls off the wall and hits the ground. How long will it be before it hits the ground?

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

Answer

The problem gives us the distance, the acceleration due to gravity, and implies that the initial velocity of the picture is zero, as it starts at rest.

We can use the appropriate motion equation to solve for the final velocity:

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

We can use our values to solve for the time. Keep in mind that the displacement will be negative because the ball is traveling in the downward direction!

$-1.5m=0\frac{m}{s}t+\frac{1}{2}(-9.8\frac{m}{s^2})t^2$

$-1.5m=-4.9\frac{m}{s^2}t^2$

$\frac{-1.5m}{-4.9\frac{m}{s^2}}=t^2$

$\sqrt{0.306s^2}=\sqrt{t^2}$

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Question

Two balls, one with mass and one with mass , are dropped from above the ground. Which ball hits the ground first?

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

Answer

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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Question

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

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

Answer

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 , the DISPLACEMENT will be . Displacement is a vector; since the direction of the distance is downward, the displacement will be negative.

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

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

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

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

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Question

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

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

Answer

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 , the DISPLACEMENT will be . Displacement is a vector; since the direction of the distance is downward, the displacement will be negative.

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

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

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

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

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Question

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

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

Answer

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

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 .

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

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

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

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

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Question

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

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

Answer

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

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 .

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

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

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

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

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Question

A 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?

Answer

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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Question

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 , what is the final velocity?

Answer

For this problem, we use the formula .

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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Question

An object moves at a constant velocity, , for some distance, . How long it in motion?

Answer

The relationship between velocity, distance, and time is:

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

We can multiply both sides by the time, then divide both sides by the velocity, to isolate the variable for time.

$v*\Delta t=\Delta x$

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

The quotient of distance and velocity will give us the time that the object was in motion.

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Question

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

Answer

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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Question

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

Answer

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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Question

A ball rolls at $0.77\frac{m}{s}$ . Assuming no outside forces, how long will it take to roll ?

Answer

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.

$\Delta t =\frac{9.23m}{0.77\frac{m}{s}}$

$\Delta t=11.99s$

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Question

A ball begins to roll with a velocity of $0.02\frac{m}{s}$ . If it has an acceleration of $0.5\frac{m}{s^2}$ and rolls for , what is the final velocity?

Answer

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

$0.5\frac{m}{s^2}=\frac{v_2-0.02\frac{m}{s}}{8s}$

$4\frac{m}{s}=v_2-0.02\frac{m}{s}$

$4.02\frac{m}{s}=v_2$

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Question

A ball is thrown vertically with a velocity of . What is its velocity at the highest point in the throw?

Answer

When examining vertical motion, the vertical velocity will always be zero at the highest point. At this point, the acceleration from gravity is working to change the motion of the ball from positive (upward) to negative (downward). This change is represented by the x-axis on a velocity versus time graph. As the ball changes direction, its velocity crosses the x-axis, momentarily becoming zero.

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