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High School Physics : Using Motion Equations

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Example Question #1 : Using Motion Equations

A potted plant falls off a window sill from 3 meters above the ground. How long does it take before the plant hits the ground?

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

Possible Answers:

0.782s

0.656 s

0.818 s

0.723 s

1.2 s

Correct answer:

0.782s

Explanation:

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

Using the equation above and the given values, we can solve for the time. First we need to find the change in distance.

$\Delta y=y_f-y_i=0m-3m=-3m$

The plant travels $\small -3$ meters in the downward direction. Now we can plug in the given values from the question to solve for time.

$-3m= (0\frac{m}{s})+\frac{1}{2}(-9.8\frac{m}{s^2})t^2$

$-3m = (-4.9\frac{m}{s^2})t^2$

$\frac {-3m}{-4.9\frac{m}{s^2}}=t^2$

0.612s^2=t^2

$\sqrt{0.612s^2}=\sqrt{t^2}$

0.782s=t

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Example Question #2 : Using Motion Equations

Walter throws a disc from 1.5 meters above the ground with purely horizontal motion. How long will it be before the disc hits the ground?

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

Possible Answers:

0.343 s

0.002 s

0.053 s

0.553 s

1.03 s

Correct answer:

0.553 s

Explanation:

The horizontal motion will not affect the time the disc is in the air. Time will be determined by the rate at which the disc falls: the acceleration due to gravity.

We can solve for the time using the equation below and the values given in the question.

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

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

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

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

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

0.306s^2=t^2

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

0.553s=t

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Example Question #3 : Using Motion Equations

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

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

Possible Answers:

$14\frac{m}{s}$

$28\frac{m}{s}$

$9\frac{m}{s}$

$12\frac{m}{s}$

$13.2\frac{m}{s}$

Correct answer:

$14\frac{m}{s}$

Explanation:

We are given the initial velocity, acceleration, and the initial and final distances.

$\Delta y=y_f-y_i=0m-10m=-10m$

Using the equation below and the given values, we can solve for the final velocity.

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

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

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

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

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

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Example Question #4 : Using Motion Equations

A 2.9kg crate slides across a floor for before coming to rest 10m from its original position.

What is the initial velocity of the crate?

Possible Answers:

$2\frac{m}{s}$

$50\frac{m}{s}$

$4\frac{m}{s}$

$8\frac{m}{s}$

$1\frac{m}{s}$

Correct answer:

$4\frac{m}{s}$

Explanation:

The problem gives us distance, final velocity, and change in time. We can use these values in the equation below to solve for the initial velocity.

$\Delta x=\frac{(v_f+v_i)}{2}\Delta t$

Plug in our given values and solve.

$10m=\frac{(0\frac{m}{s}+v_i)}{2}5s$

Divide both sides by $\small 5$ .

$\frac{10m}{5s}=\frac{v_i}{2}$

$2\frac{m}{s}=\frac{v_i}{2}$

Multiply both sides by $\small 2$ .

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

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Example Question #5 : Using Motion Equations

A 2.9kg crate slides across a floor for before coming to rest 10m from its original position.

What is the net acceleration on the crate?

Possible Answers:

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

$-10\frac{m}{s}$

$-0.8\frac{m}{s^2}$

$-4\frac{m}{s}$

$-2.32\frac{m}{s^2}$

Correct answer:

$-0.8\frac{m}{s^2}$

Explanation:

All of the equations regarding acceleration require an initial velocity. We will need to solve for the initial velocity using the given variables. The problem gives us distance, final velocity, and change in time. We can use these values in the equation below to solve for the initial velocity.

$\Delta x=\frac{(v_f+v_i)}{2}\Delta t$

Plug in our given values and solve.

$10m=\frac{(0\frac{m}{s}+v_i)}{2}5s$

$\frac{10m}{5s}=\frac{v_i}{2}$

$2\frac{m}{s}=\frac{v_i}{2}$

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

We can use a linear motion equation to solve for the acceleration, using the velocity we just found. We now have the distance, time, and initial velocity.

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

Plug in the given values to solve for acceleration.

$10m=(4\frac{m}{s}) (5s)+\frac{1}{2}a(5s)^2$

$10m=(20m)+\frac{1}{2}a(25s^2)$

$-10m=\frac{1}{2}a(25s^2)$

-20m=a(25s^2)

$\frac{-20m}{25s^2}=a$

$-0.8\frac{m}{s^2}=a$

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Example Question #6 : Using Motion Equations

A ball starts rolling at $1.5\frac{m}{s}$ . It accelerates at a constant rate of $2\frac{m}{s^2}$ for 5.23s . What is the final velocity?

Possible Answers:

$8.22\frac{m}{s}$

$11.96\frac{m}{s}$

$10.46\frac{m}{s}$

$4.12\frac{m}{s}$

$1.5\frac{m}{s}$

Correct answer:

$11.96\frac{m}{s}$

Explanation:

To solve for the final velocity, remember that the relationship between velocity, acceleration, and time is v_f=v_i+at .

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

$v_f=1.5\frac{m}{s}+(2\frac{m}{s^2}*5.23s)$

$v_f=1.5\frac{m}{s}+10.46\frac{m}{s}$

$v_f=11.96\frac{m}{s}$

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Example Question #7 : Using Motion Equations

A book, starting at rest, falls off of a table. What is its velocity after 2.1s in motion?

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

Possible Answers:

$-10.26\frac{m}{s}$

$-7.7\frac{m}{s}$

$-4.67\frac{m}{s}$

$-11.9\frac{m}{s}$

$-20.58\frac{m}{s}$

Correct answer:

$-20.58\frac{m}{s}$

Explanation:

We can solve this question using the equation for acceleration in terms of velocity:

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

We know our initial velocity (zero, since we start from rest), time, and the acceleration of gravity. Use these values to isolate the variable for the final velocity.

$-9.8\frac{m}{s^2}=\frac{v_2-0\frac{m}{s}}{2.1s}$

$(-9.8\frac{m}{s^2})(2.1s)=v_2$

$-20.58\frac{m}{s}=v_2$

Note that the final velocity is negative, since the book is traveling in the downward direction.

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Example Question #8 : Using Motion Equations

A car traveling along a highway moves at initial velocity $\small 25\frac{m}{s}$ before it begins to accelerate. If it accelerates for $\small 15s$ at $\small 2\frac{m}{s^{2}}$ , what is the final velocity of the car?

Possible Answers:

$20\frac{m}{s}$

$2\frac{m}{s}$

$42\frac{m}{s}$

$55\frac{m}{s}$

$30\frac{m}{s}$

Correct answer:

$55\frac{m}{s}$

Explanation:

Use the kinematic equation:

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

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

$v_f=25\frac{m}{s}+(2\frac{m}{s^2})(15s)$

$v_f=25\frac{m}{s}+30\frac{m}{s}$

$v_f=55\frac{m}{s}$

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Example Question #81 : Motion And Mechanics

A particle is traveling north at $\small 850\frac{m}{s}$ from an initial position. After traveling $\small 35000m$ from the initial position, the particle begins accelerating north at $\small 12\frac{m}{s^{2}}$ for $\small 15s$ . What is the final distance between the particle and the initial position?

Possible Answers:

49.1km

52.0km

55.1km

44.3km

48.7km

Correct answer:

49.1km

Explanation:

Utilize the kinematic equation:

$x = v_{o}t + \frac{1}{2}at^{2}$

The particle's motion can be broken into two parts: the initial distance and the distance traveled during acceleration. The initial distance is given.

x_o=35000m

The distance during acceleration can be found using the kinematic formula and given values for the initial velocity, acceleration, and time.

$x_a = (850\frac{m}{s})(15s) + \frac{1}{2}(12\frac{m}{s^2})(15s)^{2}$

x_a=(12750m)+(1350m)

x_a=14100m

Add the two distances together.

x_o+x_a=35000m+14100m=49100m

Convert the final answer to kilometers.

$49100m*\frac{1km}{1000m}=49.1km$

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Example Question #81 : Linear Motion

An American football kicker kicks a field goal from 58.5m in front of the goal post. The ball was in the air for 3.45s and landed 4.50m behind the goal post.

Given this information, what was the total initial velocity and angle with which the ball was kicked?

Possible Answers:

$24.9\frac{m}{s}\ \text{at}\ 47.2^o\ \text{above horizontal}$

$35.2\frac{m}{s}\ \text{at}\ 46.5^o\ \text{above horizontal}$

$35.2\frac{m}{s}\ \text{at}\ 38.2^o\ \text{above horizontal}$

$18.3\frac{m}{s}\ \text{at}\ 38.2^o\ \text{above horizontal}$

$24.9\frac{m}{s}\ \text{at}\ 42.8^o\ \text{above horizontal}$

Correct answer:

$24.9\frac{m}{s}\ \text{at}\ 42.8^o\ \text{above horizontal}$

Explanation:

Although this question involves several steps, when you break it down we can see that it is a problem involving kinematics in two dimensions.

First, begin by writing down what we know:

t=3.45s

Distance to goal post: x_1=58.5m

Distance past goal post: x_2=4.50m

What we want to know:

v_0=?

$\theta=?$

To begin, we need to calculate each of the vector components of the velocity. We can start with the horizontal component. We know that velocity is equal to distance divided by time, and that horizontal velocity, v_x , will not change because there is no acceleration in the horizontal direction. We will need to find the total distance that the ball travels in order to solve.

$v_x=\frac{x}{t}$

We will need to find the total distance that the ball travels in order to solve.

x=x_1+x_2

x=58.5m+4.50m

x=63.0m

Use this distance and the given time to find the horizontal velocity.

$v_x=\frac{63.0m}{3.45s}$

$v_x=18.3\frac{m}{s}$

Now let's find the initial vertical velocity, $v_{oy}$ . Because we are assuming that there is nothing except for gravity influencing the ball, we can say that the ball spends half of the time reaching the peak of its trajectory, where the vertical velocity will momentarily be zero. With that information, we can solve for the initial vertical velocity:

$v_y=v_{oy}+a(\frac{1}{2}t)$

We use only half the given time because we are only taking the time from when the ball is kicked to when it reaches the top of its trajectory (which will be half of its flight). As we stated above, velocity will be zero at the top of the trajectory. We are using a negative value for the acceleration due to gravity because gravity points downward, which in this case is the negative direction. Use the given values to solve for the initial vertical velocity.

$v_{oy}=v_y-a(\frac{1}{2}t)$

$v_{oy}=0\frac{m}{s}-(-9.81\frac{m}{s^2})(\frac{1}{2}*3.45s)$

$v_{oy}=16.9\frac{m}{s}$

Now that we have both directional components of the initial velocity, we can use the Pythagorean theorem to solve for the total initial velocity.

${v_o}^2={v_x}^2+{v_o_y}^2$

$v_o=\sqrt{{v_x}^2+{v_o_y}^2}$

$v_o=\sqrt{(18.3\frac{m}{s})^2+(16.9\frac{m}{s})^2}$

$v_o=24.9\frac{m}{s}$

Now to find the angle, we use trigonometry. In a triangle formed by the maximum height of the ball, the ground, and the trajectory, tangent of the angle will be equal to the vertical leg of the triangle divided by the horizontal leg of the triangle.

$tan\theta=\frac{v_{oy}}{v_x}$

$\theta=tan^{-1}(\frac{v_{oy}}{v_x})$

$\theta=tan^{-1}(\frac{16.9}{18.3})$

$\theta=42.8^0$

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High School Physics : Using Motion Equations

Study concepts, example questions & explanations for High School Physics

All High School Physics Resources

Example Questions

Example Question #1 : Using Motion Equations

Example Question #2 : Using Motion Equations

Example Question #3 : Using Motion Equations

Example Question #4 : Using Motion Equations

Example Question #5 : Using Motion Equations

Example Question #6 : Using Motion Equations

Example Question #7 : Using Motion Equations

Example Question #8 : Using Motion Equations

Example Question #81 : Motion And Mechanics

Example Question #81 : Linear Motion

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