Physics - 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?

$55\frac{m}{s}$

$42\frac{m}{s}$

$20\frac{m}{s}$

$30\frac{m}{s}$

$2\frac{m}{s}$

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.

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

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

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

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

What is the initial velocity of the crate?

$4\frac{m}{s}$

$2\frac{m}{s}$

$1\frac{m}{s}$

$8\frac{m}{s}$

$50\frac{m}{s}$

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 $\small 2$ .

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

An object starts at rest, and then accelerates at a rate of $2\frac{m}{s^2}$ for . What is the final velocity?

$10\frac{m}{s}$

$2\frac{m}{s}$

$0.4\frac{m}{s}$

$2.5\frac{m}{s}$

$7\frac{m}{s}$

Explanation

The formula relating velocity and acceleration is:

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

We are given values for the acceleration and time. We can also conclude that the initial velocity is zero. Using these values, we can solve for the final velocity.

$2\frac{m}{s^2}=\frac{v_f-0\frac{m}{s}}{5s}$

$v_f=2\frac{m}{s^2}*5s$

$v_f=10\frac{m}{s}$

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

What is the net acceleration on the crate?

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

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

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

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

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

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.

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

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

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

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

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

An egg falls from a nest in a tree that is tall. A girl, away, runs to catch the egg. If she catches it right at the moment before it hits the ground, how fast does she need to run?

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

$17.22\frac{m}{s}$

$9.83\frac{m}{s}$

$1.33\frac{m}{s}$

$0.755\frac{m}{s}$

More information is needed to solve

Explanation

The important thing to recognize here is that the amount of time the egg is falling will be equal to the amount of time the girl is running.

Our first step will be to find the time that the egg is in the air.

We know it starts from rest above the ground, and we know the gravitational acceleration. Its total displacement will be , since it falls in the downward direction. We can use the appropriate motion equation to solve for the time:

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

Use the given values in the formula to solve for the time.

$-2.8m=(0\frac{m}{s})t+\frac{1}{2}(-9.8\frac{m}{s^2})t^2$

$-2.8m=\frac{1}{2}(-9.8\frac{m}{s^2})t^2$

$-2.8m=(-4.9\frac{m}{s^2})t^2$

$\frac{-2.8m}{-4.9\frac{m}{s^2}}=t^2$

Now that we have the time, we can use it to find the speed of the girl. Her speed will be determined by the distance she travels in this amount of time.

$v=\frac{d}{t}$

Use our values for her distance and the time to solve for her velocity.

$v=\frac{13m}{0.755s}$

$v=17.22\frac{m}{s}$

A dog lying on the front porch sees a car drive past at $13\frac{m}{s}$ . If the dog catches up with the car in , how quickly must he accelerate?

$48\frac{m}{s^2}$

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

There is insufficient information to solve.

$1.86\frac{m}{s^2}$

$7.2\frac{m}{s^2}$

Explanation

First we need to know how long this chase takes. The dog doesn't help us there, but we know that the car is going along at a constant velocity and will therefore cover the same distance in the same amount of time.

Since the velocity is constant, we can say $\Delta x =v*t$ .

Plug in our given values.

$\Delta x =v*t$

$7m=13\frac{m}{s}*t$

$\frac{7m}{13\frac{m}{s}}=t$

Plug that into our equation for the dog. We can use the full distance equation for this: $\Delta x=v_it+\frac{1}{2}at^2$

Remember the dog starts at rest.

Plug in our given information.

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

$7m=0\frac{m}{s}*0.54s+\frac{1}{2}a(0.54s)^2$

$7m=\frac{1}{2}a(0.54s)^2$

$7m=\frac{1}{2}a*0.2916s^2$

$\frac{7m}{0.1458s^2}=a$

$48\frac{m}{s^2}=a$

Michael throws a ball vertically in the air. If he throws it with an initial velocity of $0.44\frac{m}{s}$ , how high will it go?

Explanation

To solve this problem we will need to use the appropriate kinematics equation:

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

Remember that when an object is thrown vertically, it will have a velocity of zero at the highest point. We can use this as our final velocity. The initial velocity is given, and the acceleration will be the acceleration due to gravity. Using these values, we can solve for the displacement when the ball is at its peak.

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

$(0\frac{m}{s})^2=(0.44\frac{m}{s})^2+2(-9.8\frac{m}{s^2})\Delta x$

$0\frac{m^2}{s^2}=(0.1936\frac{m^2}{s^2})+(-19.6\frac{m}{s^2})\Delta x$

$-0.1936\frac{m^2}{s^2}=(-19.6\frac{m}{s^2})\Delta x$

$\frac{-0.1936\frac{m^2}{s^2}}{-19.6\frac{m}{s^2}}=\Delta x$

$0.01m=\Delta x$

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?

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.

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

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

Add the two distances together.

Convert the final answer to kilometers.

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

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

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

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

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

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

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

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

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:

Distance to goal post:

Distance past goal post:

What we want to know:

$\theta=?$

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, , 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.

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.

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

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

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

$\theta=42.8^0$

An American football kicker kicks a field goal from in front of the goal post. The ball was in the air for and landed behind the goal post. If the ball just cleared the goal post by , how far above the ground is the goal post?

Explanation

To solve this problem, we need to calculate each of the vector components of the velocity. Our plan is to find the time it takes for the ball to horizontally reach the goal post. Once we have that time, we can apply that time to the vertical component to see at what height the ball is at this point. We know that velocity is equal to distance divided by time, and that horizontal velocity, , 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.

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

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

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

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

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

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

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

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

Now that we have our velocities, let's see how long it takes for the ball to reach the goal post. We will use as that time.

$v_x=\frac{x_1}{t_p}$

$t_p=\frac{x_1}{v_x}$

$t_p=\frac{58.5m}{13.8\frac{m}{s}}$

Now that we have our time, let's use a different equation to see how high the ball is at that time. We know the time, initial vertical velocity, and acceleration. Use these to find the final vertical distance, or height.

$y=v_{oy}t_p+\frac{1}{2}a{t_p}^2$

$y=(16.9\frac{m}{s})(3.20s)+\frac{1}{2}(-9.81\frac{m}{s^2})(3.20s)^2$

This gives us the height of the ball when it is just over the goal post. We know from the question that there is a gap of between the ball and the height of the goal post. To find that height, we take the ball's height at this moment and subtract the height of the gap.

Using Motion Equations

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