Linear Motion - AP Physics C Electricity & Magnetism

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Question

A man runs with a velocity described by the function below.

What is the function for his acceleration?

Answer

The function for acceleration is the derivative of the function for velocity.

$a(t)= \frac{dv(t)}{dt}$

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Question

A man runs with a velocity as described by the function below.

How far does he travel in 1 minute?

Answer

Distance is given by the integral of a velocity function. For this question, we will need to integrate over the interval of 0s to 60s.

$x(t) = \int_{0}^{60}v(t),dt$

$x(60)=\int_{0}^{60}(8t+4),dt = (4(60)^2+4(60))-(4(0)^2+4(0))$

$x(60)=4(3600)+240 = 14640\ m$

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Question

A particle traveling in a straight line accelerates uniformly from rest to $50:\uptext{cm/s}$ in $5:\uptext{s}$ and then continues at constant speed for an additional for an additional $3:\uptext{s}$ . What is the total distance traveled by the particle during the $8:\uptext{s}$ ?

Answer

First off we have to convert $50:\uptext{cm/s}$ to meters per second.

$\frac{50:\uptext{cm}}{s} \cdot \frac{1:\uptext{m}}{100\uptext{cm}} = 0.5:\uptext{m/s}$

Next we have to calculate the distance the object traveled the first 5 seconds, when it was starting from rest. We are given time, initial speed to be . The acceleration of the object at this time can be calculated using:

, substituting the values, we get:

$0.5:\uptext{m/s} = a(5s), a = 0.1:m/s^2$

Next, we use the distance equation to find the distance in the first 5 seconds:

$x_1 = v_0t + \frac{1}{2}at^2$

because the initial speed is .

If we plug in , , we get: $x_1 = 1.25:\uptext{m}$

Next we have to find the distance the the object travels at constant speed for 3 seconds. We can use the equation:

in this case is not equal to , it is equal to $0.5:\uptext{m}$ and $t=3:\uptext{s}$

Plugging in the equation, we get

Adding and we get the total distance to be $2.75:\uptext{m}$

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Question

A man runs with a velocity described by the function below.

What is the function for his acceleration?

Answer

The function for acceleration is the derivative of the function for velocity.

$a(t)= \frac{dv(t)}{dt}$

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Question

A man runs with a velocity as described by the function below.

How far does he travel in 1 minute?

Answer

Distance is given by the integral of a velocity function. For this question, we will need to integrate over the interval of 0s to 60s.

$x(t) = \int_{0}^{60}v(t),dt$

$x(60)=\int_{0}^{60}(8t+4),dt = (4(60)^2+4(60))-(4(0)^2+4(0))$

$x(60)=4(3600)+240 = 14640\ m$

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Question

A particle traveling in a straight line accelerates uniformly from rest to $50:\uptext{cm/s}$ in $5:\uptext{s}$ and then continues at constant speed for an additional for an additional $3:\uptext{s}$ . What is the total distance traveled by the particle during the $8:\uptext{s}$ ?

Answer

First off we have to convert $50:\uptext{cm/s}$ to meters per second.

$\frac{50:\uptext{cm}}{s} \cdot \frac{1:\uptext{m}}{100\uptext{cm}} = 0.5:\uptext{m/s}$

Next we have to calculate the distance the object traveled the first 5 seconds, when it was starting from rest. We are given time, initial speed to be . The acceleration of the object at this time can be calculated using:

, substituting the values, we get:

$0.5:\uptext{m/s} = a(5s), a = 0.1:m/s^2$

Next, we use the distance equation to find the distance in the first 5 seconds:

$x_1 = v_0t + \frac{1}{2}at^2$

because the initial speed is .

If we plug in , , we get: $x_1 = 1.25:\uptext{m}$

Next we have to find the distance the the object travels at constant speed for 3 seconds. We can use the equation:

in this case is not equal to , it is equal to $0.5:\uptext{m}$ and $t=3:\uptext{s}$

Plugging in the equation, we get

Adding and we get the total distance to be $2.75:\uptext{m}$

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Question

A $\small 200g$ ball is thrown horizontally from the top of a $\small 40m$ high building. It has an initial velocity of $\small 14\frac{m}{s}$ and lands on the ground $\small 40m$ away from the base of the building. Assuming air resistance is negligible, which of the following changes would cause the range of this projectile to increase?

I. Increasing the initial horizontal velocity

II. Decreasing the mass of the ball

III. Throwing the ball from an identical building on the moon

Answer

Relevant equations:

$y_f - y_o = v_{oy}t +\frac{1}{2}a_yt^2$

$R_x = v_{ox}t$

Choice I is true because $v_{ox}$ is proportional to the range , so increasing $v_{ox}$ increases if $\small t$ is constant. This relationship is given by the equation:

$R_x = v_{ox}t$

Choice II is false because the motion of a projectile is independent of mass.

Choice III is true because the vertical acceleration on the moon would be less. Decreasing increases the time the ball is in the air, thereby increasing if $v_{ox}$ is constant. This relationship is also shown in the equation:

$R_x = v_{ox}t$

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Question

Water emerges horizontally from a hole in a tank above the ground. If the water hits the ground from the base of the tank, at what speed is the water emerging from the hole? (Hint: Treat the water droplets as projectiles.)

Answer

To understand this problem, we have to understand that the water has a x-velocity and a y-veloctiy. The x-velocity never changes.

First we want to find the time it took for the water to hit the ground. We can use this equation: $d_y = v_0t + \frac{1}{2}at^2$

We know that the y-velocity is 0 to start with, acceleration is and .

Substituting into the equation, we get:

$1.5 = \frac{1}{2}(9.8)t^2\ \ t = \sqrt{\frac{1.5}{0.5\cdot9.8}}, t = 0.55:\uptext{s}$

Next we have to substitute the time into the equation for the x component

We know that and , so we can conclude that:

$v_x = \frac{2.1}{0.55} = 3.80:m/s$

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Question

A $\small 200g$ ball is thrown horizontally from the top of a $\small 40m$ high building. It has an initial velocity of $\small 14\frac{m}{s}$ and lands on the ground $\small 40m$ away from the base of the building. Assuming air resistance is negligible, which of the following changes would cause the range of this projectile to increase?

I. Increasing the initial horizontal velocity

II. Decreasing the mass of the ball

III. Throwing the ball from an identical building on the moon

Answer

Relevant equations:

$y_f - y_o = v_{oy}t +\frac{1}{2}a_yt^2$

$R_x = v_{ox}t$

Choice I is true because $v_{ox}$ is proportional to the range , so increasing $v_{ox}$ increases if $\small t$ is constant. This relationship is given by the equation:

$R_x = v_{ox}t$

Choice II is false because the motion of a projectile is independent of mass.

Choice III is true because the vertical acceleration on the moon would be less. Decreasing increases the time the ball is in the air, thereby increasing if $v_{ox}$ is constant. This relationship is also shown in the equation:

$R_x = v_{ox}t$

Compare your answer with the correct one above

Question

Water emerges horizontally from a hole in a tank above the ground. If the water hits the ground from the base of the tank, at what speed is the water emerging from the hole? (Hint: Treat the water droplets as projectiles.)

Answer

To understand this problem, we have to understand that the water has a x-velocity and a y-veloctiy. The x-velocity never changes.

First we want to find the time it took for the water to hit the ground. We can use this equation: $d_y = v_0t + \frac{1}{2}at^2$

We know that the y-velocity is 0 to start with, acceleration is and .

Substituting into the equation, we get:

$1.5 = \frac{1}{2}(9.8)t^2\ \ t = \sqrt{\frac{1.5}{0.5\cdot9.8}}, t = 0.55:\uptext{s}$

Next we have to substitute the time into the equation for the x component

We know that and , so we can conclude that:

$v_x = \frac{2.1}{0.55} = 3.80:m/s$

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Question

A guillotine blade weighing is accelerated upward into position at a rate of $2.0\frac{m}{s^2}$ .

What is the the approximate mass of the guillotine blade?

Answer

The force of gravity on the blade is , which is the same as $400\frac{kg\cdot m}{s^2}$

This unit relationship comes from Newton's second law.

is the mathematical expression of Newton's second law. The units for force must be a product of the units for mass and the units for acceleration.

Solve the expression by plugging in known values.

$400 \frac{kg\cdot m}{s^2}=m \cdot 9.8\frac{m}{s^2}$

$\frac{400}{9.8}kg=mass\approx41kg$

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Question

A guillotine blade weighing is accelerated upward into position at a rate of $2.0\frac{m}{s^2}$ .

What is the tension on the rope pulling the blade, while it is accelerating into position?

Answer

The tension in the rope is the sum of the forces acting on it. If one considers that the net force on an object must equal the mass of the object times the acceleration of the object, the net force on the object must be the force due to tension from the rope minus the force due to gravity.

$F_{T}-F_{g}=ma$

Rearrange the equation.

$F_{T}=ma+F_{g}$

Plug in known values.

$F_T=41kg\cdot2\frac {m}{s^2} + 400N$

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Question

An object is moving in two dimensions. Its vertical motion relative to the horizontal motion is described by the equation $\small y=ax^2+bx +c$ . Its motion in the horizontal direction is described by the equation $\small x=v_{x}t$ . What is the object's velocity is the direction in terms of its horizontal position $\small x$ ?

Answer

The y velocity is thetime derivative of the $\small y$ position, and not the $\small x$ derivative. In order to find it, use the chain rule:

$\frac{dy}{dt}=\left(\frac{dy}{dx}\right) \frac{dx}{dt}$

$\frac{dy}{dx}= 2ax + b$

Of course, $\frac{dx}{dt} =v_{x}$

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Question

Two objects moving in one dimension created the following velocity vs. time graph:

From the graph above, what is true about the two objects at time $\small t=5s$ ?

Answer

Since this is a graph of velocity and not position, the curves intersect where the velocities match. Since we do not know the starting position, we do not know where the objects are relative to one another.

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Question

Two objects moving in one dimension created the following velocity vs. time graph:

From the graph above, which object has traveled a greater distance from its starting position when $\small t=5s$ ?

Answer

Since this is a graph of velocity vs. time, its integral is distance travelled. We can estimate the integral by looking at the area under the curves. Since Bbject 1 has a greater area under its velocity curve, it has covered a greater distance. Its velocity is greater that Object 2's for the entire time, so it makes sense that it will travel farther.

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Question

Atwood's machine consists of two blocks connected by a string connected over a
pulley as shown. What is the acceleration of the blocks if their masses are and .

Assume the pulley has negligible mass and friction.

Answer

From the force diagram above, we can see that tension is pulling up on both sides of the string and gravity is pulling down on both blocks. With this information we can write 2 force equations:

If we add the two equations together, we get:

where

Solving for , we get

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Question

A guillotine blade weighing is accelerated upward into position at a rate of $2.0\frac{m}{s^2}$ .

What is the the approximate mass of the guillotine blade?

Answer

The force of gravity on the blade is , which is the same as $400\frac{kg\cdot m}{s^2}$

This unit relationship comes from Newton's second law.

is the mathematical expression of Newton's second law. The units for force must be a product of the units for mass and the units for acceleration.

Solve the expression by plugging in known values.

$400 \frac{kg\cdot m}{s^2}=m \cdot 9.8\frac{m}{s^2}$

$\frac{400}{9.8}kg=mass\approx41kg$

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Question

A guillotine blade weighing is accelerated upward into position at a rate of $2.0\frac{m}{s^2}$ .

What is the tension on the rope pulling the blade, while it is accelerating into position?

Answer

The tension in the rope is the sum of the forces acting on it. If one considers that the net force on an object must equal the mass of the object times the acceleration of the object, the net force on the object must be the force due to tension from the rope minus the force due to gravity.

$F_{T}-F_{g}=ma$

Rearrange the equation.

$F_{T}=ma+F_{g}$

Plug in known values.

$F_T=41kg\cdot2\frac {m}{s^2} + 400N$

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Question

An object is moving in two dimensions. Its vertical motion relative to the horizontal motion is described by the equation $\small y=ax^2+bx +c$ . Its motion in the horizontal direction is described by the equation $\small x=v_{x}t$ . What is the object's velocity is the direction in terms of its horizontal position $\small x$ ?

Answer

The y velocity is thetime derivative of the $\small y$ position, and not the $\small x$ derivative. In order to find it, use the chain rule:

$\frac{dy}{dt}=\left(\frac{dy}{dx}\right) \frac{dx}{dt}$

$\frac{dy}{dx}= 2ax + b$

Of course, $\frac{dx}{dt} =v_{x}$

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Question

Two objects moving in one dimension created the following velocity vs. time graph:

From the graph above, what is true about the two objects at time $\small t=5s$ ?

Answer

Since this is a graph of velocity and not position, the curves intersect where the velocities match. Since we do not know the starting position, we do not know where the objects are relative to one another.

Compare your answer with the correct one above

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