Linear Motion

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AP Physics C: Electricity and Magnetism › Linear Motion

Questions 1 - 10

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 $8:\uptext{s}$ ?

$2.75:\uptext{m}$

$1.10:\uptext{m}$

$0.85:\uptext{m}$

$2.20:\uptext{m}$

$1.65:\uptext{m}$

Explanation

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 $2.75:\uptext{m}$

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

I and III

I and II

I, II, and III

I only

II only

Explanation

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:

$R_x = v_{ox}t$

$2.75:\uptext{m}$

$1.10:\uptext{m}$

$0.85:\uptext{m}$

$2.20:\uptext{m}$

$1.65:\uptext{m}$

Explanation

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

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

, 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 $2.75:\uptext{m}$

I. Increasing the initial horizontal velocity

II. Decreasing the mass of the ball

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

I and III

I and II

I, II, and III

I only

II only

Explanation

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.

$R_x = v_{ox}t$

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?

Explanation

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.

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

You drive your car from your house all the way to your school which is 50km away. After you are done with classes you drive back through the same route and park exactly where you had your car at the beginning of the day. By the end of the day, what were the distance and displacement of your motion?

$\text{Distance}=100km,\ \text{Displacement}=0km$

$\text{Distance}=50km,\ \text{Displacement}=50km$

$\text{Distance}=-50km,\ \text{Displacement}=50km$

$\text{Distance}=0km,\ \text{Displacement}=-100km$

$\text{Distance}=0km,\ \text{Displacement}=100km$

Explanation

This question tests your conceptual understanding of distance as a scalar quantity vs your understanding of displacement as a vector quantity.

Distance measures the total length that was traveled in a given motion, and does not care about the direction since it is a scalar value. In your day, you traveled 50km on your way to school and 50km on your way back home. In total you traveled 100km, so that is your distance.

Displacement is a vector quantity that measures the change in position. It cares about your final and initial positions, taking into account the direction of the change in position. In this scenario you started and ended your motion exactly at the same position, so overall at the end of the day your car did not change position at all. Therefore your displacement was 0m.

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?

Explanation

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

Rearrange the equation.

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

Plug in known values.

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

$\text{Distance}=100km,\ \text{Displacement}=0km$

$\text{Distance}=50km,\ \text{Displacement}=50km$

$\text{Distance}=-50km,\ \text{Displacement}=50km$

$\text{Distance}=0km,\ \text{Displacement}=-100km$

$\text{Distance}=0km,\ \text{Displacement}=100km$

Explanation

This question tests your conceptual understanding of distance as a scalar quantity vs your understanding of displacement as a vector quantity.

A car undergoes acceleration according to the given function. If the threshold for serious injury or fatality for a human undergoing horizontal acceleration is 60 g ees (1 gee = 10 meters per second per second), how long would a human be able to withstand riding in this car?