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AP Physics 1 : Fundamentals of Displacement, Velocity, and Acceleration

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Example Question #577 : Ap Physics 1

$27.8\frac{m}{s}$ is the minimum speed needed for an airplane to take off. The airplane is accelerating at a constant $3.3\frac{m}{s^{2}}$ . What length does the runway need to be?

Possible Answers:

772m

117 m

386m

400 m

111 m

Correct answer:

117 m

Explanation:

The correct answer is 117 m . Because we don't have time, we should use an equation that does not require a time variable, like

$v^{2}=v_{o}^{2} + 2ax$

where is the final velocity, $v_{o}$ is the initial velocity, is the acceleration, and is the distance traveled. We are given , $v_{o}$ , and . Therefore, we can put these values into our equation to solve for :

$(27.8\frac{m}{s})^{2} = (0\frac{m}{s})^{2} + (2)(3.3\frac{m}{s^{2}})x$

$772.8 \frac{m}{s}= (6.6\frac{m}{s^{2}})x$

x=117m

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Example Question #31 : Fundamentals Of Displacement, Velocity, And Acceleration

A car is moving with a velocity given by $v(t)=3t^2+5t+4\ m/s$ after being ejected by a connon at an initial velocity of $100\ m/s$ .

When is the acceleration equal to $125\frac{m}{s^2}$ ?

Possible Answers:

24s

27.5s

20 s

17.5 s

15 s

Correct answer:

20 s

Explanation:

In order to find the acceleration, you must first take the derivative of the velocity:

$v(t)=3t^2+5t+4\frac{m}{s}$

$a(t)=v'(t)=6t+5\frac{m}{s^2}$

The initial velocity does not affect the acceleration! It would only affect the position function! Afterwards, just plug in and solve for t:

6t+5=125

6t=120

t=20s

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Example Question #579 : Ap Physics 1

You drive 3 mi east from your home past the grocery store. As you pass the store, you drive another 4 mi eastward to the post office. From the post office, you drive 8 mi westward. Then, you drive 1 mi east back home. Assume you do not stop and are traveling at a constant speed. The trop takes you 30 min total. What was your average speed? Your average velocity? (average speed, average velocity)

Possible Answers:

$0.533\frac{m}{s}$ , $0\frac{m}{s}$

$14.3\frac{m}{s}$ , $14.3\frac{m}{s}$

$32\frac{m}{s}$ , $32\frac{m}{s}$

$32\frac{m}{s}$ , $0\frac{m}{s}$

$14.3 \frac{m}{s}$ , $0\frac{m}{s}$

Correct answer:

$14.3 \frac{m}{s}$ , $0\frac{m}{s}$

Explanation:

The correct answer is $14.3\frac{m}{s}$ , $0\frac{m}{s}$ . Average speed is:

$\frac{distance}{time}$ , in $\frac{meters(m)}{second(s)}$

We traveled a total of 16 mi . Convert this to meters to get 25749.5 m

Then we must convert 30 min to

$30 min (\tfrac{60 s}{1 min})=1800 s$

Then divide to get your average speed. This gives us $14.3\frac{m}{s}$

Velocity, on the other hand, is $\frac{displacement}{time}$ in $\frac{m}{s}$ .

Displacement is the the final distance from the initial position. Displacement is considered a state function. Only the final position matters; it doesn't matter how you get there. For example, if you start at point A and drive in a circle that covers 100 mi , and end up at point A again, your displacement is 0 m . This is because your final position is the same as your initial position. So in this problem, because your final position is the same as your initial position, your displacement is 0 m .

So $\frac{0 m}{1800 s}=0\frac{m}{s}$ for the average velocity.

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Example Question #580 : Ap Physics 1

On your way home from school, you drive 3 mi until you pass John's house. Once you pass John's house you drive another 2 mi in the same direction until you reach the park, where you have soccer practice. The trip takes you 7 min . What is your average speed? Your average velocity? (Average speed, average velocity).

Possible Answers:

$14.0\frac{m}{s}$ , $12.0\frac{m}{s}$

$22.4\frac{m}{s}$ , $22.4\frac{m}{s}$

$22.4\frac{m}{s}$ , $0\frac{m}{s}$

$0.833\frac{m}{s}$ , $0.833\frac{m}{s}$

$13.4\frac{m}{s}$ , $13.4\frac{m}{s}$

Correct answer:

$22.4\frac{m}{s}$ , $22.4\frac{m}{s}$

Explanation:

The correct answer is $22.4\frac{m}{s}$ , $22.4\frac{m}{s}$

Remember average speed is $\frac{distance}{time}$ , also $\frac{meters(m)}{second(s)}$

By converting miles to meters and minutes to seconds , we get $\frac{8046m}{360s}=22.4 \frac{m}{s}$

This is also true for average velocity. Average velocity is $\frac{displacement}{time}$ , $\frac{meters(m)}{second(s)}$

Displacement is the distance of the final position from the initial position. Because we do not change direction, the displacement is equal to the distance. That means that the average velocity will be equal to the average speed (time remains constant).

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Example Question #31 : Fundamentals Of Displacement, Velocity, And Acceleration

A car drives 100 m in 1 m at constant velocity and without changing direction. What is the car's acceleration?

Possible Answers:

$100\frac{m}{s^{2}}$

$1.67\frac{m}{s^{2}}$

$16.7\frac{m}{s^{2}}$

$0\frac{m}{s^{2}}$

$7.0\frac{m}{s^{2}}$

Correct answer:

$0\frac{m}{s^{2}}$

Explanation:

Average acceleration is equal to $\frac{(velocity_{2}-velocity_{1})}{time}$ , where $V_{2}$ is the final velocity and $V_{1}$ is the initial velocity.

If the velocity is constant, the final and initial velocities are equal. Let's say the car is traveling at $30\frac{m}{s}$ constantly.

$30\frac{m}{s}- 30\frac{m}{s}=0\frac{m}{s}$ . So $\frac{0\frac{m}{s}}{\frac{60}{s}}= 0\frac{m}{s^{2}}$

When velocity is constant, acceleration is always equal to $0\frac{m}{s^{2}}$

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Example Question #541 : Newtonian Mechanics

A car accelerates to $75\frac{km}{h}$ from rest in 5 s . Assume the car moves in a straight line. What is the magnitude of the car's acceleration?

Possible Answers:

$4.17\frac{m}{s^{2}}$

$15.0\frac{m}{s^{2}}$

$14.3\frac{m}{s^{2}}$

$3.32\frac{m}{s^{2}}$

$20.8\frac{m}{s^{2}}$

Correct answer:

$4.17\frac{m}{s^{2}}$

Explanation:

Average acceleration is equal to $\frac{(velocity_{2}-velocity_{1})}{time}$ where $V_{2}$ is the final velocity and $V_{1}$ is the initial velocity.

In this problem, we know our initial velocity is $0\frac{km}{h}$ (starting from rest). The final velocity is $75\frac{km}{h}$ . We must convert these units to $\frac{meters}{second}$ .

$75 km\frac{1000m}{1 km}=75000m$

$60 min\frac{60 s}{1 min}=3600s$

Using these values, we can calculate the final velocity to be $\frac{75000m}{3600s}=20.8\frac{m}{s}$

Using this number, we can finally calculate the magnitude of acceleration.

$({20.8\frac{m}{s}}- 0\frac{m}{s})(5s)=4.17\frac{m}{s^{2}}$

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Example Question #33 : Fundamentals Of Displacement, Velocity, And Acceleration

A car at a stop sign must cross an intersection that is 20.0m long. If the car accelerates at $2.5\frac{m}{s^2}$ constantly, how long will it take for the car to cross the intersection?

Possible Answers:

8.0s

16.0s

9.0s

4.47s

4.0s

Correct answer:

4.0s

Explanation:

We can solve this problem using the kinematic equation:

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

Where is the displacement, v_o is the initial velocity, is the acceleration, and is time. We are given values for all but the time. Therefore, we may plug these values into the above equation and solve for .

$20.0m=\left(0\frac{m}{s}*t\right)+\left(\frac{1}{2}*2.5\frac{m}{s}^2*t^2\right)$

20.0 m=t^2

t=4.0s

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Example Question #31 : Fundamentals Of Displacement, Velocity, And Acceleration

A ball is thrown upward at $3.0\frac {m}{s}$ from a tower. Ignore air resistance. How far does the ball travel after 1 s ?

Possible Answers:

7.9 m

1.9m

-1.9 m

6.8 m

-6.8 m

Correct answer:

-1.9 m

Explanation:

The correct answer is -1.9m . We can use the kinematics equation:

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

We are given the values of $v_{o}$ , , and (initial velocity, time, and acceleration). Therefore we can solve for to find the distance traveled by the ball.

$y= (3\frac{m}{s})(1s) + (\frac{1}{2})(9.8\frac{m}{s^{2}})(1.0s)^{2}$

y= (3m)+(-4.9m)

y=-1.9m

is negative because the position of the ball is 1.9m below its initial point.

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Example Question #35 : Fundamentals Of Displacement, Velocity, And Acceleration

From the top of a mountain, you throw a rock downward at $3.0\frac{m}{s}$ . What is the rock's displacement after 2 s ?

Possible Answers:

19.6 m

-25.6 m

-19.6 m

-15.8m

15.8 m

Correct answer:

-25.6 m

Explanation:

To solve this problem, we can use the kinematics equation:

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

Where $v_{o}$ is the initial velocity, is time, and is acceleration. We are given all of these variables, so we can plug them in for .

$y = (-3.0\frac{m}{s})(2.0s) + (\frac{1}{2})(-9.8\frac{m}{s^{2}})(2.0s)^{2}$

y=-25.6m

In this problem, $v_{o}$ and are negative because the rock is initially being thrown downhill and is accelerating downwards.

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Example Question #32 : Fundamentals Of Displacement, Velocity, And Acceleration

If the driver of a $1200\ kg$ car traveling at $32\ \frac{m}{s}$ slams on the brakes, and it takes $2.0\ s$ for the car to come to a complete stop, what is the coefficient of kinetic friction between the car's tires and the asphalt that the car is traveling on?

Possible Answers:

0.54

1.2

1.6

2.4

0.85

Correct answer:

1.6

Explanation:

The first step in doing this problem is figuring out how many forces are acting on the car when the driver slams the brakes. Since there is nothing pushing the car in the direction that it was originally traveling, the only force that is acting on the car is the force of friction. This means that the net force is simply just the force of kinetic friction, net force is also equal to mass multiplied by acceleration, so this equation appears as follows:

$F_{k}=ma$

where $F_{k}$ is the force of kinetic friction.

The force of friction is also equal to normal force multiplied by the coefficient of kinetic friction for a certain surface. This equation appears as follows:

$F_{k}=\mu _{k}N$

where $F_{k}$ is the force of kinetic friction, $\mu _{k}$ is the coefficient of kinetic friction, and N is normal force (also equal to mass x gravity).

These 2 equations can be connected in order to figure out the coefficient of kinetic friction.

$ma=\mu _{k}mg$

mass can be cancelled from both sides. Acceleration is found as follows:

$a=\frac{\Delta V}{\Delta t}=\frac{32\ \frac{m}{s}}{2\ s}=16\ \frac{m}{s^{2}}$

By plugging in the given values, $\mu _{k}$ is found to be:

$16=\mu _{k}*9.81$

$\mu _{k}=1.63$

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AP Physics 1 : Fundamentals of Displacement, Velocity, and Acceleration

Study concepts, example questions & explanations for AP Physics 1

All AP Physics 1 Resources

Example Questions

Example Question #577 : Ap Physics 1

Example Question #31 : Fundamentals Of Displacement, Velocity, And Acceleration

Example Question #579 : Ap Physics 1

Example Question #580 : Ap Physics 1

Example Question #31 : Fundamentals Of Displacement, Velocity, And Acceleration

Example Question #541 : Newtonian Mechanics

Example Question #33 : Fundamentals Of Displacement, Velocity, And Acceleration

Example Question #31 : Fundamentals Of Displacement, Velocity, And Acceleration

Example Question #35 : Fundamentals Of Displacement, Velocity, And Acceleration

Example Question #32 : Fundamentals Of Displacement, Velocity, And Acceleration

All AP Physics 1 Resources

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