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AP Physics 1 : Linear Motion and Momentum

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Example Question #311 : Linear Motion And Momentum

The position of a particle can be described by:

s(t)=2t^2-3t+12

The position is measured in , and the time is in seconds . What is the average velocity for the particle in the time interval [0,4] ?

Possible Answers:

$10 \frac{m}{s}$

$3 \frac{m}{s}$

$2 \frac{m}{s}$

$5 \frac{m}{s}$

$4 \frac{m}{s}$

Correct answer:

$5 \frac{m}{s}$

Explanation:

You can think of the average velocity of a particle as the difference between the final and initial states. Conventionally, we call this slope. Therefore, we need to plug in our two initial times into the position function, and use the equation for slope between two points to determine the average velocity.

s(4)=2(4^2)-3(4)+12=32

s(0)=2(0^2)-3(0)+12=12

Therefore, our two coordinates are: (0,12),(4,32)

Now, we use the slope equation:

$\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\frac{32-12}{4-0}=\frac{20}{4}=5\frac{m}{s}$

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Example Question #311 : Linear Motion And Momentum

A 55kg woman has a jet pack that has a mass of 5kg . She ignites the jet pack and it accelerates her at a rate of $5\frac{m}{s^2}$ . Determine the time it took to reach a height of 15m .

Possible Answers:

3.98s

None of these

2.45s

6.67s

3.00s

Correct answer:

2.45s

Explanation:

Using the kinematic equation:

d=.5at^2+v_0t+d_0

15=.5*5*t^2+0*t+0

t=2.45s

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Example Question #581 : Ap Physics 1

A 55kg woman has a jet pack that has a mass of 5kg . She ignites the jet pack and it accelerates her at a rate of $5\frac{m}{s^2}$ . Determine her velocity at the moment she reaches a height of 15m .

Possible Answers:

$24.50\frac{m}{s}$

None of these

$12.25\frac{m}{s}$

Impossible to determine

$15\frac{m}{s}$

Correct answer:

$12.25\frac{m}{s}$

Explanation:

Using the kinematic equation:

d=.5at^2+v_0t+d_0

15=.5*5*t^2+0*t+0

t=2.45s

Using

v_f=v_i+a*t

v_f=0+5*2.45

$v_f=12.25\frac{m}{s}$

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Example Question #41 : Fundamentals Of Displacement, Velocity, And Acceleration

A 55kg woman has a jet pack that has a mass of 5kg . She ignites the jet pack and it accelerates her at a rate of $5\frac{m}{s^2}$ . Determine her kinetic energy at a height of 20m .

Possible Answers:

3388J

3985J

1562J

None of these

8875J

Correct answer:

8875J

Explanation:

$E_{total}=E_{kinetic}+E_{potential}$

$E_{total}=.5mv_f^2+mgh_f$

E_f-E_i=W

Initially she is at rest.

Combining equations:

W=.5mv_f^2+mgh_f

W=F*d

F*d=.5mv_f^2+mgh_f

ma*d=.5mv_f^2+mgh_f

a*d=.5v_f^2+gh_f

Solving for v_f

$\sqrt{a*d-g*h_f}=v_f$

Plugging in values:

$\sqrt{5*20-(-9.8)*20}=v_f$

$v_f=17.2\frac{m}{s}$

Using KE=.5mv^2

KE=8875J

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Example Question #41 : Fundamentals Of Displacement, Velocity, And Acceleration

A woman with a jet pack is 50m above the ground. She drops a ball. If she waits until after she dropped the ball to accelerate downward to try and catch it, how fast will she have to accelerate to reach it at a height of 10m ?

Possible Answers:

$254\frac{m}{s^2}$

None of these

$108\frac{m}{s^2}$

$18.6\frac{m}{s^2}$

$81\frac{m}{s^2}$

Correct answer:

$108\frac{m}{s^2}$

Explanation:

Distance fallen by ball:

$.5at_{ball}^2=d_{ball}$

Distance fallen by woman:

$.5at_{woman}^2=d_{woman}$

Both distances need to be 40m

If it fell

Since she waits 2 s to drop, $t_{ball}-2s=t_{woman}$

Combining equations:

$.5at_{ball}^2=d_{ball}$

Solving for $t_{ball}$

$t_{ball}=\sqrt{\frac{2*d_{ball}}{a}}$

Acceleration will be gravity

$t_{ball}=2.86s$

$t_{woman}=.86s$

$.5at_{woman}^2=d_{woman}$

$a_{woman}=108\frac{m}{s^2}$

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Example Question #316 : Linear Motion And Momentum

Four strings are used to hold up a .50kg puppet, motionless. Determine the tension in each string.

Possible Answers:

16.4N

4.90N

.1225N

None of these

1.225N

Correct answer:

1.225N

Explanation:

$F_{net}=F_{gravity}+T_1+T_2+T_3+T_4$

Assuming all tensions will be equal

T_1=T_2=T_3=T_4

$F_{net}=F_{gravity}+4T_1$

The puppet is not moving

$0=F_{gravity}+4T_1$

Using $F_{gravity}=mg$

0=mg+4T_1

Solving for T_1

$T_1=\frac{-mg}{4}$

Plugging in values

$T_1=\frac{-.50*-9.8}{4}$

T_1=1.225N

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Example Question #317 : Linear Motion And Momentum

A man with a jet pack is 40m above the ground. He drops a ball. If he waits until 1.5s after he dropped the ball to accelerate downward to try and catch it, how fast will he have to accelerate to reach it at a height of ?

Possible Answers:

$25\frac{m}{s^2}$

$51\frac{m}{s^2}$

None of these

$18\frac{m}{s^2}$

$104\frac{m}{s^2}$

Correct answer:

$51\frac{m}{s^2}$

Explanation:

$t_{man}=t_{ball}-1.5s$

In order for the ball to reach a height of , it would need to fall a distance of 35m . Using

d_f=d_i+v_i*t+.5at^2

The initial velocity is zero, and the initial distance can be taken as zero as well.

35=0+0*t+.5*9.8*t^2

Solving for time

t=2.67s

This is the time the ball takes to fall to a height of .

The man with the jet pack will have:

2.67-1.5=1.17s

Again, using

d_f=d_i+v_i*t+.5at^2

Again,

Initial velocity is zero, and the initial distance can be taken as zero as well.

35=0+0*1.17+.5*a*1.17^2

$a=51\frac{m}{s^2}$

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Example Question #311 : Linear Motion And Momentum

A car is stopped at a red light. When the light turns green, it accelerates uniformly at a rate of $2 \frac{m}{s^{2}}$ . How fast is the car traveling after seconds?

Possible Answers:

$10 \frac{m}{s}$

$0.2 \frac{m}{s}$

$5 \frac{m}{s}$

$1 \frac{m}{s}$

$20 \frac{m}{s}$

Correct answer:

$20 \frac{m}{s}$

Explanation:

Acceleration is defined as a change in velocity over time:

$a = \frac{\Delta v}{t} = \frac{v_{f}-v_{i}}{t}$

Since the car is initially at rest at the stop light, $v_{i}=0$ . The acceleration is given as $2\frac{m}{s^{2}}$ and the problem is asking for the velocity at t=10s . Therefore,

$2\frac{m}{s^{2}} = \frac{\Delta v}{10s}$

$\Delta v=20\frac{m}{s}$

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Example Question #311 : Linear Motion And Momentum

A train travels 200km in hour. What is the train's average speed in units of meters per second?

Possible Answers:

$55.6 \frac{m}{s}$

$556 \frac{m}{s}$

$200,000 \frac{m}{s}$

$5.56 \frac{m}{s}$

$200 \frac{m}{s}$

Correct answer:

$55.6 \frac{m}{s}$

Explanation:

Average velocity is defined as distance over time. Therefore, the train is traveling at an average rate of $200 \frac{km}{h}$ . However, this problem asks for the answer to be in units of $\frac{m}{s}$ . Here is an efficient way to convert the units using dimensional analysis:

$\frac{200[km]}{1[h]}=\frac{1000[m]}{1[km]}=\frac{1[h]}{60[min]}=\frac{1[min]}{60[s]}$

$\frac{200\times1000[m]}{60\times60[s]} = 55.6\frac{m}{s}$

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Example Question #41 : Fundamentals Of Displacement, Velocity, And Acceleration

Madison enters an elevator on the ground floor of a building. She selects the button for the fifth floor and the elevator doors close. The elevator accelerates for seconds until it reaches a constant velocity for seconds, and then decelerates for seconds before stopping at the fifth floor.

What direction is the acceleration seconds into Madison's elevator ride?

Possible Answers:

The acceleration is in the upward direction.

The acceleration is in the downward direction.

The acceleration is to the left.

The acceleration is zero.

The acceleration is to the right.

Correct answer:

The acceleration is zero.

Explanation:

At t=10 seconds, the elevator is traveling at a constant velocity. When the velocity is constant, acceleration is zero. This can be confirmed using the equation for acceleration:

$a=\frac{\Delta v}{t}$

When there is no change in (constant velocity), then the numerator is zero. Thus, v=0 .

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AP Physics 1 : Linear Motion and Momentum

Study concepts, example questions & explanations for AP Physics 1

All AP Physics 1 Resources

Example Questions

Example Question #311 : Linear Motion And Momentum

Example Question #311 : Linear Motion And Momentum

Example Question #581 : Ap Physics 1

Example Question #41 : Fundamentals Of Displacement, Velocity, And Acceleration

Example Question #41 : Fundamentals Of Displacement, Velocity, And Acceleration

Example Question #316 : Linear Motion And Momentum

Example Question #317 : Linear Motion And Momentum

Example Question #311 : Linear Motion And Momentum

Example Question #311 : Linear Motion And Momentum

Example Question #41 : Fundamentals Of Displacement, Velocity, And Acceleration

All AP Physics 1 Resources

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