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AP Physics 1 : AP Physics 1

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Example Questions

Example Question #44 : 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:

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

None of these

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

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

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

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 #551 : Newtonian Mechanics

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

Possible Answers:

4.90N

None of these

1.225N

16.4N

.1225N

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 #45 : Fundamentals Of Displacement, Velocity, And Acceleration

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:

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

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

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 #555 : Newtonian Mechanics

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:

$20 \frac{m}{s}$

$1 \frac{m}{s}$

$0.2 \frac{m}{s}$

$5 \frac{m}{s}$

$10 \frac{m}{s}$

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 #312 : 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:

$200 \frac{m}{s}$

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

$55.6 \frac{m}{s}$

$5.56 \frac{m}{s}$

$556 \frac{m}{s}$

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 #313 : Linear Motion And Momentum

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 zero.

The acceleration is in the downward direction.

The acceleration is to the right.

The acceleration is to the left.

The acceleration is in the upward direction.

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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Example Question #591 : Ap Physics 1

An object travels with a velocity v_1 for a period of time t_1 . Its velocity is instantaneously changed to a second velocity v_2 at which it travels for a period of time t_2 .

What is the correct expression for the object's average velocity during the entire two-part motion?

Possible Answers:

$\frac{(v_1t_1)+(v_2t_2)}{2}$

$\frac{(v_1+v_2)}{t_1+t_2}$

$\frac{(v_1)+(v_2)}{t_1-t_2}$

$\frac{(v_1t_1)+(v_2t_2)}{t_1+t_2}$

$\frac{(v_1t_1)(v_2t_2)}{t_1t_2}$

Correct answer:

$\frac{(v_1t_1)+(v_2t_2)}{t_1+t_2}$

Explanation:

Average velocity can be found by taking the total distance traveled in the two part motion and dividing by the total time of the two part motion as shown by the equation for velocity:

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

Where is velocity, is displacement, and is time.

Our average velocity then would be:

$v_a_v= \frac{x_1+x_2}{t_1+t_2}$

While we were not given the displacements for either part, we can solve for them by rearranging the velocity equation:

x_1=(v_1)(t_1), x_2=(v_2)(t_2)

Substituting for the unknown displacements with equivalent terms comprised of the known velocities and times we find that the average velocity is:

$\frac{(v_1t_1)+(v_2t_2)}{t_1+t_2}$

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Example Question #51 : Fundamentals Of Displacement, Velocity, And Acceleration

In the traditional coordinate system, an object is moving downwards and slowing down.

What is the sign of the object's velocity?

Possible Answers:

Positive

Negative

It depends.

Neutral

It's impossible to tell.

Correct answer:

Negative

Explanation:

In a traditional coordinate system: left and down are considered negative directions while up and right are considered positive directions. The question states that the object is moving down, so regardless of its acceleration, its velocity is negative at the time described.

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Example Question #52 : Fundamentals Of Displacement, Velocity, And Acceleration

In the traditional coordinate system, an object is moving downwards and slowing down.

What is the sign of the object's acceleration?

Possible Answers:

Positive

Negative

Neutral

It's impossible to tell.

It depends.

Correct answer:

Positive

Explanation:

In a traditional coordinate system: left and down are considered negative directions while up and right are considered positive directions. Though the object is moving down (a negative direction) it is stated that it is slowing down, which means it is accelerating upwards in a positive direction.

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Example Question #53 : Fundamentals Of Displacement, Velocity, And Acceleration

In the traditional coordinate system, a car is speeding up while moving to the left.

What is the sign of the car's acceleration?

Possible Answers:

It depends

It's impossible to tell.

Negative

Positive

Neutral

Correct answer:

Negative

Explanation:

In a traditional coordinate system: left and down are considered negative directions while up and right are considered positive directions. Though the car is said to be speeding up, it is speeding up while moving in a negative direction. Therefor, the car's velocity is becoming more and more negative, meaning it has a negative acceleration.

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AP Physics 1 : AP Physics 1

Study concepts, example questions & explanations for AP Physics 1

All AP Physics 1 Resources

Example Questions

Example Question #44 : Fundamentals Of Displacement, Velocity, And Acceleration

Example Question #551 : Newtonian Mechanics

Example Question #45 : Fundamentals Of Displacement, Velocity, And Acceleration

Example Question #555 : Newtonian Mechanics

Example Question #312 : Linear Motion And Momentum

Example Question #313 : Linear Motion And Momentum

Example Question #591 : Ap Physics 1

Example Question #51 : Fundamentals Of Displacement, Velocity, And Acceleration

Example Question #52 : Fundamentals Of Displacement, Velocity, And Acceleration

Example Question #53 : Fundamentals Of Displacement, Velocity, And Acceleration

All AP Physics 1 Resources

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