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AP Physics C: Mechanics : Understanding Distances

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Example Questions

Example Question #1 : Understanding Distances

An object starts from rest and reaches a velocity of $10\frac{m}{s}$ after accelerating at a constant rate for four seconds. What is the distance traveled in this time?

Possible Answers:

20m

10m

2.5m

Correct answer:

20m

Explanation:

Since the acceleration is constant in this problem, we can apply the kinematics given equation to calculate the distance:

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

First, we need to calculate the acceleration.

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

Plug in our velocity and time values to find the acceleration.

$a=\frac{10\frac{m}{s}-0\frac{m}{s}}{4s}=2.5\frac{m}{s^2}$

Now we can return to the kinematics equation and solve for the distance traveled:

$\Delta x=(0\frac{m}{s})(4s)+\frac{1}{2}(2.5\frac{m}{s^2})(4s)^2$

$\Delta x=20m$

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Example Question #1 : Ap Physics C

At the moment a car is passed by another car constantly traveling at $10\frac{m}{s}$ , it begins to accelerate at $5\frac{m}{s^2}$ . In how many seconds does this car catch up to and pass the other car?

Possible Answers:

Correct answer:

Explanation:

First, find the displacement equations for both cars. Car 1 will be the car that is initially stationary; car 2 will be the car traveling with constant velocity.

Car 1:

a_1(t) = 5

$v_1(t) = \int a_1(t) dt = 5t$

$x_1(t) = \int v_1(t)dt = 2.5t^2$

Car 2:

v_2(t) = 10

$x_2(t) = \int v_2(t)dt = 10t$

Now, set those displacement equations equal to each other and solve for .

x_1(t) = x_2(t)

2.5t^2 = 10t

t = 4

The accelerating car will catch and pass the car traveling at constant velocity after 4 seconds.

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Example Question #3 : Mechanics Exam

An object moving along a line has a displacement equation of x(t) = 5x^2 - 40x , where is in seconds. For what value of is the object stationary?

Possible Answers:

Correct answer:

Explanation:

Use the fact that {x}'(t) = v(t) to find the velocity equation, and then solve for when v = 0 .

x(t) = 5x^2 - 40x

Take the derivative of the displacement equation.

x'(t)=v(t)=10t-40

Set the velocity equal to zero.

v(t) = 0 = 10t - 40

Solve for the time.

$t = \frac{40}{10} = 4s$

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Example Question #1 : Understanding Distances

An object is moving on a line and has displacement equation of x(t) = 6(t - 9)^3 , where is in seconds. At what value of is the object not accelerating?

Possible Answers:

10s

11s

Correct answer:

Explanation:

Take the second derivative of x(t) , to find the acceleration function a(t) . Then find the value of where a(t) = 0 . The fact that it is simply and not or inside that quantity means that using the Chain Rule becomes decidedly easier.

x(t) = 6(t - 9)^3

v(t) = x'(t) = 18(t - 9)^2

a(t) = v'(t) = 36(t - 9)

Set the acceleration fuction equal to zero and solve for the time.

a(t)=0=36(t-9)

t = 9s

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Example Question #2 : Ap Physics C

A projectile is launched out of a cannon or launch tube with zero air resistance or friction. At what angle (in degrees) should the projectile be launched to maximize the distance it travels?

Possible Answers:

$45^\circ$

$30^\circ$

$90^\circ$

$75^\circ$

$15^\circ$

Correct answer:

$45^\circ$

Explanation:

The projectile travels the farthest when the vertical component of its velocity matches the sum of its horizontal component and whatever the wind/friction adds or subtracts. If the wind has no effect, then a 45-degree angle will be the best because the horizontal and vertical components of the velocity will create a right isosceles triangle (remember special triangles).

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AP Physics C: Mechanics : Understanding Distances

Study concepts, example questions & explanations for AP Physics C: Mechanics

All AP Physics C: Mechanics Resources

Example Questions

Example Question #1 : Understanding Distances

Example Question #1 : Ap Physics C

Example Question #3 : Mechanics Exam

Example Question #1 : Understanding Distances

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All AP Physics C: Mechanics Resources

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