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AP Physics C: Mechanics : Calculating Motion in One Dimension

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

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Example Questions

Example Question #11 : Mechanics Exam

A man runs with a velocity as described by the function below.

v(t) = 8t + 4

How far does he travel in 1 minute?

Possible Answers:

7430 m

12100 m

14640 m

23140 m

3320 m

Correct answer:

14640 m

Explanation:

Distance is given by the integral of a velocity function. For this question, we will need to integrate over the interval of 0s to 60s.

$x(t) = \int_{0}^{60}v(t)\,dt$

$x(60)=\int_{0}^{60}(8t+4)\,dt = (4(60)^2+4(60))-(4(0)^2+4(0))$

$x(60)=4(3600)+240 = 14640\ m$

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Example Question #11 : Ap Physics C

A man runs with a velocity described by the function below.

v(t) = 8t + 4

What is the function for his acceleration?

Possible Answers:

a(t)=0

a(t)=4

a(t)=8

a(t)=4t^2+4t

a(t)=8t

Correct answer:

a(t)=8

Explanation:

The function for acceleration is the derivative of the function for velocity.

$a(t)= \frac{dv(t)}{dt}$

v(t) = 8t + 4

a(t)=v'(t)=8

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Example Question #11 : Mechanics Exam

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 $8\:\uptext{s}$ ?

Possible Answers:

$2.75\:\uptext{m}$

$2.20\:\uptext{m}$

$1.65\:\uptext{m}$

$0.85\:\uptext{m}$

$1.10\:\uptext{m}$

Correct answer:

$2.75\:\uptext{m}$

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:

v = v_0 + at , 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:

$x_1 = v_0t + \frac{1}{2}at^2$

v_0t = 0 because the initial speed is .

If we plug in a = 0.1 , t = 5 , 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:

x_2 = v_0t

v_0 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 x_2 = 1.5m

Adding x_1 and x_2 we get the total distance to be $2.75\:\uptext{m}$

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AP Physics C: Mechanics : Calculating Motion in One Dimension

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

All AP Physics C: Mechanics Resources

Example Questions

Example Question #11 : Mechanics Exam

Example Question #11 : Ap Physics C

Example Question #11 : Mechanics Exam

All AP Physics C: Mechanics Resources

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