AP Physics C: Mechanics 1.2 Displacement Velocity and Acceleration Questions

Relate displacement, average and instantaneous velocity, and acceleration using vectors, time functions, derivatives, integrals, and initial conditions.

Exam points

calculate displacement from initial and final positions or from a definite velocity integral
interpret average velocity or acceleration from changes over a stated time interval
differentiate position and velocity functions to obtain instantaneous kinematic quantities
integrate acceleration or velocity with the correct initial condition to recover motion functions
find extrema or direction changes and distinguish total distance from net displacement

Covered syllabus points

1.2.A—Describe a change in an object’s position 1.2.B—Describe the average velocity and acceleration of an object 1.2.C—Describe the instantaneous position, velocity, and acceleration of an object as a function of time

[Maximum number: 9]

Table for Question 1 — AP Physics C: Mechanics

Mech. 1. A new sports car is undergoing acceleration tests to determine its specifications. The following data on speed v versus time t are recorded for the car as it accelerates from rest along a straight track.

(a)

Answer each of the following for the time period $t=0 \mathrm{~s}$ to $t=10 \mathrm{~s}$ .

[ 1 ]

(i)

Does the speed of the car increase, decrease, or stay the same? Increase Decrease Stay the same Justify your answer.

[ 1 ]

(b)

The equation for the speed v of the car as a function of time t found from the graph is $v(t)=-0.3 t^{2}+7 t$ , where v is in meters per second and t is in seconds.

(d) Derive an expression for the acceleration of the car a(t) as a function of time t.

[ 2 ]

(d) 2 points For beginning with an indication that the acceleration is the derivative of velocity $a(t)=\frac{d v}{d t}$ For correctly evaluating the derivative $a(t)=\frac{d}{d t}\left(-0.3 t^{2}+7 t\right)=-0.6 t+7$

(c)

The equation for the speed v of the car as a function of time t found from the graph is $v(t)=-0.3 t^{2}+7 t$ , where v is in meters per second and t is in seconds.

(e) Calculate the position of the car as a function of time, x(t), assuming that the car starts from rest at the origin of a coordinate system.

[ 3 ]

(e) 3 points For an indication that the displacement is the integral of the velocity $\Delta x=\int_{t_{0}}^{t} v(t) d t$ For attempting to integrate with appropriate limits or constant of integration $\Delta x=\int_{0}^{t}\left(-0.3 t^{2}+7 t\right) d t$ For the correct answer $x(t)=-0.1 t^{3}+3.5 t^{2}$

(d)

The equation for the speed v of the car as a function of time t found from the graph is $v(t)=-0.3 t^{2}+7 t$ , where v is in meters per second and t is in seconds.

(f) Calculate the distance traveled by the car between $t=2 \mathrm{~s}$ and $t=8 \mathrm{~s}$ .

[ 3 ]

(f) 3 points For recognizing that the distance is the change in position $\Delta x=x(8)-x(2)$ For using the expression from part (e) to calculate x(2) and x(8) $x(2)=13.2 \mathrm{~m}$ $x(8)=172.8 \mathrm{~m}$ For the correct answer with units $\Delta x=159.6 \mathrm{~m}$ Alternate solution For using the integral of the speed and recognizing that the distance is the change in position For using the correct limits or constant of integration $\Delta x=\int_{2}^{8}\left(-0.3 t^{2}+7 t\right) d t$ $\Delta x=\frac{-0.3 t^{3}}{3}+\left.\frac{7 t^{2}}{2}\right|_{2} ^{8}=-51.2+224-(-0.8+14)$ For the correct answer with units $\Delta x=159.6 \mathrm{~m}$

[Maximum number: 7]

Scientists have created a new type of lightweight foam and are performing experiments to investigate the properties of the foam. The mass of Cart A is 1000 kg and the mass of Cart B is 2000 kg. A piece of foam with negligible mass is attached to the front of Cart A, as shown. Cart A moves with a constant speed toward Cart B, which is initially at rest. At time $t=0 \mathrm{~s}$ , the foam connected to Cart A makes contact with Cart B. The foam remains in contact with Cart B for 0.5 s, after which the carts separate and both carts move with constant velocities.

Figure for Question 2 — AP Physics C: Mechanics

For $0 \leq t \leq 0.50 \mathrm{~s}$ , the velocity v of Cart A can be described by the function $v(t)=64 t^{3}-48 t^{2}+5$ .

Calculate the magnitude of the maximum net force acting on Cart A during this interval.

On the following grid, draw a smooth curve of the magnitude of the force acting on Cart A as a function of time. Clearly indicate the value of the maximum force on the vertical axis.

The foam is removed from the front of Cart A and the experiment is repeated. The carts collide, with both Cart A and Cart B having the same initial and final velocities as in the original collision. The time intervals during which the carts are in contact are different in the collision with the foam and the collision without the foam. In the collision without the foam, Cart A is in contact with Cart B for a shorter duration than in the original collision, when the foam was present.

For the original collision when the foam is present, the magnitude of the average net force exerted on Cart B is $F_{1}$ . For the collision without the foam, the magnitude of the average net force exerted on Cart B is $F_{2}$ .

$\begin{aligned} a(t)&=\frac{dv}{dt}=192t^2-96t,\\ \frac{da}{dt}&=384t-96=0 \quad\Longrightarrow\quad t=0.25\ \mathrm{s}. \end{aligned}$

At (t=0.25\ \mathrm{s}), (a=-12\ \mathrm{m\,s^{-2}}). With (m=1000\ \mathrm{kg}),
$F_{\max}=m|a|=(1000)(12)=12\,000\ \mathrm{N}.$

For the graph,
$|F(t)|=1000\,|192t^2-96t|=96\,000t(1-2t), \qquad 0\le t\le0.50,$
so it is a smooth downward-opening curve through (0\ \mathrm N) at (t=0) and (t=0.50\ \mathrm s), with maximum (12\,000\ \mathrm N) at (t=0.25\ \mathrm s).

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Covered syllabus points

Question 1

Question (a)

Question (i)

Question (b)

Question (c)

Question (d)

Question 2