AP Calculus AB: 4.2 Straight-Line Motion: Connecting Position, Velocity, and Acceleration – Exam Style questions with Answer- MCQ

Question

The graph of

y=f(x)

 on the interval 

0<x<5

is shown above. Which of the following could be the graph of 

y=f(x)

 ?

A

B

C

D

▶️Answer/Explanation

Ans:A

The graph of

f

 indicates that f

f

is increasing from x

=0

 to x=2, then decreasing from x

=2

 to x=3, and then increasing from x=3 to x=5. Therefore, the graph of

f

 should be positive from x=0 to x=2, negative from

x=2

to

x=3

, and positive from x

=3

to 

x=5

. This graph is the only one that has this behavior, so it could be the graph of

f

. Some other features of the graph of

f

support this conclusion. Since f is not differentiable at

x=2

, the graph of

f

should not be defined at 

x=2

. Since f

f

has a local minimum at 

x=3

and is differentiable there, 

f(3)

should equal 

0

. This graph is consistent with those observations.

Question

The equation \(y=2e^{6x}-5 \) is a particular solution to which of the following differential equations?

A y6y30=0

B 2y12y+5=0

C y′′5y6y=0

D y′′2y+y+5=0

▶️Answer/Explanation

Ans:A

Correct. One way to verify that a function is a solution to a differential equation is to check that the function and its derivatives satisfy the differential equation. The differential equation in this option involves y and y. The correct derivative must be computed and the algebra correctly done to verify that the differential equation is satisfied.

\(y{}’=12e^{6x}\)

\(y{}’-6y-30=12e^{6x}-6(2e^{6x}-5)-30=0\)

Question

A particle moves along the x-axis so that its position at time t is given by \(x(t)=t^{2}-6t+5\).For what value of t is the velocity of the particle zero?

A 1

B 2

C 3

D 4

E 5

▶️Answer/Explanation

Ans:C

Question

An object moves along a straight line so that at any time t 0t9, its position is given by \(x(t)=7+6t-t^{2}\). For what value of t is the object at rest?

A t=3

B t=6

C \( t=\frac{13}{2}\)

D t=7

▶️Answer/Explanation

Ans:A

The object is at rest when its velocity equals 0. Because the object’s velocity is given by x(t)=62t, it is at rest at t=3.

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