Question 1

A cyclist's position is given by s(t), measured in meters. Which expression gives the cyclist's average velocity from t = 3 to t = 8?

Accepted Answer

$\frac{s(8)-s(3)}{8-3}$ — The average velocity is change in position divided by change in time: $\frac{s(8)-s(3)}{8-3}$. Option B uses the wrong denominator and reverses the position change, option C is a ratio rather than a rate of change, and option D reverses only the time difference.

Question 2

For $s(t)=t^2+3t$, what is the instantaneous velocity at t = 2?

Accepted Answer

7 meters per second — The average velocity from 2 to $2+h$ is $\frac{s(2+h)-s(2)}{h}=7+h$. As h approaches 0, $7+h$ approaches 7, so the instantaneous velocity is 7 meters per second. The other values do not result from the simplified difference quotient.

Question 3

Which statement best explains why the expression $\frac{s(a+h)-s(a)}{h}$ can be used to find an instantaneous rate even though it is undefined when h = 0?

Accepted Answer

The expression is evaluated only for nonzero h values near 0, and its approaching value is identified. — The expression is used for nonzero h values increasingly close to 0, and the limit of those average rates gives the instantaneous rate. Option B incorrectly claims the quotient becomes defined, option C confuses instantaneous and average rate, and option D ignores the essential division by the time change.

Question 4

![Figure](https://getlykke-assets.s3.amazonaws.com/wiki-quiz-figures/c14bedc7-55ee-4bd3-99ff-a9a537a06a66/ap-exam-style/7fe9b45e5aaa-F1.png)

The graph of the function f is shown in the figure above. Which of the following statements about f is true?

Accepted Answer

$\lim_{x 	o a} f(x) = 2$ — The limit as x approaches 'a' from both sides (left and right) equals 2.
This means $\lim_{x 	o a} f(x) = 2$ exists and is well-defined.

For option A: The limits at a and b are not equal (limit at b doesn't exist).
For option C: The limit at b doesn't exist because left and right limits differ.
For option D: The limit at a is 2, not 1.
For option E: The limit at a does exist (it's 2).

Therefore, only statement B is correct.

Question 5

The position of a particle moving to the right on the x-axis is given by $x(t)$, where $x(t)$ is measured in centimeters and $t$ is measured in seconds for $0 \leq t \leq 50$. If $y = x(t)$ is a linear function, which of the following would most likely give the best estimate of the speed of the particle, in centimeters per second, at time $t = 10$ seconds?

Accepted Answer

The slope of the graph of $y = x(t)$ — - For a linear function $x(t)$, the speed is constant and equal to its slope.
- Option D gives the exact value of the speed (slope) at any time $t$ in $[0,50]$.

Why other options are incorrect:
- A) $x(10)$ gives position, not speed
- B) $\frac{x(10)}{10}$ gives average speed from $0$ to $10$, not instantaneous speed
- C) $x(11)-x(9)$ gives average speed over $[9,11]$, which equals the constant speed but is less direct than D

Therefore, option D is the most appropriate choice.

Question 6

![Figure](https://getlykke-assets.s3.amazonaws.com/wiki-quiz-figures/c14bedc7-55ee-4bd3-99ff-a9a537a06a66/ap-exam-style/082261a954eb-F1.png)

A model rocket leaves the ground at time $t=0$ and travels straight up from the ground. The height, in feet, of the rocket above the ground is given by $y(t)$, where $t$ is measured in seconds for $0 \leq t \leq 120$. Values of $y(t)$ for selected values of $t$ are given in the table above. Of the following values of $t$, at which value would the velocity of the rocket most likely be greatest based on the data in the table?

Accepted Answer

$t=80$ — – Velocity can be estimated by the change in height ($\Delta y$) over change in time ($\Delta t$)
– From the table data, the greatest height difference occurs between $t=60$ and $t=80$ for $t=20$ till $t=80$
– Therefore, the rocket’s velocity is highest around $t=80$

Analysis of the options:
• A) $t=20$ – Early stage, velocity still increasing
• B) $t=40$ – Velocity greater than at $t=20$ but not maximum
• C) $t=60$ – Velocity increasing but not yet peak
• D) $t=80$ – Shows the largest $\Delta y/\Delta t$, indicating maximum velocity

The data shows increasing velocity with time, making $t=80$ the most likely point of maximum velocity among the options.

Time interval (seconds)	Approximate rate (liters per second)	Signed accumulation (liters)
$0$ to $60$	$4$	$4(60)=240$
$60$ to $90$	$5$	$5(30)=150$
$90$ to $120$	$3$	$3(30)=90$
$120$ to $135$	$-1$	$-1(15)=-15$
$135$ to $150$	$-2$	$-2(15)=-30$

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6.1 Exploring Accumulations of Change

6.1 Exploring Accumulations of Change

From rate to accumulation

Net change versus total change

Skill focus and retrieval check

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