AP Calculus AB Rates of Change in Context and Straight-Line Motion Worksheet Solutions (PDF)

AP Calculus AB math worksheet solutions: Rates of Change in Context and Straight-Line Motion – Maths PDF

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This rates motion answers for AP Calculus AB covers rates of change in context and straight-line motion (Common Core AP CALC AB 4.1, AP CALC AB 4.2, AP CALC AB 4.3), with a free PDF to print.

This answer key gives a full worked solution for every problem in the rates of change worksheet. Compare each step, not only the final answer, because AP graders award credit for a correct setup and clear justification. Final answers are in bold. Redo any problem you missed before moving on to the test.

AP Calculus AB · Common Core State Standards: AP CALC AB 4.1, AP CALC AB 4.2, AP CALC AB 4.3

Rates motion answers: exercises 1 to 10

Exercise 1 – Interpreting a cost derivative

The units of $C^{\prime}$ are dollars per flyer.

Final answer: When 500 flyers are printed, the cost is increasing at about \$0.12 per additional flyer.

Exercise 2 – Interpreting a temperature derivative

The units are °F per hour, and the sign is negative.

Final answer: At 3 hours the temperature is falling at 4.5 °F per hour.

Exercise 3 – Velocity and direction

$v(t)=2t-6$. At rest when $2t-6=0$: $t=3$.

For $0\le t < 3$, $v < 0$: moving left. For $t > 3$, $v > 0$: moving right.

Final answer: $v(t)=2t-6;\ \text{at rest at }t=3;\ \text{left on }[0,3),\ \text{right for }t > 3$.

Exercise 4 – Velocity and acceleration

$v(t)=3t^{2}-12t+5$, so $v(2)=12-24+5=-7$.

$a(t)=6t-12$, so $a(2)=0$.

Final answer: $v(2)=-7\text{ ft/s},\ a(2)=0\text{ ft/s}^{2}$.

The point t = 2 is an inflection point: a(2) = 0.

Exercise 5 – Speeding up or slowing down

$v=6(t-1)(t-2)$ and $a=12t-18$.

$v(0.5)=6(-0.5)(-1.5)=4.5 > 0$, $a(0.5)=-12 < 0$: opposite signs, slowing down.
$v(1.5)=6(0.5)(-0.5)=-1.5 < 0$, $a(1.5)=0$: the acceleration is zero, so the speed is momentarily constant.
$v(3)=12 > 0$, $a(3)=18 > 0$: same sign, speeding up.

Velocity (green) and acceleration (orange).

Exercise 6 – Reading a velocity graph

At $t=2$ and again at $t=6.8$, where the last segment crosses zero: $-2+2.5(t-6)=0$.
Velocity is negative from $t=2$ to $t=6.8$.
Speed $|v|$ is $4$ at $t=0$, the largest value.
Acceleration is the slope of $v$: $\frac{0-4}{2-0}=-2\text{ ft/s}^{2}$.

Exercise 7 – Sign chart for direction

$v=0$ at $t=2$ and $t=5$. Test values: $v(0)=10 > 0$, $v(3)=-2 < 0$, $v(6)=4 > 0$.

Final answer: $\text{right on }[0,2),\ \text{left on }(2,5),\ \text{right for }t > 5$.

Exercise 8 – Average versus instantaneous velocity

$s(1)=3$, $s(3)=15$, so the average velocity is $\frac{15-3}{2}=6$ m/s.

$v(t)=2t+2$, so $v(1)=4$ and $v(3)=8$ m/s.

Final answer: $6\text{ m/s average;}\ 4\text{ and }8\text{ m/s instantaneous}$.

Exercise 9 – Estimating velocity from a table

$\frac{s(3)-s(1)}{3-1}=\frac{39-5}{2}=17$.

Final answer: $v(2)\approx 17\text{ ft/s}$.

Exercise 10 – Area of a circle

$A^{\prime}(r)=2\pi r$, so $A^{\prime}(5)=10\pi\approx 31.4$.

Final answer: $10\pi\approx 31.4\text{ ft}^{2}\text{ per ft}$: at a radius of 5 feet, each extra foot of radius adds about 31.4 square feet of area.

Exercises 11 and up

Exercise 11 – Volume of a cube

$V^{\prime}(s)=3s^{2}$, so $V^{\prime}(4)=48$.

Final answer: $48\text{ in}^{3}\text{ per inch}$.

Exercise 12 – Population growth

$P^{\prime}(t)=20-t$.

$P^{\prime}(10)=10$: the population is growing by 10 people per year at year 10. $P^{\prime}(20)=0$: at year 20 growth stops (the population is at its maximum of 700).

Final answer: $10\text{ people/year and }0\text{ people/year}$.

Exercise 13 – Marginal revenue

$R^{\prime}(x)=120-0.8x$, so $R^{\prime}(50)=120-40=80$.

Final answer: $80\text{ dollars per skateboard}$.

At 50 skateboards, the 51st board adds about \$80 to the revenue.

Exercise 14 – Oscillating spring

$v=3\cos t$, so $v\left(\frac{\pi}{3}\right)=\frac{3}{2}$. $a=-3\sin t$, so $a\left(\frac{\pi}{3}\right)=-\frac{3\sqrt{3}}{2}$.

Final answer: $v=\frac{3}{2}\text{ cm/s},\ a=-\frac{3\sqrt{3}}{2}\text{ cm/s}^{2}$.

Exercise 15 – Bacteria growth

$N^{\prime}(t)=200\cdot 0.3e^{0.3t}=60e^{0.3t}$.

$N^{\prime}(5)=60e^{1.5}\approx 268.9$.

Final answer: $N^{\prime}(5)\approx 268.9\text{ bacteria/hour}$.

Exercise 16 – Total distance

$v=6t^{2}-30t+24=6(t-1)(t-4)$ changes sign at $t=1$ and $t=4$.

Positions: $s(0)=0$, $s(1)=11$, $s(4)=-16$, $s(5)=-5$.

Distances: $11+27+11=49$.

Final answer: $49\text{ feet}$.

Turning points at t = 1 and t = 4.

Exercise 17 – Draining tank

Chain rule: $V^{\prime}(t)=1000\left(1-\frac{t}{20}\right)\left(-\frac{1}{20}\right)=-50\left(1-\frac{t}{20}\right)$.

$V^{\prime}(10)=-50\cdot 0.5=-25$.

Final answer: $-25\text{ gallons per minute}$.

At 10 minutes the tank is losing 25 gallons per minute.

Exercise 18 – Pauses and speeding up

$v=4t^{3}-24t^{2}+36t=4t(t-3)^{2}$. At rest at $t=0$ and $t=3$. The factor $(t-3)^{2}$ never changes sign, so $v\ge 0$ for all $t\ge 0$: it never changes direction.

$a=12t^{2}-48t+36=12(t-1)(t-3)$. Speed increases where $v$ and $a$ have the same sign: $v > 0$ and $a > 0$ for $0 < t < 1$ and $t > 3$.

Acceleration signs for t > 0 (velocity is positive except at t = 3).

Final answer: $\text{at rest at }t=0,3;\ \text{no direction change;}\ \text{speeding up on }(0,1)\text{ and }(3,\infty)$.

Exercise 19 – Displacement versus distance

$v=3t^{2}-6t=3t(t-2)$, so the cart turns around at $t=2$. Positions: $s(0)=0$, $s(2)=-4$, $s(3)=0$.

Displacement: $s(3)-s(0)=0$. Distance: $|-4-0|+|0-(-4)|=8$.

Final answer: $\text{displacement }0\text{ m;}\ \text{distance }8\text{ m}$.

Exercise 20 – Model rocket

$v(t)=-32t+96$. At the top, $v=0$: $t=3$ and $s(3)=-144+288+80=224$ ft.

Ground: $s(t)=0$ gives $t^{2}-6t-5=0$, so $t=3+\sqrt{14}\approx 6.74$ s. Then $v\approx -32(6.74)+96\approx -119.7$.

Final answer: $\text{maximum height }224\text{ ft;}\ v\approx -119.7\text{ ft/s at impact}$.

Exercise 21 – Estimating a rate from a table

$S^{\prime}(2)\approx\frac{S(3)-S(1)}{3-1}=\frac{44-8}{2}=18$.

Final answer: $S^{\prime}(2)\approx 18$.

At 2 months, sales are growing at about 18 thousand dollars per month.

Key facts behind this rates motion answers

Key facts

A derivative has units of output per input, such as dollars per pencil or feet per second.
Velocity is s′(t), acceleration is v′(t), and speed is the absolute value of velocity.
Speed increases when velocity and acceleration have the same sign.
Total distance adds the distances between turning points, where velocity changes sign.

How to use this rates motion answers

Cover each solution, solve the exercise again, then uncover it and compare every line. If your answer differs, find the first step that changed and redo the exercise from there.

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