Unit 4 · Topic 4.2 · about 30 minutes · 7 parts · 13 scored checks, 3 of them multiple choice

AP Calculus AB 4.2: Straight-Line Motion: Connecting Position, Velocity, and Acceleration

Q: Is velocity the same as speed?

No. Velocity has a sign that gives the direction; speed is its absolute value and is never negative. A velocity of -4 meters per second is a speed of 4 meters per second in the negative direction.

Q: Does a negative acceleration mean the particle is slowing down?

Only when the velocity is positive. If the velocity is negative too, the velocity moves further from 0 and the particle speeds up. Compare the signs of v(t) and a(t).

Q: How can a particle be at rest and still accelerating?

At rest means v(t) = 0 at that instant, but the velocity can still be changing. A ball at the top of a toss has velocity 0 for an instant and is accelerating downward the whole time.

Q: How do I tell whether a particle is moving toward the origin?

Compare the signs of x(t) and v(t). If they are opposite, the particle is moving toward the origin; if they are the same, it is moving away.

Q: Should I differentiate by hand or use the calculator's numerical derivative?

Either works on a calculator question. Write the derivative you are evaluating, such as a(3) = v'(3), then give the value to three decimal places. Without a calculator, differentiate by the rules.

Built by Tanner Crow from College Board's course framework · Every answer key independently checked

Unit 4: 10 to 15% of the multiple-choice section

What you'll be able to do

Use derivatives to find the velocity, speed and acceleration of a particle moving along a line, and use their signs to describe its direction and whether it is speeding up or slowing down.

Key vocabulary

Term	What it means	The exam trap
Rectilinear motion	Motion along a straight line, described by a single position function of time.	Read which direction the problem calls positive before you interpret the sign of any answer.
Position, $x (t)$	The particle's signed distance from the origin at time $t$ .	Position is where the particle is, not how far it has traveled; a particle can come back to the same position.
Velocity, $v (t)$	The rate of change of position, $v (t) = x^{'} (t)$ , whose sign gives the direction of motion.	Velocity and speed have the same units, so the units alone will not tell you which one a question wants.
Speed	How fast the particle moves regardless of direction, $\| v (t) \|$ .	A question can ask for the speed at an instant or for whether the speed is increasing; those are different questions with different work.
Acceleration, $a (t)$	The rate of change of velocity, $a (t) = v^{'} (t) = x^{''} (t)$ .	Its units carry time twice, such as meters per second per second; write both.

Two graphs side by side for t from 0 to 4 seconds. Left: the position x(t) = t cubed - 6t squared + 9t, which rises from (0, 0) to (1, 4), falls to (3, 0) and rises to (4, 4). Right: the velocity v(t) = 3t squared - 12t + 9, which starts at 9, is 0 at t = 1 and t = 3, is negative between them with v(2) = -3, and ends at 9. Dashed lines at t = 1 and t = 3 join the graphs.

Part 1 Think first The top of a toss

Guess before you learn. A wrong guess still makes the answer stick.

Predict first

A ball is tossed straight up. Its height after $t$ seconds is $h (t) = 1 + 12 t - 5 t^{2}$ meters, and it reaches its highest point, 8.2 meters, at $t = 1.2$ seconds. At that instant, at the very top, what is the ball's acceleration?

In short. For a particle moving along a line, velocity is the derivative of position and acceleration is the derivative of velocity, and the signs of the two describe how the particle moves.

$v (t) = x^{'} (t)$ and $a (t) = v^{'} (t) = x^{''} (t)$ , in units such as meters, meters per second and meters per second per second.
Speed is $| v (t) |$ . The sign of $v (t)$ gives the direction, and the particle changes direction only where $v (t)$ changes sign.
The speed is increasing when $v (t)$ and $a (t)$ have the same sign, and decreasing when their signs are opposite.

Part 2 Learn it Three short ideas, each checked as you go

2a. Velocity and acceleration are derivatives

A particle moving along a straight line is in rectilinear motion. Its position $x (t)$ is where it is at time $t$ : its signed distance from a fixed point, the origin. The problem says which way is positive, usually right on a horizontal line and up on a vertical one.

The velocity is the rate of change of position, and the acceleration is the rate of change of velocity:

v (t) = x^{'} (t), a (t) = v^{'} (t) = x^{''} (t) .

So from a position function you get the other two by differentiating once, then again.

Units come from the rates. With $x$ in meters and $t$ in seconds, velocity is in meters per second and acceleration in meters per second per second.

The graphs at the top show $x (t) = t^{3} - 6 t^{2} + 9 t$ . Differentiating, $v (t) = 3 t^{2} - 12 t + 9$ and $a (t) = 6 t - 12$ . At $t = 4$ the particle is at $x (4) = 4$ meters, moving at $v (4) = 9$ meters per second, with acceleration $a (4) = 12$ meters per second per second. Three numbers about one instant, and each answers a different question.

Quick check · Calculate

No calculator. A cart moves along a straight track. Its position at time $t$ seconds is $s (t) = 2 t^{3} - 9 t^{2} + 12 t + 1$ feet. What is the velocity of the cart, in feet per second, at $t = 3$ ?

Your answer

2b. Direction, rest and speed

The sign of the velocity tells you which way the particle is going. If $v (t) > 0$ it moves in the positive direction, and if $v (t) < 0$ it moves in the negative direction.

It moves toward the origin when $x (t)$ and $v (t)$ have opposite signs, and away from it when they have the same sign.

When $v (t) = 0$ the particle is at rest for that instant. The ball in Think first was at rest at the top, and its acceleration was still $- 10$ . A particle changes direction only where $v (t)$ changes sign. A velocity can touch 0 and stay on the same side; then the particle pauses and carries on the way it was going.

If a continuous velocity has opposite signs at two times, the Intermediate Value Theorem gives a time between them when $v (t) = 0$ , even when no table row shows it.

Speed is how fast, without the direction: speed is $| v (t) |$ , so it is never negative. Velocities of $- 3$ and $3$ meters per second are the same speed in opposite directions.

On the graphs at the top, $v (t) = 3 (t - 1) (t - 3)$ . It is positive before $t = 1$ , negative between 1 and 3, and positive after 3, so the particle changes direction at both times, and the position graph turns at both. At $t = 2$ , $v (2) = - 3$ : the particle is moving in the negative direction at a speed of 3 meters per second.

Quick check · Multiple choice

No calculator. A particle moves along a line with velocity $v (t) = (t - 1) (t - 3)^{2}$ for $0 < t < 5$ . At which times does the particle change direction?

2c. Speeding up or slowing down

A positive acceleration does not mean the particle is speeding up. Acceleration is the rate of change of velocity, and whether the speed $| v (t) |$ grows depends on the velocity and the acceleration together.

If $v (t)$ and $a (t)$ have the same sign, the speed is increasing: the particle is speeding up.
If they have opposite signs, the speed is decreasing: the particle is slowing down.

Here is why. With $v (t) < 0$ and $a (t) < 0$ , the velocity is negative and getting more negative, so it moves away from 0 and its absolute value grows. With $v (t) < 0$ and $a (t) > 0$ , the velocity is negative and rising toward 0, so its absolute value shrinks.

Take the particle at the top again, with $v (t) = 3 (t - 1) (t - 3)$ and $a (t) = 6 t - 12$ . At $t = 1.5$ , $v = - 2.25$ and $a = - 3$ : the same sign, so it is speeding up. At $t = 2.5$ , $v = - 2.25$ again but $a = 3$ : opposite signs, so it is slowing down. The velocity is the same at both times, and the sign of the acceleration alone would have called the second one speeding up.

Quick check · Sort it

No calculator. Each card gives a particle's velocity $v$ , in meters per second, and acceleration $a$ , in meters per second per second, at one instant. Sort each card by what the particle's speed is doing at that instant.

Speeding up

Slowing down

Common mistakes

The mistake	Why it is wrong	Do this instead
Saying the particle changes direction at every time where $v (t) = 0$ .	$v (t) = 0$ means the particle is at rest for an instant. It turns around only if the velocity has opposite signs on the two sides of that time.	Check the sign of $v (t)$ just before and just after each zero, and say the particle changes direction only where the sign changes.
Saying a particle is speeding up because its acceleration is positive, or slowing down because it is negative.	The acceleration is the rate of change of velocity, not of speed. A negative velocity with a negative acceleration has a growing speed.	Compare the signs of $v (t)$ and $a (t)$ at the time asked: the same sign means speeding up, opposite signs mean slowing down.
Giving a negative speed, or calling a larger velocity a larger speed.	Speed is $\| v (t) \|$ , so it is never negative, and $v = - 8$ is faster than $v = - 5$ although $- 8 < - 5$ .	Take the absolute value of the velocity for speed, and give the direction separately, from the sign of $v (t)$ .

Part 3 Watch it done A particle that turns around twice

Watch each step, then do the last one yourself.

A particle moves along the $x$ -axis. Its position at time $t$ seconds is $x (t) = t^{3} - 9 t^{2} + 24 t$ meters, for $0 \leq t \leq 6$ . Find its velocity and acceleration, the times it is at rest, the intervals on which it moves left, and its speed at $t = 2.5$ .

Step 1 · Differentiate twice
$v (t) = x^{'} (t) = 3 t^{2} - 18 t + 24$ meters per second, and $a (t) = v^{'} (t) = 6 t - 18$ meters per second per second.
Step 2 · Find when it is at rest
$v (t) = 3 (t^{2} - 6 t + 8) = 3 (t - 2) (t - 4)$ , so $v (t) = 0$ at $t = 2$ and $t = 4$ .
Step 3 · Test the sign of the velocity on each interval
$v (1) = 9 > 0$ , $v (3) = - 3 < 0$ and $v (5) = 9 > 0$ . So $v (t) > 0$ on $0 < t < 2$ and $4 < t < 6$ , and $v (t) < 0$ on $2 < t < 4$ .
Step 4 · Read the direction
The particle moves left on $2 < t < 4$ and right on $0 < t < 2$ and $4 < t < 6$ . Its velocity changes sign at $t = 2$ and at $t = 4$ , so it changes direction at both times, at $x (2) = 20$ and $x (4) = 16$ meters.
Why this step?
Why test a value in each interval instead of stopping once you know where $v (t) = 0$ ?
A zero of $v$ only says the particle is at rest for an instant. It changes direction only if the velocity has opposite signs on the two sides, and that has to be checked: $v (t) = (t - 2)^{2}$ is 0 at $t = 2$ and positive on both sides, so a particle with that velocity never turns around.
Your step · Calculate

No calculator. Finish it. For the particle with position $x (t) = t^{3} - 9 t^{2} + 24 t$ meters and velocity $v (t) = 3 t^{2} - 18 t + 24$ meters per second, what is its speed, in meters per second, at $t = 2.5$ ?

Your answer

Part 4 Try it A bead on a wire and an elevator

Try it · Calculate

Graphing calculator required. A bead slides along a straight wire. Its velocity at time $t$ seconds is $v (t) = e^{\sin t} - 1.5$ centimeters per second. Find the acceleration of the bead, in centimeters per second per second, at $t = 2$ . Use your calculator in radian mode, and give your answer to three decimal places.

Your answer

Try it · Decide and justify

No calculator. An elevator moves along a straight vertical shaft. Its velocity $v (t)$ is in feet per second, with up positive. At $t = 1$ second, $v (1) = - 8$ , and at $t = 2$ seconds, $v (2) = - 5$ . A rider says the elevator was moving faster at $t = 2$ than at $t = 1$ , because its velocity was greater. Is the rider right?

Which reason supports your answer?

Part 5 Explain it Write it the way the exam asks

On the exam free response is handwritten, so work on paper first. Nothing you type here leaves this page. Then compare with the model answer and score each point.

Free response is 50% of the exam score. This one is shaped like Question 2, from free-response Part A (graphing calculator required).

Free-response practice · Question 2 style

Graphing calculator required. A particle moves along the $x$ -axis. Its position at time $t$ seconds is $x (t) = 3 \sin t + 0.4 t^{2}$ meters, for $0 \leq t \leq 5$ . Use your calculator in radian mode, and give decimal answers to three decimal places.

(a) Find the velocity of the particle at time $t = 3$ . At $t = 3$ , is the particle moving toward the origin or away from it? Give a reason for your answer.

(b) Find the acceleration of the particle at time $t = 3$ . Is the speed of the particle increasing or decreasing at $t = 3$ ? Give a reason for your answer.

Your answer stays on this page. It is never sent or saved.

Model answer

(a) $v (t) = x^{'} (t) = 3 \cos t + 0.8 t$ , so $v (3) = 3 \cos 3 + 2.4 = - 0.570$ meters per second. Also $x (3) = 3 \sin 3 + 3.6 = 4.023$ meters. The particle is moving toward the origin, because $x (3) > 0$ and $v (3) < 0$ : it is to the right of the origin and moving left.

(b) $a (3) = v^{'} (3) = 0.377$ meters per second per second (the numerical derivative of $v$ at $t = 3$ , or $- 3 \sin 3 + 0.8$ ). The speed is decreasing at $t = 3$ , because $v (3) < 0$ and $a (3) > 0$ have opposite signs.

(c) Solving $v (t) = 3 \cos t + 0.8 t = 0$ on the calculator gives $t = 2.197$ and $t = 3.505$ . $v (t) > 0$ for $0 < t < 2.197$ (for example $v (1) = 2.421$ ), $v (t) < 0$ for $2.197 < t < 3.505$ (for example $v (3) = - 0.570$ ), and $v (t) > 0$ for $3.505 < t < 5$ (for example $v (4) = 1.239$ ). The velocity changes sign at both times, so the particle changes direction at $t = 2.197$ and at $t = 3.505$ .

Score your own answer. This is a self-check, not a grade. Tick each point your answer makes.

Writes $v (3) = x^{'} (3) = 3 \cos 3 + 0.8 (3)$ and gets $- 0.570$ (or $- 0.569$ ) meters per second.
Says the particle is moving toward the origin, because $x (3) = 4.023 > 0$ and $v (3) < 0$ .
Gets $a (3) = v^{'} (3) = 0.377$ (or 0.376) meters per second per second, and says the speed is decreasing because $v (3)$ and $a (3)$ have opposite signs.
Sets $v (t) = 0$ and finds $t = 2.197$ (or 2.196) and $t = 3.505$ .
Justifies both answers in (c) by the velocity changing sign at each time, not by $v (t) = 0$ alone.

Part 6 Check yourself A cyclist on a straight path

The three case questions use the same table. Answer them from it.

A cyclist rides along a straight east-west path. Her velocity at time $t$ seconds is $v (t)$ meters per second, where positive values mean she is riding east. The velocity is a differentiable function of $t$ . Some values are in the table.

Velocity of the cyclist
$t$ (seconds)	$v (t)$ (meters per second)
0	2
2	5
5	4
8	$- 2$
10	$- 3$

Case question 1 · Calculate

No calculator. Use the data in the table to estimate the cyclist's acceleration at $t = 6.5$ seconds, in meters per second per second.

Your answer

Case question 2 · Multiple choice

No calculator. The cyclist's acceleration at $t = 8$ is $a (8) = - 0.4$ meters per second per second. At $t = 8$ , is her speed increasing or decreasing?

Case question 3 · Decide and justify

No calculator. Must there be a time between $t = 5$ and $t = 8$ seconds when the cyclist is at rest?

Which reason supports your answer?

The rest stand on their own and do not use the case.

Check yourself · Spot the error

No calculator. A particle's position is $x (t) = t^{4} - 4 t^{3} + 2 t$ meters at time $t$ seconds. A student finds its acceleration at $t = 3$ . Tap the first line with a mistake.

Check yourself · Match

No calculator. A particle's position is $x (t) = 2 t^{3} - 9 t^{2}$ meters at time $t$ seconds. Each value is in meters, meters per second or meters per second per second. Tap a quantity at $t = 1$ , then its value.

Part 7 Keep it Bring back earlier ideas

Motion problems stand on the second derivative from Unit 3 and on what a derivative needs from Unit 2. Pulling them back up is what makes them last.

Bring it back · Calculate

No calculator. Let $f (x) = x^{4} - 2 x^{3} + 5 x$ . Find $f^{''} (2)$ .

Your answer

Bring it back · Multiple choice

No calculator. Let $g (x) = | x - 3 | + 1$ . Which statement about $g$ at $x = 3$ is true?

In three lines

Velocity is the derivative of position, and acceleration is the derivative of velocity.
Speed is the absolute value of velocity, and the particle changes direction only where velocity changes sign.
Speed increases when velocity and acceleration share a sign, and decreases when their signs are opposite.

Exam strategy for this topic

Section I Part A, Multiple Choice (calculator not permitted): 29 questions in 62 minutes.
Section I Part B, Multiple Choice (graphing calculator required): 13 questions in 38 minutes.
Section II Part A, Free Response (graphing calculator required): 2 questions in 30 minutes.
Section II Part B, Free Response (calculator not permitted): 4 questions in 60 minutes.

On a motion question, write $v (t) = x^{'} (t)$ or $a (t) = v^{'} (t)$ before any number, and back every claim about direction or speed with a sign: the sign of $v$ at that time, or the signs of both $v$ and $a$ . When you use the calculator, write the equation you solve, such as $v (t) = 0$ , or the derivative you evaluate, then the result to three decimal places with its units.

Questions students ask

Is velocity the same as speed?

No. Velocity has a sign that gives the direction; speed is its absolute value and is never negative. A velocity of $- 4$ meters per second is a speed of 4 meters per second in the negative direction.

Does a negative acceleration mean the particle is slowing down?

Only when the velocity is positive. If the velocity is negative too, the velocity moves further from 0 and the particle speeds up. Compare the signs of $v (t)$ and $a (t)$ .

How can a particle be at rest and still accelerating?

At rest means $v (t) = 0$ at that instant, but the velocity can still be changing. A ball at the top of a toss has velocity 0 for an instant and is accelerating downward the whole time.

How do I tell whether a particle is moving toward the origin?

Compare the signs of $x (t)$ and $v (t)$ . If they are opposite, the particle is moving toward the origin; if they are the same, it is moving away.

Should I differentiate by hand or use the calculator's numerical derivative?

Either works on a calculator question. Write the derivative you are evaluating, such as $a (3) = v^{'} (3)$ , then give the value to three decimal places. Without a calculator, differentiate by the rules.

Course alignment, for teachers

AP Calculus AB topic 4.2, Unit 4: Contextual Applications of Differentiation.

Skill 1.E: Apply appropriate mathematical rules or procedures, with and without technology.

This lesson maps the one essential knowledge statement of the topic to the part that teaches each one: 1 is checked by a scored question. The map is in the lesson file.

What you'll be able to do

Key vocabulary

Part 1 Think first The top of a toss

Part 2 Learn it Three short ideas, each checked as you go

2a. Velocity and acceleration are derivatives

2b. Direction, rest and speed

2c. Speeding up or slowing down

Part 3 Watch it done A particle that turns around twice

Part 4 Try it A bead on a wire and an elevator

Part 5 Explain it Write it the way the exam asks

Part 6 Check yourself A cyclist on a straight path

Part 7 Keep it Bring back earlier ideas

Exam strategy for this topic

Questions students ask

Past exam questions on this topic