Straight-line motion: position, velocity, acceleration

AP Calculus AB· AP Calculus AB CED — Contextual Applications of Differentiation· 14 min read

1. Position and Instantaneous Velocity★★☆☆☆⏱ 4 min

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Position of a particle at time $t$ is given by the function $s (t)$ , which returns the coordinate of the particle on the straight axis. Time is always non-negative ( $t \geq 0$ ) in AP motion problems, and displacement between $t = a$ and $t = b$ is $s (b) - s (a)$ , the net change in position, distinct from total distance traveled.

📘 Definition

Instantaneous Velocity

$v (t)$

The instantaneous rate of change of position with respect to time, equal to the first derivative of the position function.

Example:

If $s (t)$ is position in cm, $v (t)$ has units of cm/s.

v(t) = s'(t) = \frac{ds}{dt}

The sign of velocity tells you direction of motion: $v (t) > 0$ means the particle moves in the positive direction (right/up), $v (t) < 0$ means it moves in the negative direction (left/down), and $v (t) = 0$ means the particle is instantaneously at rest. Common exam questions ask for velocity at a specific time or for all times when the particle is at rest.

📐 Worked Example

The position of a particle moving along the x-axis is given by $s (t) = t^{3} - 6 t^{2} + 9 t + 2$ for $t \geq 0$ , where $s$ is measured in centimeters and $t$ in seconds. (a) Find the velocity of the particle at $t = 2$ seconds. (b) At what times $t > 0$ is the particle at rest?

1
Velocity is the first derivative of position, so differentiate using the power rule:
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$v(t) = s'(t) = 3t^2 - 12t + 9$
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For part (a), substitute $t = 2$ into $v (t)$ :
4
$v(2) = 3(2)^2 - 12(2) + 9 = 12 - 24 + 9 = -3 \text{ cm/s}$
5
For part (b), the particle is at rest when $v (t) = 0$ , so set the derivative equal to zero:
6
$3t^2 - 12t + 9 = 0$
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Factor the quadratic and solve:
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$3(t-1)(t-3) = 0$
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Both solutions are positive, so the particle is at rest at $t = 1$ s and $t = 3$ s.

Exam tip:

Always include units in your final answer for FRQ motion problems. AP graders dock 1 point for missing or incorrect units on contextual questions, so add units to every final answer by default.

2. Acceleration and Speeding Up/Slowing Down★★☆☆☆⏱ 3 min

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Acceleration is the instantaneous rate of change of velocity with respect to time, making it the second derivative of the original position function.

📘 Definition

Instantaneous Acceleration

$a (t)$

The derivative of velocity, or the second derivative of position, that describes how velocity is changing over time.

a(t) = v'(t) = \frac{dv}{dt} = s''(t) = \frac{d^2s}{dt^2}

A key conceptual skill tested heavily on the AP exam is determining if a particle is speeding up or slowing down. The rule is: if $v (t)$ and $a (t)$ have the same sign (both positive or both negative), the particle is speeding up; if they have opposite signs, the particle is slowing down.

📐 Worked Example

For the particle with position $s (t) = t^{3} - 6 t^{2} + 9 t + 2$ ( $t \geq 0$ , units cm, s), find the acceleration at $t = 1.5$ seconds, and determine if the particle is speeding up or slowing down at that time.

1
We already found velocity from the previous example: calculate $v (1.5)$ :
2
$v(1.5) = 3(2.25) - 12(1.5) + 9 = -2.25 \text{ cm/s}$
3
Velocity is negative at $t = 1.5$ . Acceleration is the derivative of velocity, so differentiate $v (t)$ :
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$a(t) = 6t - 12 \text{ cm/s}^2$
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Substitute $t = 1.5$ to find acceleration:
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$a(1.5) = 6(1.5) - 12 = -3 \text{ cm/s}^2$
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Acceleration is also negative. Since $v (t)$ and $a (t)$ have the same sign (both negative), the particle is speeding up at $t = 1.5$ seconds.

Exam tip:

Never assume that negative acceleration means the particle is slowing down. The sign of acceleration alone only describes how velocity is changing; you always need to compare the sign of acceleration to the sign of velocity to check if speed is increasing or decreasing.

3. Speed and its Rate of Change★★★☆☆⏱ 4 min

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Velocity and speed are often confused, but they are distinct quantities. Velocity is a one-dimensional vector that includes direction encoded in its sign. Speed is a non-negative scalar that has no direction, equal to the absolute value of velocity.

📘 Definition

Speed

$Speed (t)$

Non-negative magnitude of velocity, which describes how fast the particle is moving regardless of direction.

Example:

Speed is always positive or zero, never negative.

\text{Speed}(t) = |v(t)|

AP Calculus AB also tests the rate of change of speed, which can be derived using the chain rule for absolute value, for $v (t) \neq = 0$ :

\frac{d}{dt}|v(t)| = \frac{v(t) \cdot v'(t)}{|v(t)|} = \frac{v(t)a(t)}{|v(t)|}

The sign of this derivative confirms the earlier speeding up/slowing down rule: when $v$ and $a$ have the same sign, the derivative of speed is positive (speed increasing), and when opposite, it is negative (speed decreasing).

📐 Worked Example

A particle moving along the x-axis has velocity $v (t) = 4 cos (2 t)$ at time $t \geq 0$ , where velocity is measured in m/s. (a) What is the speed of the particle at $t = \frac{π}{3}$ ? (b) What is the rate of change of speed at that time?

1
For part (a), first calculate velocity at the given time:
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$v\left(\frac{\pi}{3}\right) = 4\cos\left(\frac{2\pi}{3}\right) = 4\left(-\frac{1}{2}\right) = -2 \text{ m/s}$
3
Speed is the absolute value of velocity, so:
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$\left|v\left(\frac{\pi}{3}\right)\right| = |-2| = 2 \text{ m/s}$
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For part (b), first find acceleration by differentiating velocity:
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$a(t) = v'(t) = -8\sin(2t) \implies a\left(\frac{\pi}{3}\right) = -4\sqrt{3} \text{ m/s}^2$
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Substitute into the derivative of speed formula:
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$\frac{d}{dt}|v(t)| = \frac{v(t) a(t)}{|v(t)|} = \frac{(-2)(-4\sqrt{3})}{|-2|} = 4\sqrt{3} \text{ m/s}^2$

Exam tip:

Circle the word "speed" every time it appears in a question, to remind yourself to take the absolute value of velocity. It is one of the most common avoidable mistakes on AP motion questions.

4. AP-Style Practice Worked Examples★★★☆☆⏱ 3 min

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📐 Worked Example

A particle moves along the x-axis with position function $s (t) = e^{t} - 3 t$ for $t \geq 0$ . What is the speed of the particle at $t = 1$ ? A) $e - 3$ B) $3 - e$ C) $e$ D) $3$

1
Speed is the absolute value of velocity, and velocity is the first derivative of position. Differentiate $s(t):
2
$v(t) = s'(t) = e^t - 3$
3
Substitute $t = 1$ to get $v (1) = e - 3 \approx - 0.282$ , which is negative.
4
Take the absolute value to get speed: $∣ e - 3∣ = 3 - e$ , since $e < 3$ . The correct answer is B.

📐 Worked Example

A marble moves along a straight horizontal track with position given by $s (t) = t^{4} - 10 t^{3} + 24 t^{2} + 5$ for $t \geq 0$ , where $s$ is in meters and $t$ in seconds. (a) Find the velocity function $v (t)$ , and calculate $v (2)$ . (b) Find all non-negative values of $t$ for which the particle is at rest. (c) Determine if the particle is speeding up or slowing down at $t = 1.5$ . Justify your answer.

1
Part (a): Velocity is the first derivative of position:
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$v(t) = s'(t) = 4t^3 - 30t^2 + 48t$
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Substitute $t = 2$ :
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$v(2) = 4(8) - 30(4) + 48(2) = 8 \text{ m/s}$
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Part (b): The particle is at rest when $v (t) = 0$ . Factor $v (t)$ :
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$2t(2t^2 - 15t + 24) = 0$
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Solve for roots using the quadratic formula:
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$t = 0, \frac{15 \pm \sqrt{33}}{4}$
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All solutions are non-negative, so the particle is at rest at $t = 0$ , $t \approx 2.31$ s, and $t \approx 5.19$ s.
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Part (c): Calculate acceleration and values at $t = 1.5$ :
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$a(t) = 12t^2 - 60t + 48, \quad v(1.5)=18 \text{ (positive)}, \quad a(1.5)=-15 \text{ (negative)}$
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Since $v (t)$ and $a (t)$ have opposite signs, the particle is slowing down at $t = 1.5$ .

5. Common Pitfalls

Wrong move:

Concluding that a particle changes direction at $t = a$ just because $v (a) = 0$ .

Why:

Students memorize "particle changes direction when velocity is zero" but forget that velocity must change sign at that point; a particle can stop momentarily without reversing direction.

Correct move:

Always test the sign of $v (t)$ on intervals on either side of any $v (t) = 0$ solution before concluding direction changes.

Wrong move:

Stating that a particle is slowing down because acceleration is negative.

Why:

Students associate "negative acceleration" with "deceleration" and assume it means slowing down regardless of velocity's sign.

Correct move:

Always write down the sign of both $v (t)$ and $a (t)$ for the given time, then check for same (speeding up) or opposite (slowing down) signs.

Wrong move:

Writing velocity as the answer when asked for acceleration.

Why:

When working quickly, students forget that acceleration is the derivative of velocity, so they stop at the first derivative of position.

Correct move:

Write $s (t) \to v (t) = s^{'} \to a (t) = v^{'} = s^{''}$ at the top of every motion problem to label derivatives clearly.

Wrong move:

Keeping negative time solutions when solving for when the particle is at rest.

Why:

Students solve the equation $v (t) = 0$ but forget the context that $t \geq 0$ for all motion problems.

Correct move:

Cross out any negative solutions for $t$ immediately after solving, since time starts at $t = 0$ .

Wrong move:

Forgetting to take the absolute value of velocity when calculating speed.

Why:

Students are used to working with velocity, so they just copy the velocity value (including the negative sign) as the answer for speed.

Correct move:

Add a step to take absolute value immediately after calculating velocity whenever speed is requested.

6. Quick Reference Cheatsheet

Category	Formula	Notes
Position	$s (t)$	Gives coordinate of particle at time $t \geq 0$ ; sign gives position relative to origin.
Instantaneous Velocity	$v (t) = s^{'} (t) = \frac{d s}{d t}$	Sign gives direction: $v > 0$ = positive direction, $v < 0$ = negative direction, $v = 0$ = at rest.
Instantaneous Acceleration	$a (t) = v^{'} (t) = s^{''} (t) = \frac{d ^{2} s}{d t ^{2}}$	Sign tells if velocity is increasing ( $a > 0$ ) or decreasing ( $a < 0$ ).
Speed	$Speed (t) = ∣ v (t) ∣$	Non-negative scalar with no direction; always take absolute value of velocity.
Rate of Change of Speed	$\frac{d}{d t} ∣ v (t) ∣ = \frac{v ( t ) a ( t )}{∣ v ( t ) ∣}$	Valid for $v (t) \neq = 0$ ; positive = speed increasing, negative = speed decreasing.
Speeding Up / Slowing Down	Same sign of $v, a$ = speeding up; opposite = slowing down	Never use acceleration sign alone to make this determination.
Change of Direction	Occurs when $v (t)$ changes sign at $t = a$	$v (a) = 0$ is required but not sufficient; must confirm sign change.

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2023 · AB
FRQ part on speeding up/slowing down
2022 · AB
MCQ speed calculation

Going deeper

official syllabusAP Calculus AB Course and Exam DescriptionOfficial College Board curriculum document

What's Next

This topic is the foundation for all motion problems you will encounter for the rest of the AP Calculus AB course. Immediately next, you will learn how to use integration to find position from velocity and calculate total distance traveled from velocity, a common FRQ topic that relies entirely on your ability to connect derivatives of position, velocity, and acceleration. Without mastering the relationships between these three functions from this subtopic, solving integral motion problems will be extremely difficult, as you will not be able to correctly interpret the quantities you calculate. This topic also builds core conceptual understanding of derivatives as rates of change, which is central to all contextual applications of differentiation, including related rates.