AP®MathsCollege BoardCalculus BCStudy GuidesUnit 4: Contextual Applications of DifferentiationRates of Change & Related RatesMotion in a Straight Line

Motion in a Straight Line (College Board AP® Calculus BC): Study Guide

Written by: Jamie Wood

Reviewed by: Dan Finlay

Updated on 6 May 2026

Introduction to straight line motion

What is straight line motion?

Straight line motion models how objects move in a straight line, with respect to time
- This may be described as motion 'along the $x$ -axis'
- The straight line will have a positive and a negative direction
  - On the $x$ -axis this will be the usual positive and negative directions
  - If the question doesn't specify, you can choose the positive and negative directions
  - Just be consistent once you've made a choice!

What terminology do I need to be aware of?

Particle
- A particle is the general term used for an object
- A particle is assumed to be the 'size' of a single point
  - So you don't need to worry about its 3D dimensions
Time $t$
- Time is usually measured in seconds ( $s$ )
- Displacement, velocity and acceleration are all functions of time $t$
- 'Initial' or 'Initially' means 'when $t = 0$ '
Displacement $s$
- $s$ is the usual notation for displacement
  - For motion along the $x$ -axis, $x$ may be used instead of $s$
- Displacement is usually measured in feet ( $ft$ ) or meters ( $m$ )
  - Sometimes questions may not state a particular unit, and will simply use "units"
- The displacement of a particle is its distance relative to a fixed point
  - The fixed point may be (but is not always) the particle’s initial position
- Displacement will be zero, $s = 0$ , when the object is at the fixed point
- Otherwise the displacement will be
  - positive if the particle is in the positive direction from the fixed point
  - or negative if it is in the negative direction from the fixed point

Distance $d$
- Distance is the magnitude of displacement
- Use of the word distance could refer to
  - the distance traveled by a particle
  - the (straight line) distance the particle is from a particular point
- Be careful not to confuse displacement with distance
  - Consider a bus starting and ending its journey at a bus depot,
  - its displacement will be zero when it returns to the depot
  - but the distance the bus has travelled will be the length of the route
- Distance is always positive (or zero)
Velocity $v$
- Velocity is usually measured in feet or meters per second
- The velocity of a particle is the rate of change of its displacement at time $t$
  - Velocity will be positive if the particle is moving in the positive direction
  - Or negative if it is moving in the negative direction
- If the particle is stationary, that means the velocity is zero, $v = 0$
  - '(Instantaneously) at rest' also means that $v = 0$
Speed
- Speed is the magnitude (i.e. absolute value or modulus) of the velocity
- For a particle moving in a straight line
  - speed is the 'velocity ignoring the direction'
  - if $v = 4$ , speed = $|4| = 4$
  - if $v = - 6$ , speed = $|- 6| = 6$
Acceleration $a$
- Acceleration is usually measured in feet or meters per second squared
  - That is the same as feet or meters per second per second
- The acceleration of a particle is the rate of change of its velocity at time $t$
- Acceleration can be positive or negative
  - but the sign alone cannot fully describe the particle’s motion
- If velocity and acceleration have the same sign
  - then the particle is accelerating (speeding up)
- if velocity and acceleration have different signs
  - then the particle is decelerating (slowing down)
- At times when the acceleration is zero, $a = 0$ ,
  - the particle is moving with constant velocity
- In all cases the direction of motion is determined by the sign (+ or -) of the velocity
- There is no special term for the magnitude of acceleration
  - "The magnitude of the acceleration" is simply used instead

Worked Example

A particle is moving along the $x$ -axis. At time $t$ seconds, the velocity of the particle is given by $v (t) = {(t^{2} - 5)}^{3}$ inches per second, and the acceleration of the particle is given by $a (t) = 6 t {(t^{2} - 5)}^{2}$ inches per second per second.

Is the speed of the particle increasing, decreasing, or neither at time $t = 2$ seconds? Give a reason for your answer.

Answer:

Find the values of the velocity and acceleration by substituting $t = 2$ into the expressions

$v (2) = {(2^{2} - 5)}^{3} = - 1$ inches per second

$a (2) = 6 (2) {(2^{2} - 5)}^{2} = 12$ inches per second per second

Consider the signs

$v (2) < 0 a (2) > 0$

The speed is decreasing at $t = 2$ seconds because the signs are different for the velocity and acceleration

Velocity & acceleration as derivatives

What is velocity as a derivative?

Velocity is the rate of change of displacement
- Differentiate displacement to get velocity
- $v = \frac{d s}{d t}$
Velocity is the slope of a displacement-time graph

What is acceleration as a derivative?

Acceleration is the rate of change of velocity
- Differentiate velocity to get acceleration
- $a = \frac{d v}{d t}$
Acceleration is the slope of a velocity-time graph
This means that acceleration is also the rate of change of the rate of change of displacement
- Differentiate displacement twice to get acceleration
- $a = \frac{d^{2} s}{d t^{2}}$

Worked Example

$t$ (seconds)	0	2	4	6	8
$v (t)$ (meters per second)	5	2	−3	−1	4

A particle moves along the $x$ -axis. The velocity of the particle at time $t$ is given by a differentiable function $v$ , where $v (t)$ is measured in meters per second and $t$ is measured in seconds. Selected values of $v (t)$ are given in the table above.

(a) Justify why there must be at least two times, for $0 < t < 8$ , when the particle is at rest.

(b) Approximate $a (5)$ , the acceleration of the particle at time $t = 5$ seconds. Show the computations that lead to your answer, and indicate units of measure.

Answer:

(a)

The particle is at rest when $v (t) = 0$

You need to use the intermediate value theorem

First, state why the function is continuous

$v (t)$ is differentiable ⇒ $v (t)$ is continuous

Then, find two pairs of values of $t$ for which the values of $v (t)$ have different signs

Note there are many possible pairs

$v (2) = 2 > 0$ and $v (4) = - 3 < 0$

$v (6) = - 1 < 0$ and $v (8) = 4 > 0$

Therefore, by the intermediate value theorem, there is at least one value $t_{1}$ for $2 < t_{1} < 4$ and at least one value $t_{2}$ for $6 < t < 8$ such that $v (t_{1}) = v (t_{2}) = 0$

Therefore, there are at least two times when the particle is at rest

(b)

Acceleration is the rate of change of the velocity

$a (5) = v' (5)$

Find the average rate of change using the two closest values to $t = 5$

$\begin{array}{rcl} v' (5) & \approx & \frac{v (6) - v (4)}{6 - 4} \\ = & \frac{- 1 - (- 3)}{2} \\ = & 1 \end{array}$

Include the units

The acceleration of the particle at time $t = 5$ seconds is approximately 1 meter per second per second

Worked Example

The displacement from the origin of a particle, $P$ , as it travels along the $x$ -axis is given by $s_{P} = 3 \sin (\frac{1}{2} t)$ .

The displacement from the origin of a second particle, $Q$ , as it travels along the $x$ -axis is given by $s_{Q} = \frac{1}{8} t^{3} - \frac{7}{4} t^{2} + 5 t$ .

$s_{P}$ and $s_{Q}$ are measured in meters and $t$ is measured in seconds for $0 \leq t \leq 10$ .

(a) Determine which particle is furthest from the origin at $t = 5$ .

(b) At $t = 5$ , determine if the particles are moving closer together or further apart. Explain your reasoning in your working.

Answer:

(a)

Find the displacement of each particle at $t = 5$

$s_{P} = 3 \sin (\frac{1}{2} \cdot 5) = 1.795416 . . .$ meters

$s_{Q} = \frac{1}{8} {(5)}^{3} - \frac{7}{4} {(5)}^{2} + 5 (5) = - 3.125$ meters

$P$ is 1.795... meters away from the origin in the positive direction, while $Q$ is 3.125 meters away from the origin in the negative direction

At $t = 5$ particle $Q$ is the furthest from the origin.

(b)

This question is about the direction of motion, so we need to find the velocity

Find expressions for the velocities by differentiating the expressions for the displacements

$v_{P} = \frac{d s_{P}}{d t} = \frac{3}{2} \cos (\frac{1}{2} t)$

$v_{Q} = \frac{d s_{Q}}{d t} = \frac{3}{8} t^{2} - \frac{7}{2} t + 5$

Find the velocities when $t = 5$

$v_{P} = \frac{3}{2} \cos (\frac{1}{2} \cdot 5) = - 1.201715 . . .$ meters per second

$v_{Q} = \frac{3}{8} {(5)}^{2} - \frac{7}{2} (5) + 5 = - 3.125$ meters per second

Consider the positions and velocities of the two particles

$P$ is on the positive side (right) of the $x$ axis (1.795... meters), moving with a negative velocity, so back towards the origin (moving to the left)

$Q$ is on the negative side (left) of the $x$ axis (-3.125... meters), moving with a negative velocity, so further away from the origin (moving to the left)

$\overset{3.125}{\leftarrow} \overset{1.2017}{\leftarrow} Q P$

So both particles are moving to the left (negative $x$ direction) but $Q$ is moving faster, and its position is further to the left, so $Q$ is "escaping" from $P$

Therefore at $t = 5$ , the particles are moving further apart

(c)

Find expressions for the accelerations by differentiating the expressions for the velocities

$a_{P} = \frac{d v_{P}}{d t} = - \frac{3}{4} \sin (\frac{1}{2} t)$

$a_{Q} = \frac{d v_{Q}}{d t} = \frac{3}{4} t - \frac{7}{2}$

Find the acceleration of each when $t = 5$

$a_{P} = - \frac{3}{4} \sin (\frac{1}{2} \cdot 5) = - 0.448854 . . .$ meters per second squared

$a_{Q} = \frac{3}{4} (5) - \frac{7}{2} = 0.25$ meters per second squared

Whilst the acceleration of $Q$ is the largest (as it is positive whereas $P$ 's is negative), it is $P$ whose acceleration has the greatest absolute value (magnitude)

At $t = 5$ , particle $P$ has the greatest magnitude of acceleration.

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Motion in a Straight Line (College Board AP® Calculus BC): Study Guide

Introduction to straight line motion

What is straight line motion?

What terminology do I need to be aware of?

Worked Example

Velocity & acceleration as derivatives

What is velocity as a derivative?

What is acceleration as a derivative?

Worked Example

Worked Example

Unlock more, it's free!

Join the 100,000+ Students that ❤️ Save My Exams

Author: Jamie Wood

Reviewer: Dan Finlay