Motion with Calculus

Move between displacement, velocity and acceleration by differentiating and integrating, find when a particle is at rest, work out distance travelled, and sketch motion graphs.

In the exam for Edexcel Further Pure Maths 4PM1, IGCSE Additional Maths 0606 · 26 real past-paper questions below

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Displacement, velocity, acceleration

A phone's map knows where you are every second. From that it can work out how fast you go and how quickly you speed up. Calculus does the same for a particle moving along a straight line.

Displacement $s$ is the position measured from a fixed point O; it can be negative. Velocity $v$ is how fast $s$ changes. Acceleration $a$ is how fast $v$ changes.

Formula

v = \frac{d s}{d t}

$s$ — displacement from O, in metres (m).
$v$ — velocity, in m/s. Positive means moving in the positive direction; negative means moving back.
$t$ — time, in seconds (s).

Formula

a = \frac{d v}{d t}

$a$ — acceleration, in m/s². It is the gradient of velocity.

Worked example

1
A particle has displacement $s = t^{3} - 6 t^{2} + 9 t + 2$ . Find when it is at rest, and its acceleration at $t = 3$ .
2
$v = \frac{d s}{d t} = 3 t^{2} - 12 t + 9$ — Differentiate s to get v.
3
$3 (t - 1) (t - 3) = 0$ — At rest means $v = 0$ . Factorise.
4
$t = 1 and t = 3$ — It stops twice: once to turn back, then to turn forward again.
5
$a = \frac{d v}{d t} = 6 t - 12, a = 6 (3) - 12 = 6$ — Differentiate v, then put in $t = 3$ . The unit is m/s².

Rule

Learn these. No board prints them. Speed is the size of velocity, so speed is never negative.

Key idea

Galileo rolled balls down sloping ramps around 1600. He found the distance grows with the square of the time: twice the time, four times the distance. That is constant acceleration.

Key idea

Differentiate to go down: s → v → a.

Integrate to go back up

Often you are given the acceleration and must find the velocity, or given the velocity and must find the displacement. Integration undoes differentiation, so integrate to go back up.

Formula

v = \int a d t

$a$ — the acceleration, in terms of $t$ .
$v$ — the velocity. It comes with + c, found from a fact you are given.

Formula

s = \int v d t

$s$ — the displacement. A new + c, found from another given fact.

The question tells you the facts that fix each c. Learn what the words mean:

The question says	It means
“Initially” or “when it starts”	$t = 0$
“At rest” or “stops”	$v = 0$
“At O” or “returns to O”	$s = 0$
“Starts from rest at O”	$v = 0$ and $s = 0$ when $t = 0$

Worked example

1
A particle has acceleration $a = 2 t - 7$ . Initially its velocity is 10 m/s and it is at O. Find $v$ , the times it is at rest, and $s$ .
2
$v = t^{2} - 7 t + c$ — Integrate a. Do not forget + c.
3
$10 = 0 - 0 + c, c = 10$ — Initially means $t = 0$ , and then $v = 10$ .
4
$v = t^{2} - 7 t + 10 = (t - 2) (t - 5)$ — So it is at rest when $t = 2$ and $t = 5$ .
5
$s = \frac{t ^{3}}{3} - \frac{7 t ^{2}}{2} + 10 t + k$ — Integrate v. A new constant k.
6
$s = \frac{t ^{3}}{3} - \frac{7 t ^{2}}{2} + 10 t$ — It starts at O, so $s = 0$ when $t = 0$ . That gives $k = 0$ .

Marks usually go: one for integrating, one for putting the given fact into your expression with its + c, and one for the answer. So always show the line with c.

Common mistake

Assuming c = 0. It is only 0 when the given fact makes it 0. In this example c was 10.

Key idea

Integrate to go up: a → v → s. Each step brings a new constant, and the question gives a fact to find it.

Distance is not displacement

Displacement only cares where the particle ends up. Distance counts every metre it moves, forwards and back. If the particle turns round, the two are different.

A particle turns round when $v = 0$ and $v$ changes sign. So split the journey at those times.

Steps

1
1. Find the times when $v = 0$ inside the time interval.
2. Find $s$ at the start, at each of those times, and at the end.
3. Find the size of each change in $s$ , and add them.

Worked example

1
Use $s = \frac{t ^{3}}{3} - \frac{7 t ^{2}}{2} + 10 t$ from the last step. Find the distance travelled in the first 4 seconds.
2
$v = 0 at t = 2$ — Inside 0 to 4, the particle turns round at $t = 2$ only.
3
$s (0) = 0, s (2) = \frac{8}{3} - 14 + 20 = \frac{26}{3}$ — Where it is at the start and when it turns.
4
$s (4) = \frac{64}{3} - 56 + 40 = \frac{16}{3}$ — Where it is at the end.
5
$\frac{26}{3} + \frac{16}{3} - \frac{26}{3} = \frac{26}{3} + \frac{10}{3} = 12$ — Out by 26/3 m, then back by 10/3 m. Distance = 12 m.

Its displacement after 4 seconds is only $\frac{16}{3}$ m, because it came part of the way back.

Example

The distance between the two places where it is at rest is $∣ s (5) - s (2) ∣ = \frac{25}{6} - \frac{26}{3} = \frac{9}{2}$ m. The same answer comes from $\int_{2}^{5} v d t$ .

Common mistake

Integrating $v$ from 0 to 4 in one go. That gives displacement, $\frac{16}{3}$ , not distance.

Key idea

Distance: split where $v = 0$ , then add sizes. Displacement: just the end value minus the start value.

Motion with e

The same rules work when $s$ , $v$ or $a$ uses $e$ , sin or cos. Use the standard derivatives and integrals you already know.

Worked example

1
A particle starts from rest at O. Its velocity is $v = 8 - 8 e^{- \frac{t}{2}}$ . Find $a$ , the time when $v = 6$ , and $s$ .
2
$a = \frac{d v}{d t} = 4 e^{- \frac{t}{2}}$ — The 8 differentiates to 0. Then $- 8 \times (- \frac{1}{2}) = 4$ .
3
$8 - 8 e^{- \frac{t}{2}} = 6 \Rightarrow e^{- \frac{t}{2}} = \frac{1}{4}$ — Set $v = 6$ and rearrange.
4
$- \frac{t}{2} = ln \frac{1}{4} = - ln 4, t = 2 ln 4$ — Take ln of both sides. Leave it exact.
5
$s = 8 t + 16 e^{- \frac{t}{2}} + c$ — Integrate. $- 8$ divided by $- \frac{1}{2}$ is 16.
6
$0 = 0 + 16 + c, s = 8 t + 16 e^{- \frac{t}{2}} - 16$ — At O when $t = 0$ , and $e^{0} = 1$ , so $c = - 16$ .

Notice that $e^{- \frac{t}{2}}$ gets closer and closer to 0 as $t$ grows. So $v$ gets closer to 8 m/s but never reaches it. A question may ask you to explain this.

Common mistake

Thinking the constant is 0 because the particle starts at O. Here $e^{0} = 1$ , so the constant is −16.

Key idea

Starting conditions always mean $t = 0$ . Remember that $e^{0} = 1$ and $cos 0 = 1$ .

Motion graphs

Rule

This step is for 0606 and 4037.

You may need to sketch, or read, five graphs against time $t$ . Two links connect them:

The gradient of the s–t graph is the velocity. The gradient of the v–t graph is the acceleration.
The area under the v–t graph is the displacement. Area below the t-axis counts as negative.

Worked example

1
Describe the graphs for $v = t^{2} - 7 t + 10$ from the last steps, for $0 \leq t \leq 6$ .
2
$v = (t - 2) (t - 5)$ — v–t: a U-shaped curve. It starts at 10, crosses the t-axis at 2 and 5, and is lowest at $t = 3.5$ , where $v = - 2.25$ .
3
$∣ v ∣$ — Speed–time: the same curve, but the part between 2 and 5 is flipped above the axis. Speed is never negative.
4
$a = 2 t - 7$ — a–t: a straight line. It starts at −7 and crosses the t-axis at 3.5, where v is lowest.
5
$s = \frac{t ^{3}}{3} - \frac{7 t ^{2}}{2} + 10 t$ — s–t: rises to a peak of 26/3 at $t = 2$ , falls to 25/6 at $t = 5$ , then rises again.
6
$distance$ — Distance–time: like s–t, but it never goes down. From 2 to 5 it keeps rising instead of falling.

Example

Constant acceleration from rest: the v–t graph is a straight line through the origin. If $s = 20$ m at $t = 5$ , the triangle area gives $\frac{1}{2} \times 5 \times v = 20$ , so $v = 8$ m/s, and $a$ = gradient = $\frac{8}{5} = 1.6$ m/s².

Common mistake

Drawing speed below the t-axis. Speed is the size of velocity, so reflect the negative parts up.

Key idea

Where $v$ changes sign, the s–t graph has a peak or a dip. Where $a$ changes sign, the v–t graph has a peak or a dip.

Which method? Read the question

Read what you are given and what you are asked for. That tells you which way to go.

You see	You do
Given $s$ , find $v$ or $a$	Differentiate (once for v, twice for a)
Given $a$ , find $v$ ; given $v$ , find $s$	Integrate, + c, then use the given fact to find c
“At rest” or “changes direction”	Solve $v = 0$
“Maximum velocity”	Solve $a = 0$
“Distance travelled”	Split where $v = 0$ ; add the sizes of each change in $s$
“Displacement” at a time	Put the time into $s$
Sketch a graph (0606 / 4037)	Gradient and area links; speed and distance never go below the axis

Common mistake

Mixing up the letters. Velocity is not acceleration: at rest means $v = 0$ , not $a = 0$ .

Key idea

Differentiate to go down the chain s → v → a. Integrate to come back up, and use the facts given to find each + c.

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