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hscscience Maths Adv · Y12
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Module 6 · L14 of 15 ~35 min ⚡ +95 XP available

Motion Applications

Calculus was invented to describe motion. The connections between position, velocity, and acceleration, differentiate to go down the chain, integrate to go back up, are the most direct application of everything in this module. This is where calculus stops being abstract.

Today's hook, When a SpaceX Falcon 9 booster lands itself, its flight computer integrates acceleration to get velocity, integrates velocity to get position, and compares both to the target 60 times per second. The mathematics is exactly what you do on paper, just at millisecond precision.
0/5QUESTS
1

Get oriented

Set up the kinematics chain and key terms for motion.

Worksheets

Practise this lesson

Three printable worksheets that build from foundations to mastery, or build your own from any module’s questions.

Build Foundations & guided practice Apply Application practice Master Mastery challenge Build custom Build your own from any module question
01
Think First, your gut answer
+5 XP warm-up

A car accelerates from rest with $a(t) = 2t$ m/s². How would you find its velocity after 3 seconds? And how far has it travelled in those 3 seconds? Write your gut answer first.

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02
The kinematics chain
+5 XP to read

There are three quantities and only two operations. Memorise the direction of each and the rest is just executing the calculus.

Differentiation moves down the chain: position → velocity → acceleration. Integration moves up: acceleration → velocity → position. Each integration introduces one constant, determined by an initial condition.

Differentiate down the chain; integrate to go back up.
$$v = \frac{dx}{dt},\quad a = \frac{dv}{dt},\quad x = \int v\,dt,\quad v = \int a\,dt$$
Displacement vs distance
Displacement $= \int v\,dt$ (can be negative). Distance $= \int |v|\,dt$ (always positive). Split the integral where $v$ changes sign.
Initial conditions
Each integration introduces a constant $C$. Use initial conditions $v(0)$ and $x(0)$ to determine them. Two integrations → two constants.
Particle at rest
A particle is momentarily at rest when $v(t) = 0$. This is where you split the integral for distance calculations.
03
What you'll master
Know

Key facts

  • $v = \frac{dx}{dt}$, $a = \frac{dv}{dt}$
  • $x = \int v\,dt$, $v = \int a\,dt$
  • Distance $= \int |v(t)|\,dt$
Understand

Concepts

  • The relationship between position, velocity, and acceleration
  • Why displacement and distance travelled differ
  • How initial conditions determine constants
Can do

Skills

  • Find velocity from acceleration
  • Find displacement from velocity
  • Calculate total distance travelled
04
Key terms
Position $x(t)$Where the particle is at time $t$, measured from the origin.
Velocity $v(t)$Rate of change of position: $v = \frac{dx}{dt}$. Positive = moving right/up.
Acceleration $a(t)$Rate of change of velocity: $a = \frac{dv}{dt} = \frac{d^2x}{dt^2}$.
DisplacementNet change in position: $\Delta x = \int_{t_1}^{t_2} v\,dt$. Can be negative.
Distance travelledTotal path length: $\int_{t_1}^{t_2} |v(t)|\,dt$. Always non-negative.
At restWhen $v(t) = 0$, use these times to split distance integrals.
2

Link the three quantities

Move between position, velocity and acceleration by calculus.

05
Position, velocity, and acceleration
core concept

For a particle on a straight line, the three quantities form a chain connected by differentiation and integration. The key insight is that integration moves up the chain, and each step needs an initial condition to pin the constant.

Going up (integration):

$$v(t) = \int a(t)\,dt + C_1 \qquad x(t) = \int v(t)\,dt + C_2$$

Displacement vs distance: These are not the same when the particle reverses direction.

  • Displacement $= x(t_2) - x(t_1) = \int_{t_1}^{t_2} v(t)\,dt$, can be negative
  • Distance $= \int_{t_1}^{t_2} |v(t)|\,dt$, always non-negative

To calculate distance: find where $v(t) = 0$ in the interval, split the integral there, and use absolute values.

SpaceX Rocket Landing. When a Falcon 9 booster returns to Earth, its flight computer continuously calculates position, velocity, and acceleration. It integrates acceleration to get velocity, integrates velocity to get position, and compares against the target landing pad, 60 times per second. If velocity is too high, thrust increases (more upward acceleration). Australian aerospace company Gilmour Space is developing identical guidance systems. The calculus is exactly what you do on paper, executed at millisecond precision.

$v = \frac{dx}{dt}$ and $a = \frac{dv}{dt} = \frac{d^2x}{dt^2}$; To find $v$ from $a$: integrate $a(t)$, use $v(0)$ to find $C_1$

Pause, copy the kinematics chain $v = \dfrac{dx}{dt}$, $a = \dfrac{dv}{dt} = \dfrac{d^2x}{dt^2}$ and the integration direction (integrate $a$ with initial condition to find $v$; integrate $v$ to find $x$) into your book.

Did you get this? True or false: the total distance a particle travels equals the absolute value of its displacement.

3

Work through examples

Follow three problems, including total distance and initial conditions.

Worked examples · 3 applications

PROBLEM 1 · FIND POSITION FROM ACCELERATION

A particle has $a(t) = 6t - 4$ m/s². Initially at rest at the origin. Find its position at $t = 2$.

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$v(t) = \int (6t-4)\,dt = 3t^2 - 4t + C_1$. $v(0)=0 \implies C_1 = 0$
Integrate $a$ to get $v$; use $v(0) = 0$ (at rest).
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PROBLEM 2 · TOTAL DISTANCE TRAVELLED

A particle has velocity $v(t) = t^2 - 4t + 3$ m/s. Find the total distance from $t = 0$ to $t = 3$.

1
$v(t) = (t-1)(t-3)$. Zeroes at $t = 1$ and $t = 3$
Factor to find when particle reverses. On $[0,1]$: $v \geq 0$. On $[1,3]$: $v \leq 0$.
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PROBLEM 3 · WITH INITIAL CONDITIONS

A particle has $a(t) = 2$ m/s². At $t = 0$: $v = 5$ m/s, $x = 10$ m. Find $x(t)$.

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$v(t) = \int 2\,dt = 2t + C_1$. $v(0) = 5 \implies C_1 = 5$
Integrate $a = 2$; apply $v(0) = 5$.
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4

Avoid the common traps

Meet the three errors that cost marks, starting with distance versus displacement.

Quick check: A particle has $v(t) = 12 - 2t$ m/s. What is the distance it travels before coming to rest?

Common errors · 3 traps that cost marks

Trap 01
Confusing displacement and distance
$\int v\,dt$ gives displacement (signed, can be zero or negative). Distance requires $\int |v|\,dt$. If you use displacement when asked for distance, your answer is wrong even if your integration is correct.
Trap 02
Forgetting to split the integral
For distance, you must find where $v = 0$ and split the integral there. Integrating $v$ as one piece over an interval where it changes sign will give you displacement, not distance.
Trap 03
Wrong number of initial conditions
Going from acceleration to position requires two integrations, so you need two initial conditions: $v(0)$ and $x(0)$. Using only one will leave a constant undetermined.

Odd one out: Three of these steps are needed to find the total distance a particle travels from $t=0$ to $t=4$ given $v(t) = t^2 - 5t + 6$. Which one is not?

Total distance needs A, C and D. The maximum of $v(t)$ is the fastest the particle goes, which never enters the distance calculation.

5

Drill the problems

Work the activities on position, velocity and acceleration.

Activities · apply what you know

1

A particle has $a(t) = 4t - 2$ m/s², $v(0) = 3$ m/s, $x(0) = 0$. Find $x(2)$.

2

A car has $v(t) = 12 - 2t$ m/s. Find the distance travelled before it stops.

3

A ball is thrown upward with $a(t) = -10$ m/s², $v(0) = 20$ m/s, $x(0) = 0$. Find maximum height.

4

Explain the difference between displacement and distance with a concrete example.

5

How do GPS navigation systems use integration to track position from accelerometer data?

Match each operation to its result: Starting from $a(t) = 6$, $v(0) = 2$, $x(0) = 5$, which expression matches each quantity?

6

Revisit and consolidate

Revisit your first answer and confirm what stuck.

09
Revisit your thinking

For $a(t) = 2t$, $v(0) = 0$: $v(t) = \int 2t\,dt = t^2$. After 3 s: $v(3) = 9$ m/s. For displacement: $x(t) = \int t^2\,dt = \frac{t^3}{3}$. With $x(0) = 0$: $x(3) = 9$ metres. Notice how each integration step requires an initial condition to fix the constant. Without $v(0) = 0$ we couldn't determine $C_1$; without $x(0) = 0$ we couldn't determine $C_2$. This is why second-order motion problems need two initial conditions.

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7

Practise motion applications

Work the question bank and the short-answer questions.

01
Multiple choice
+5 XP per correct · +25 XP all-correct

Pick your answer, then rate your confidence.

02
Short answer
ApplyBand 43 marks

Q1. A particle has acceleration $a(t) = 6 - 2t$ m/s². At $t = 0$, $v = 1$ m/s and $x = 2$ m. Find $v(t)$ and $x(t)$, then evaluate $x(3)$. (3 marks)

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ApplyBand 54 marks

Q2. A particle moves with $v(t) = t^2 - 5t + 6$ m/s. Find: (a) when the particle is at rest, (b) displacement from $t = 0$ to $t = 4$, (c) total distance from $t = 0$ to $t = 4$. (4 marks)

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AnalyseBand 53 marks

Q3. In autonomous vehicle navigation, accelerometers measure $a(t)$ and the computer must determine position $x(t)$. (a) Explain why two integrations are needed. (b) What errors can accumulate if the accelerometer has small measurement errors? (c) How do GPS systems correct these errors? (3 marks)

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Comprehensive answers (click to reveal)

Activity 1:

1. $v(t) = 2t^2 - 2t + 3$, $x(t) = \frac{2t^3}{3} - t^2 + 3t$. $x(2) = \frac{16}{3} - 4 + 6 = \frac{22}{3}$ m.

2. Stops when $v = 0$: $t = 6$ s. Distance $= \int_0^6 (12-2t)\,dt = 36$ m.

3. Max height when $v = 0$: $t = 2$ s. $x(2) = -20 + 40 = 20$ m.

Q1 (3 marks): $v(t) = 6t - t^2 + 1$ [1]. $x(t) = 3t^2 - \frac{t^3}{3} + t + 2$ [1]. $x(3) = 27 - 9 + 3 + 2 = 23$ m [1].

Q2 (4 marks): (a) $v = (t-2)(t-3) = 0$: $t = 2, 3$ s [1]. (b) Displacement $= \frac{16}{3}$ m [1.5]. (c) Distance $= \frac{14}{3} + \frac{1}{6} + \frac{5}{6} = \frac{17}{3}$ m [1.5].

Q3 (3 marks): (a) $a \to v$ then $v \to x$ needs two integrations [1]. (b) Errors accumulate quadratically (drift) [1]. (c) GPS provides absolute position fixes that reset drift [1].