Integration in kinematics: WACE Year 12 Mathematics Methods Unit 3

Syllabus dot point

“Apply integration to straight-line motion, recovering velocity from acceleration and displacement from velocity, and distinguishing displacement from distance travelled”

8 min readUpdated 2 Jun 2026

Quick answer

Integrating acceleration gives velocity and integrating velocity gives displacement, each needing an initial condition for the constant. Displacement over $[t_1,t_2]$ is $\int v\,dt$ (signed); distance travelled is $\int |v|\,dt$ , found by splitting where $v=0$ .

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What this dot point is asking
The kinematic relationships
Recovering velocity and displacement
Why each step needs its own constant
Definite integrals avoid the constants
Displacement versus distance travelled
Exam-style questions

Syllabus placement. In the SCSA Mathematics Methods ATAR Year 12 syllabus, integrals are Unit 3 (Topic 3.2), not Unit 4. This page sits in our Unit 4 folder for historical reasons; the content is Unit 3.

What this dot point is asking

SCSA Unit 4 applies integration to motion in a straight line, reversing the differentiation links from Unit 3. This dot point asks you to recover velocity and displacement by integration and to distinguish signed displacement from total distance travelled. It is examined in both sections, with extended motion problems common in the calculator-assumed section.

The kinematic relationships

Differentiation takes displacement to velocity to acceleration; integration reverses each step.

Integration in kinematics

v(t)=\int a(t)\,dt, \qquad x(t)=\int v(t)\,dt.

Each integration introduces a constant, found from an initial condition such as

v(0)

x(0)

Recovering velocity and displacement

From acceleration to displacement

A particle has acceleration $a(t)=6t-4$ . Given $v(0)=1$ and $x(0)=2$ , find $v(t)$ and $x(t)$ .

Step 1: Integrate acceleration to find velocity

Integrating acceleration reverses the differentiation link, producing the velocity function up to an unknown constant. The initial velocity condition will fix that constant.

v(t)=\int (6t-4)\,dt = 3t^{2}-4t+c_1.

Apply $v(0)=1$ : $c_1=1$ , so $v(t)=3t^{2}-4t+1$ .

Step 2: Integrate velocity to find displacement

A second integration is needed because velocity and displacement are linked by differentiation, introducing a fresh constant $c_2$ that is independent of $c_1$ . The initial position condition fixes this new constant.

x(t)=\int (3t^{2}-4t+1)\,dt = t^{3}-2t^{2}+t+c_2.

Apply $x(0)=2$ : $c_2=2$ , so $x(t)=t^{3}-2t^{2}+t+2$ .

Final answer: $v(t)=3t^{2}-4t+1$ and $x(t)=t^{3}-2t^{2}+t+2$ .

Markers reward each integration, the use of the initial condition to find each constant, and the correctly labelled final expressions.

Why each step needs its own constant

Each integration introduces a fresh unknown constant, and the two constants are found from different pieces of information. The velocity constant $c_1$ is fixed by an initial velocity such as $v(0)$ , and the displacement constant $c_2$ is fixed by an initial position such as $x(0)$ . A frequent error is to use the position condition to fix the velocity constant; keep the two stages separate and apply the matching condition to each. If a question gives a condition at a time other than $t=0$ , substitute that time instead, the method is identical.

Definite integrals avoid the constants

When only a change is required, a definite integral sidesteps the constants entirely. The change in velocity from $t_1$ to $t_2$ is $\displaystyle\int_{t_1}^{t_2}a(t)\,dt$ , and the change in displacement is $\displaystyle\int_{t_1}^{t_2}v(t)\,dt$ , because the constant cancels on subtraction. Use the indefinite-integral-with-conditions approach when you need the full function $v(t)$ or $x(t)$ , and the definite integral when you only need the net change over an interval.

Displacement versus distance travelled

Over an interval, the change in displacement is the signed integral of velocity:

\text{displacement} = \int_{t_1}^{t_2} v(t)\,dt.

Distance travelled adds up movement regardless of direction, so it integrates the speed

|v(t)|

. To compute it, find where

v(t)=0

(where the particle reverses), split the interval there, and add the absolute values of the pieces.

Distance travelled with a direction change

A particle has velocity $v(t)=t^{2}-4$ on $[0,3]$ . Find the distance travelled.

Step 1: Find where the velocity is zero

The particle reverses direction where $v=0$ , so those times are the points at which the interval must be split. Solving $t^{2}-4=0$ gives $t=2$ (taking the positive root within the interval).

On $[0,2]$ , $v<0$ (moving backward); on $[2,3]$ , $v>0$ (moving forward).

Step 2: Compute the signed displacement on each sub-interval

Evaluating the integral over each piece gives the signed displacement there. Splitting at $t=2$ ensures each integral corresponds to motion in one direction.

\int_0^2 (t^{2}-4)\,dt = \left[\frac{t^{3}}{3}-4t\right]_0^2 = \frac{8}{3}-8 = -\frac{16}{3}.

\int_2^3 (t^{2}-4)\,dt = \left[\frac{t^{3}}{3}-4t\right]_2^3 = \left(9-12\right)-\left(\frac{8}{3}-8\right) = -3+\frac{16}{3} = \frac{7}{3}.

Step 3: Add the absolute values to find total distance

Distance is the total path length, so the signed values from each piece must be taken as absolute values before adding. The net displacement $(-\tfrac{16}{3}+\tfrac{7}{3}=-3)$ is different from the distance.

\text{distance} = \left|-\frac{16}{3}\right| + \left|\frac{7}{3}\right| = \frac{16}{3}+\frac{7}{3} = \frac{23}{3} \text{ units}.

Final answer: $\dfrac{23}{3}$ units.

Markers reward finding $t=2$ , splitting there, and adding absolute values for distance.

Common errors

Treating displacement as distance: Distance integrates $|v|$ ; you must split at the times when $v=0$ and add absolute values.
Forgetting an initial condition: Each integration needs a condition to fix its constant; using $x(0)$ for the velocity constant is a mix-up.
Wrong root for the direction change: Solve $v(t)=0$ within the interval; ignore roots outside it.
Confusing speed and velocity: Speed is $|v|$ ; velocity carries a sign indicating direction.

Exam-style questions

Questions in the style of SCSA exam questions on this dot point, each with a worked answer. They are written by ExamExplained unless tagged "Past paper"; the year shows the paper a question is modelled on.

WACE 2021-style7 marks

Calculator-assumed. A particle moves with acceleration

a(t)=6t-2

m/s

^{2}

. Initially

v(0)=-4

m/s and

x(0)=0

m. (a) Find

v(t)

. (b) Find

x(t)

. (c) Find the displacement after

3

seconds.

Show worked answer →

A two-stage integration with initial conditions.

(a) $v(t)=\displaystyle\int(6t-2)\,dt=3t^{2}-2t+c_1$ . Using $v(0)=-4$ , $c_1=-4$ , so $v(t)=3t^{2}-2t-4$ .

(b) $x(t)=\displaystyle\int(3t^{2}-2t-4)\,dt=t^{3}-t^{2}-4t+c_2$ . Using $x(0)=0$ , $c_2=0$ , so $x(t)=t^{3}-t^{2}-4t$ .

Markers reward both integrations, correct use of each initial condition, and the evaluated displacement.

WACE 2023-style5 marks

Calculator-free. A particle has velocity

v(t)=t^{2}-3t+2

m/s on

0\le t\le 3

. (a) Find the times when the particle is at rest. (b) Find the distance travelled over

[0,3]

Show worked answer →

A distance-travelled question requiring splitting.

(a) $v(t)=0$ : $t^{2}-3t+2=(t-1)(t-2)=0$ , so $t=1$ and $t=2$ .

(b) Velocity is positive on $[0,1]$ , negative on $[1,2]$ , positive on $[2,3]$ . Antiderivative $\dfrac{t^{3}}{3}-\dfrac{3t^{2}}{2}+2t$ . The three pieces give $\dfrac{5}{6}$ , $-\dfrac{1}{6}$ and $\dfrac{5}{6}$ . Distance $=\dfrac{5}{6}+\dfrac{1}{6}+\dfrac{5}{6}=\dfrac{11}{6}$ m.

Markers reward both rest times, splitting at $t=1$ and $t=2$ , and summing absolute values.