What is wrong root for the direction change?

Solve v(t)=0 within the interval; ignore roots outside it.

WAMath MethodsSyllabus dot point

How do we use integration to recover velocity and displacement from acceleration, and find distance travelled?

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

WACE Year 12 Mathematics Methods Unit 4 integration in kinematics: recovering velocity from acceleration and displacement from velocity, using initial conditions, and finding distance travelled by splitting where velocity is zero, with worked examples.

Generated by Claude Opus 4.76 min answerUpdated 2026-05-29

Reviewed by: AI editorial process; not yet individually human-reviewed

Have a quick question? Jump to the Q&A page

Jump to a section

What this dot point is asking
The kinematic relationships
Recovering velocity and displacement
Displacement versus distance travelled

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.

Recovering velocity and displacement

From acceleration to displacement

A particle with acceleration $a(t)=6t-4$

Given $v(0)=1$ and $x(0)=2$ , find $v(t)$ and $x(t)$ .

Integrate the acceleration for velocity.

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

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

Integrate the velocity for displacement.

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

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

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

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

Distance for $v(t)=t^{2}-4$ on $[0,3]$

Find where the velocity is zero. $t^{2}-4=0$ gives $t=2$ (taking the positive root in the interval).

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

Compute each signed displacement.

\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}.

The net displacement is $-\tfrac{16}{3}+\tfrac{7}{3}=-3$ units, but the distance travelled adds the magnitudes.

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

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