What is wrong root for the direction change?

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

SCSA WACE 2023-style practice: Calculator-free. A particle has velocity v(t)=t^2-3t+2 m/s on 0 t 3. (a) Find the times when the particle is at rest. (b) Find the distance travelled over [0,3].

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 t^33-3t^22+2t. The three pieces give 56, -16 and 56. Distance =56+16+56=116 m. Markers reward both rest times, splitting at t=1 and t=2, and summing absolute values.

§-Syllabus dot point

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 4.88 min answerUpdated 2026-06-02

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

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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

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 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.

Exam-style practice questions

Practice questions written in the style of SCSA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

WACE 20217 marksCalculator-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 20235 marksCalculator-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.