What is not checking the endpoints for an extreme?

In a closed interval the maximum displacement may be at an endpoint, not at a turning point.

Given s = t^2 - 4t, find the minimum displacement for t ≥ 0. [3 marks]

SingaporeAdditional MathematicsSyllabus dot point

How do we solve practical motion problems involving maximum displacement, total distance and changes of direction?

Solve kinematics problems involving maximum or minimum displacement and velocity, total distance travelled, and changes of direction

A focused answer to the O-Level A-Maths outcome on applied kinematics. Finding maximum displacement and velocity, total distance travelled allowing for direction changes, and combining differentiation and integration in motion problems.

Generated by Claude Opus 4.89 min answerUpdated 2026-06-06

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

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What this dot point is asking
The answer
Examples in context
Try this

What this dot point is asking

SEAB wants you to solve fuller motion problems: finding the maximum or minimum displacement or velocity, computing the total distance travelled when the particle changes direction, and combining differentiation and integration in a single question. These pull together everything in the kinematics topic and reward careful, structured working.

The answer

Maximum or minimum displacement

Displacement is greatest or least where velocity is zero, because that is where displacement stops increasing or decreasing. So set $v = \dfrac{ds}{dt} = 0$ , solve for $t$ , and evaluate $s$ there. In a closed interval, also check the endpoints, as the extreme may occur at the start or end.

Maximum or minimum velocity

By the same logic one level up, velocity is greatest or least where acceleration is zero. Set $a = \dfrac{dv}{dt} = 0$ , solve for $t$ , and evaluate $v$ there, checking endpoints if the interval is closed.

Total distance travelled

This is the key subtlety. The change in displacement is $\int v\,dt$ , but if the particle reverses, parts of the journey cancel. To find total distance:

Find the times when $v = 0$ (the direction changes).
Integrate $v$ over each sub-interval between consecutive such times and the endpoints.
Take the modulus of each piece and add them.

Putting the tools together

A typical problem differentiates to find velocity and acceleration, sets one of them to zero to locate a turning point, and integrates to find a distance, so fluency in moving both ways along the chain is essential.

Worked example

A particle has velocity $v = 3t^2 - 12$ m per second for $t \geq 0$ . Find the total distance travelled in the first $3$ seconds.

Step 1: Find where the velocity is zero

$3t^2 - 12 = 0$ gives $t^2 = 4$ , so $t = 2$ (taking $t \geq 0$ ). The particle changes direction at $t = 2$ .

Step 2: Integrate over the first sub-interval

\int_0^2 (3t^2 - 12)\,dt = \big[t^3 - 12t\big]_0^2 = (8 - 24) - 0 = -16.

The magnitude is $16$ m.

Step 3: Integrate over the second sub-interval

\int_2^3 (3t^2 - 12)\,dt = \big[t^3 - 12t\big]_2^3 = (27 - 36) - (8 - 24) = -9 + 16 = 7.

The magnitude is $7$ m.

Step 4: Add the magnitudes

Total distance $= 16 + 7 = 23$ m.

Examples in context

Example 1. A projectile's peak. A ball thrown straight up reaches its maximum height when its velocity is zero; setting $v = 0$ and evaluating the height there gives the peak, a direct maximum-displacement calculation.

Example 2. A shuttle bus route. A bus that drives forward, stops, reverses to a depot, and stops again covers a total distance found by splitting at each stop (where $v = 0$ ) and adding the leg distances, even though its net displacement may be small.

Try this

Q1. A particle has $v = 2t - 6$ . At what time is it instantaneously at rest? [1 mark]

Cue. $2t - 6 = 0$ , so $t = 3$ seconds.

Q2. Given $s = t^2 - 4t$ , find the minimum displacement for $t \geq 0$ . [3 marks]

Cue. $v = 2t - 4 = 0$ at $t = 2$ ; $s = 4 - 8 = -4$ m, the minimum.

Q3. A particle has $v = t - 2$ m per second. Find the total distance from $t = 0$ to $t = 4$ . [4 marks]

Cue. $v = 0$ at $t = 2$ ; $\left|\int_0^2 (t-2)\,dt\right| = 2$ and $\left|\int_2^4 (t-2)\,dt\right| = 2$ , so total $= 4$ m.

Exam-style practice questions

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

Original6 marksA particle moves in a straight line so that its displacement is

s = t^3 - 6t^2 + 9t

metres at time

t

seconds. Find the maximum displacement in the interval

0 \leq t \leq 2

Show worked answer →

Velocity: $v = \dfrac{ds}{dt} = 3t^2 - 12t + 9 = 3(t - 1)(t - 3)$ . In $[0, 2]$ , $v = 0$ at $t = 1$ .

At $t = 1$ : $s = 1 - 6 + 9 = 4$ m. Check the endpoints: $s(0) = 0$ and $s(2) = 8 - 24 + 18 = 2$ m.

The greatest of these is $4$ m, at $t = 1$ , so the maximum displacement is $4$ m.

Markers reward finding $v = 0$ inside the interval, evaluating $s$ there and at the endpoints, and selecting the maximum.

Original6 marksA particle has velocity

v = t^2 - 4t + 3

m per second. Find the total distance travelled from

t = 0

t = 3

Show worked answer →

The velocity factorises as $v = (t - 1)(t - 3)$ , so $v = 0$ at $t = 1$ and $t = 3$ . The particle changes direction at $t = 1$ .

Distance for $0 \leq t \leq 1$ : $\left|\displaystyle\int_0^1 (t^2 - 4t + 3)\,dt\right| = \left|\left[\dfrac{t^3}{3} - 2t^2 + 3t\right]_0^1\right| = \left|\dfrac{1}{3} - 2 + 3\right| = \dfrac{4}{3}$ .

Distance for $1 \leq t \leq 3$ : $\left|\left[\dfrac{t^3}{3} - 2t^2 + 3t\right]_1^3\right| = \left|(9 - 18 + 9) - \dfrac{4}{3}\right| = \dfrac{4}{3}$ .

Total distance $= \dfrac{4}{3} + \dfrac{4}{3} = \dfrac{8}{3}$ metres.

Markers reward finding where $v = 0$ , splitting the integral at the direction change, taking the modulus of each part, and adding.