KBAT: Kinematics by Integration

Q: Why is the total distance not just the value of s at the end?

Because s(t)s(t)s(t) measures displacement, position relative to OOO, and it can decrease when the particle moves backward. Total distance adds every stretch as a positive amount. Here s(5)=20s(5)=20s(5)=20 m is the displacement, but the particle went forward 20 m, back 4 m, then forward 4 m, so the distance travelled is 20+4+4=2820+4+4=2820+4+4=28 m.

Q: How do I know the velocity changes sign?

Solve v=0v=0v=0. If it has real solutions inside your time interval, the velocity may change sign there, so test a value on each side. Here v=3(t−2)(t−4)v=3(t-2)(t-4)v=3(t−2)(t−4) is zero at t=2t=2t=2 and t=4t=4t=4; checking a point in each interval shows the motion goes forward, backward, then forward.

Q: Can I earn marks if my final distance is wrong?

Yes. Add Math Paper 2 is 2 hours 30 minutes and 100 marks, and marking is analytic, method marks are awarded line by line. A correct integration, a properly fixed constant and a sound split at the rest times can score well even if an arithmetic slip changes the final figure.

What makes this a KBAT question

A routine integration question tells you exactly what to integrate: 'given $v=\dots$ , find $s$ '. A KBAT kinematics question hands you the physical story instead, an acceleration and a starting velocity, or a velocity that changes sign, and expects you to decide what to integrate, when to add the constant of integration, and whether the question wants displacement or the total distance travelled.

That is Kemahiran Berfikir Aras Tinggi, higher-order thinking: the integration itself is ordinary Form 5 work, but choosing the right antiderivative, fixing $C$ from the initial condition, finding where the particle is momentarily at rest, and splitting the motion into forward and backward stretches are steps no one lists for you. In Add Math this rewards students who know that acceleration integrates to velocity and velocity integrates to displacement, and who understand that displacement and distance stop being equal the moment the velocity changes sign.

The reading of the motion, not the antiderivative, carries the difficulty.

One worked problem, in the style of Paper 2

This is an original question written in the style of SPM Paper 2. Try it yourself before reading the solution.

Q1[9 marks]

A particle moves along a straight line and passes a fixed point $O$ . Its acceleration, $t$ seconds after passing $O$ , is $a=(6t-18)$ m s $^{-2}$ , and its velocity as it passes $O$ is $24$ m s $^{-1}$ .

(a) Show that the velocity is $v=3t^{2}-18t+24$ , and find the times at which the particle is momentarily at rest. (b) Given that the particle is at $O$ when $t=0$ , find the total distance travelled during the first $5$ seconds, and compare it with the displacement from $O$ at $t=5$ .

Show worked solution

Understand. We are given acceleration, not velocity, so the first integration takes $a$ to velocity $v$ , and a second integration takes $v$ to displacement $s$ .

Each integration produces a constant that an initial condition must fix. The phrase 'total distance' is a warning that the velocity may change sign, so we will need the times when $v=0$ .

Plan. Integrate $a$ and use $v=24$ at $t=0$ to find $v$ ; solve $v=0$ for the rest times.

Integrate $v$ with $s=0$ at $t=0$ to get $s(t)$ . Evaluate $s$ at each rest time and at $t=5$ , then add the magnitudes of the displacement in each stretch.

Execute and check. (a) Integrate the acceleration:

v=\int a\,dt=\int (6t-18)\,dt=3t^{2}-18t+C

As the particle passes $O$ , $t=0$ and $v=24$ , so $C=24$ :

v=3t^{2}-18t+24=3(t-2)(t-4)

The particle is momentarily at rest when $v=0$ , i.e. $t=2$ s and $t=4$ s.

(b) Integrate the velocity for displacement, with $s=0$ at $t=0$ :

s=\int v\,dt=\int (3t^{2}-18t+24)\,dt=t^{3}-9t^{2}+24t

Evaluate at the times that matter, the two rest times and the endpoint:

s(0)=0,\quad s(2)=8-36+48=20,\quad s(4)=64-144+96=16,\quad s(5)=125-225+120=20

Check the sign of $v$ on each interval: for $0<t<2$ , $v>0$ (e.g. $v(0)=24$ ); for $2<t<4$ , $v<0$ (e.g. $v(3)=27-54+24=-3$ ); for $4<t<5$ , $v>0$ (e.g. $v(5)=75-90+24=9$ ). So the particle moves forward, reverses, then moves forward again.

Total distance is the sum of the magnitudes of the displacement over each stretch:

\text{distance}=|s(2)-s(0)|+|s(4)-s(2)|+|s(5)-s(4)|=|20|+|-4|+|4|=28\ \text{m}

The displacement from $O$ at $t=5$ is simply $s(5)=\textbf{20 m}$ , while the total distance travelled is 28 m. They differ precisely because the particle doubled back by $4$ m between $t=2$ and $t=4$ .

Check. Independently, $\int_{0}^{5} v\,dt=[t^{3}-9t^{2}+24t]_{0}^{5}=20-0=20$ m, which agrees with the displacement and confirms the antiderivative.

The distance must be at least the displacement, and $28\ge 20$ as expected, the extra $8$ m is twice the $4$ m backward leg, which is exactly right.

Finding a sensible first step

The reliable first move is to read what you are given and decide which direction you are integrating. Here you are handed an acceleration, so the first integral produces velocity and the second produces displacement, never try to jump straight to $s$ .

Write the antiderivative immediately and attach the constant $C$ ; students lose easy marks by forgetting it or by fixing it with the wrong initial condition. The second decisive move is to notice the phrase 'total distance'.

That phrase tells you the velocity almost certainly changes sign, so before integrating for distance you should solve $v=0$ to find the times the particle stops and turns. Once you know where the particle reverses, the rest is careful arithmetic: evaluate the displacement at each of those times and at the endpoint, then add the magnitudes of the changes rather than the changes themselves.

What markers reward

Marking is analytic, so method marks are awarded line by line. On a kinematics-by-integration question a marker looks for:

The first integration $v=\int a\,dt=3t^{2}-18t+C$ , with the constant shown, not dropped.
The constant fixed correctly from $v=24$ at $t=0$ , giving $C=24$ .
The rest times found by solving $v=0$ : $t=2$ and $t=4$ .
The second integration $s=\int v\,dt=t^{3}-9t^{2}+24t$ , with $s=0$ at $t=0$ .
Recognition that total distance needs the motion split at $t=2$ and $t=4$ , not a single value of $s$ .
Displacement magnitudes added to give 28 m, kept distinct from the 20 m displacement.

How a teacher helps

Kinematics improves fastest when a student can say, before writing anything, whether they are integrating up from acceleration or down toward displacement, and why the constant matters. In a one-to-one lesson our teachers ask you to label each integration, to fix $C$ and $D$ from the stated conditions, and to test for a sign change in the velocity before you ever compute a distance.

We spend real time on the difference between displacement and distance, because that single idea decides most kinematics KBAT marks. Teachers at spmaddmath.com.my are experienced, and lessons are online and taught in English.

Message us on WhatsApp to arrange a one-hour paid class from RM50/hr at the teacher's rate.

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Frequently asked questions

Why is the total distance not just the value of

s

at the end?

Because $s(t)$ measures displacement, position relative to $O$ , and it can decrease when the particle moves backward. Total distance adds every stretch as a positive amount.

Here $s(5)=20$ m is the displacement, but the particle went forward 20 m, back 4 m, then forward 4 m, so the distance travelled is $20+4+4=28$ m.

When do I add the constant of integration, and how do I find it?

Add a constant every time you integrate. Fix it with the condition given for that stage: the velocity at $t=0$ fixes the constant in $v$ , and the position at $t=0$ fixes the constant in $s$ .

Forgetting the constant, or using a velocity condition to fix a displacement constant, is a common and costly slip.

How do I know the velocity changes sign?

Solve $v=0$ . If it has real solutions inside your time interval, the velocity may change sign there, so test a value on each side.

Here $v=3(t-2)(t-4)$ is zero at $t=2$ and $t=4$ ; checking a point in each interval shows the motion goes forward, backward, then forward.

Can I earn marks if my final distance is wrong?

Yes. Add Math Paper 2 is 2 hours 30 minutes and 100 marks, and marking is analytic, method marks are awarded line by line.

A correct integration, a properly fixed constant and a sound split at the rest times can score well even if an arithmetic slip changes the final figure.

Source:SRC-DSKP-EN SRC-FORMAT

Written by the spmaddmath.com.my editorial team.· Last updated 5 September 2026

KBAT: Kinematics Solved by Integration