KBAT: Reasoning About Motion

Q: How do I know when to split the journey to find total distance?

Solve v=0v=0v=0. Each root inside the time interval is a moment the particle is at rest and can change direction. Split the journey there, measure each leg as a positive length, and add, that sum is the total distance, which can be larger than the displacement.

Q: What is the difference between displacement and total distance here?

Displacement is the net change in position, s(3)−s(0)=0s(3)-s(0)=0s(3)−s(0)=0 m; total distance is all the ground covered, 888 m. Because the particle goes out 444 m and returns 444 m, the two differ. Read the wording carefully,'distance travelled' means the 888 m.

Q: Where does the constant of integration come from?

Integrating vvv gives s=t3−6t2+9t+Cs=t^{3}-6t^{2}+9t+Cs=t3−6t2+9t+C. You need a condition to fix CCC. Here the particle passes OOO at t=0t=0t=0, so s=0s=0s=0 when t=0t=0t=0, giving C=0C=0C=0. Always state the condition you use.

Q: How is Add Math Paper 2 marked on these questions?

Paper 2 is 2 hours 30 minutes and 100 marks, and marking is analytic, method marks are awarded line by line. Correct relationships between sss, vvv and aaa and a proper split for distance can score well even if one arithmetic step slips.

What makes this a KBAT question

A routine kinematics question might hand you a velocity function and ask for the acceleration at a given time, one differentiation and you are done. A KBAT question in Kinematics of Linear Motion asks you to reason about the motion itself: when the particle is momentarily at rest, when it reverses direction, and, the step most students miss, the difference between displacement and total distance travelled.

That is Kemahiran Berfikir Aras Tinggi, higher-order thinking: integrating $v$ to get $s$ and solving $v=0$ are familiar skills, but here you must interpret them, notice that the particle turns around, and split the journey at each rest point. In Add Math this rewards students who picture the motion, not only manipulate the formulas.

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[8 marks]

A particle moves along a straight line and passes a fixed point $O$ . Its velocity, $v$ m s $^{-1}$ , at time $t$ seconds after passing $O$ is given by $v=3t^{2}-12t+9$ .

(a) Find the initial velocity of the particle. (b) Find the time(s) at which the particle is momentarily at rest.

Show worked solution

Understand. The particle starts at $O$ when $t=0$ and its velocity is the given quadratic.

Initial velocity is the value at $t=0$ ; the particle is at rest when $v=0$ . Total distance is not the same as displacement, if the particle reverses direction inside the interval, the outward and return journeys must be measured separately.

Plan. (a) Substitute $t=0$ into $v$ .

(b) Solve $v=0$ by factorising. (c) Integrate $v$ to obtain the displacement $s$ , fix the constant from the condition that the particle passes $O$ at $t=0$ , then split the journey at each rest point in $0\le t\le 3$ and add the length of each leg.

Execute and check. (a) At $t=0$ , $v=3(0)^{2}-12(0)+9=9$ .

The initial velocity is $9$ m s $^{-1}$ .

(b) Set $v=0$ and factorise:

3t^{2}-12t+9=0\;\Rightarrow\;t^{2}-4t+3=0\;\Rightarrow\;(t-1)(t-3)=0

so $t=1$ s or $t=3$ s. The particle is momentarily at rest at these two times.

s=\int\left(3t^{2}-12t+9\right)dt=t^{3}-6t^{2}+9t+C,\qquad C=0

The particle is at rest at $t=1$ , which lies inside the interval, so it changes direction there. Evaluate the position at the ends of each leg:

s(0)=0,\qquad s(1)=1-6+9=4,\qquad s(3)=27-54+27=0

From $t=0$ to $t=1$ the velocity is positive, so the particle moves forward a distance $|s(1)-s(0)|=4$ m. From $t=1$ to $t=3$ the velocity is negative, so it moves back a distance $|s(3)-s(1)|=4$ m.

The total distance is:

\text{total distance}=|s(1)-s(0)|+|s(3)-s(1)|=4+4=8\text{ m}

Check. On $0<t<1$ , $v(0.5)=3(0.25)-6+9=3.75>0$ ; on $1<t<3$ , $v(2)=12-24+9=-3<0$ .

So there is exactly one reversal at $t=1$ . The particle returns to $O$ at $t=3$ , giving a displacement of $0$ m but a total distance of $8$ m, the two are different precisely because it turned around.

Finding a sensible first step

When a kinematics question looks unfamiliar, resist the urge to compute a single number before you know what the motion is doing. The reliable first step is to fix the three linked quantities in your mind, displacement $s$ , velocity $v$ and acceleration $a$ , and how they connect: differentiate to move from $s$ to $v$ to $a$ , integrate to move back.

Then ask the question that unlocks most KBAT motion problems: does the particle ever stop? Solve $v=0$ .

Each root is a time when the particle is momentarily at rest and may reverse direction, and those instants split the journey into legs you can measure one at a time. If the word 'distance' appears, that splitting is almost always the point of the question.

A quick number line marking the rest times keeps the signs straight before any integration begins.

What markers reward

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

Displacement, velocity and acceleration related correctly, $s\to v\to a$ by differentiation, $v\to s$ by integration.
The equation $v=0$ solved to find every time the particle is at rest, with working shown.
The constant of integration fixed from a stated condition, here $s=0$ when $t=0$ , not left as $C$ .
Recognition that total distance is not displacement when the particle reverses, so the journey is split at each rest point inside the interval.
Each leg taken as a positive length, such as $|s(1)-s(0)|$ and $|s(3)-s(1)|$ , then added.
The final answer interpreted in context with units, $8$ m travelled, distinct from a displacement of $0$ m.

How a teacher helps

Motion questions improve fastest once a student stops treating them as formula-substitution and starts picturing the journey. In a one-to-one lesson our teachers draw the number line with you, mark where $v=0$ , and ask which way the particle is moving on each stretch before any integration, so the displacement-versus-distance trap stops catching you.

We also insist on stating the condition that fixes the constant of integration, because markers reward that line. 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

How do I know when to split the journey to find total distance?

Solve $v=0$ . Each root inside the time interval is a moment the particle is at rest and can change direction.

Split the journey there, measure each leg as a positive length, and add, that sum is the total distance, which can be larger than the displacement.

What is the difference between displacement and total distance here?

Displacement is the net change in position, $s(3)-s(0)=0$ m; total distance is all the ground covered, $8$ m. Because the particle goes out $4$ m and returns $4$ m, the two differ.

Read the wording carefully,'distance travelled' means the $8$ m.

Where does the constant of integration come from?

Integrating $v$ gives $s=t^{3}-6t^{2}+9t+C$ . You need a condition to fix $C$ .

Here the particle passes $O$ at $t=0$ , so $s=0$ when $t=0$ , giving $C=0$ . Always state the condition you use.

How is Add Math Paper 2 marked on these questions?

Paper 2 is 2 hours 30 minutes and 100 marks, and marking is analytic, method marks are awarded line by line. Correct relationships between $s$ , $v$ and $a$ and a proper split for distance can score well even if one arithmetic step slips.

Source:SRC-DSKP-EN SRC-FORMAT

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