Kinematics is one of the most predictable topics in the O-Level Additional Mathematics syllabus (4049). Almost every year there is a question about a particle moving in a straight line, and almost every year students lose marks for the same three reasons: mixing up displacement and distance, forgetting the constant of integration, and ignoring the sign of velocity.
The good news is that once you internalise one diagram and four habits, kinematics becomes free marks. Here is exactly how we teach it at Dojo.
The one relationship you must know
Everything in A Math kinematics comes from this chain:
Displacement (s) → differentiate → Velocity (v) → differentiate → Acceleration (a)
And going backwards:
Acceleration (a) → integrate → Velocity (v) → integrate → Displacement (s)
In symbols:
- $v = \dfrac{ds}{dt}$
- $a = \dfrac{dv}{dt} = \dfrac{d^2s}{dt^2}$
When you go downwards you differentiate. When you go upwards you integrate, and every integration produces a constant that must be found using the information given in the question, usually something like "initially at rest" or "passes through O when t = 2".
Remember that A Math kinematics is always motion in a straight line. There are no vectors and no projectiles. That makes signs the whole game.
What the key phrases actually mean
Examiners use standard wording. Translate it immediately.
- "At instantaneous rest" or "momentarily at rest": set $v = 0$.
- "Returns to O" or "passes through the origin": set $s = 0$.
- "Maximum speed" or "minimum velocity": set $a = 0$, then check.
- "Initially": $t = 0$.
- "Decelerating": $v$ and $a$ have opposite signs, not simply $a < 0$.
- "Total distance travelled": you cannot just substitute values into $s$. More on this below.
Worked example: differentiating down the chain
A particle moves along a straight line so that its displacement from a fixed point O is $s = t^3 - 6t^2 + 9t$ metres, where $t$ is the time in seconds.
Step 1: find velocity.
$v = \dfrac{ds}{dt} = 3t^2 - 12t + 9 = 3(t-1)(t-3)$
Step 2: find when the particle is at rest.
Set $v = 0$, so $t = 1$ or $t = 3$. Always factorise if you can, because it makes the sign analysis instant.
Step 3: find acceleration.
$a = \dfrac{dv}{dt} = 6t - 12$
At $t = 1$, $a = -6\ \text{m/s}^2$. At $t = 3$, $a = 6\ \text{m/s}^2$.
Step 4: total distance in the first 4 seconds.
This is where students lose two marks. The particle changes direction at $t = 1$ and $t = 3$, so you must break the journey into legs:
- $s(0) = 0$
- $s(1) = 1 - 6 + 9 = 4$
- $s(3) = 27 - 54 + 27 = 0$
- $s(4) = 64 - 96 + 36 = 4$
Distances travelled: 4 m, then 4 m, then 4 m. Total distance = 12 m, while the net displacement is only 4 m. If you had simply written $s(4) - s(0)$ you would have answered 4 m and lost the marks.
Worked example: integrating up the chain
A particle starts from rest at O with acceleration $a = 6 - 2t\ \text{m/s}^2$.
$v = \int (6 - 2t), dt = 6t - t^2 + c_1$
Since the particle starts from rest, $v = 0$ when $t = 0$, so $c_1 = 0$ and $v = 6t - t^2$.
$s = \int (6t - t^2), dt = 3t^2 - \tfrac{1}{3}t^3 + c_2$
Since it starts at O, $s = 0$ when $t = 0$, so $c_2 = 0$.
Maximum velocity occurs when $a = 0$, that is $t = 3$, giving $v = 18 - 9 = 9\ \text{m/s}$.
Four habits that protect your marks
- Write the chain at the top of your working. Deciding "differentiate or integrate" before you touch algebra prevents the most common error.
- Never drop the constant. Write $+c$ every single time, then find it in the next line. Some questions give the condition later in the part, so read the whole stem first.
- Sketch a quick number line for velocity. Mark where $v = 0$ and the sign of $v$ in each interval. This tells you when the particle reverses and therefore how many legs the distance question has.
- Include units and context in the final line. Write "the total distance travelled is 12 m" rather than a bare number. Presentation marks in the O-Level paper reward clear conclusions.
Common traps in past papers
- Assuming negative acceleration means slowing down. A particle with $v < 0$ and $a < 0$ is speeding up.
- Using $t = 0$ conditions when the question says "when the particle first returns to O".
- Forgetting that a definite integral of velocity gives displacement, so you must split the integral at the turning points to get distance.
- Rounding too early in questions involving exponential or trigonometric velocity functions, which appear in the harder A Math kinematics parts.
Work through five past-year kinematics questions using this exact structure and the topic will stop costing you marks. It is one of the fastest scoring gains available in Paper 2.
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