Kinematics is one of the most predictable topics in the O-Level Additional Mathematics syllabus (4049). Almost every paper has a question on a particle moving in a straight line, and almost every one of them can be cracked with the same small set of moves. The problem is that students lose marks not on the calculus, but on the interpretation: distance versus displacement, sign of velocity, and describing motion in the language the mark scheme wants.
Here is how we teach it at Dojo, in the order we actually use it in the exam.
The one relationship you need
Everything comes from this chain:
Displacement (s) → differentiate → Velocity (v) → differentiate → Acceleration (a)
And going backwards:
Acceleration (a) → integrate → Velocity (v) → integrate → Displacement (s)
So:
- $v = \dfrac{ds}{dt}$
- $a = \dfrac{dv}{dt} = \dfrac{d^2s}{dt^2}$
- $s = \int v , dt$ and $v = \int a , dt$
When you integrate, you get a constant $+c$. You must use the initial condition given in the question (usually "initially at O" meaning $s = 0$ when $t = 0$, or "initial velocity is 5 m/s") to find it. Forgetting $+c$ is the single most common way students throw away marks in this topic.
Translating exam phrasing into maths
Half the battle is decoding the wording. Learn these by heart:
| Exam phrase | What you write |
|---|---|
| Instantaneously at rest | $v = 0$ |
| At O / at the origin | $s = 0$ |
| Changes direction | $v = 0$ and $v$ changes sign |
| Maximum or minimum velocity | $a = 0$ |
| Moving with constant velocity | $a = 0$ for a range of $t$ |
| Returns to starting point | $s = 0$ again for $t > 0$ |
| Deceleration | $a$ and $v$ have opposite signs |
Notice that "at rest" and "at the origin" are completely different conditions. Mixing them up is a guaranteed zero for that part.
Worked example 1: differentiating
A particle moves in a straight line so that its displacement from O is $s = t^3 - 6t^2 + 9t$ metres, $t \ge 0$ seconds.
(a) Find the times when the particle is instantaneously at rest.
$v = \dfrac{ds}{dt} = 3t^2 - 12t + 9 = 3(t-1)(t-3)$
Set $v = 0$: $t = 1$ or $t = 3$ seconds.
(b) Find the total distance travelled in the first 4 seconds.
This is where marks are won. Because the particle changes direction at $t = 1$ and $t = 3$, you cannot just compute $s(4) - s(0)$. Split the journey at every time where $v = 0$:
- $s(0) = 0$
- $s(1) = 1 - 6 + 9 = 4$
- $s(3) = 27 - 54 + 27 = 0$
- $s(4) = 64 - 96 + 36 = 4$
Distances travelled: $|4 - 0| + |0 - 4| + |4 - 0| = 4 + 4 + 4 = 12$ m.
Compare that with the displacement, which is only 4 m. The examiner is testing whether you know the difference.
(c) Find the acceleration when the particle is at rest for the second time.
$a = 6t - 12$. At $t = 3$, $a = 6$ m/s².
Worked example 2: integrating
A particle starts from O with velocity 5 m/s and moves so that $a = 2t - 6$ m/s².
$v = \int (2t - 6), dt = t^2 - 6t + c$. When $t = 0$, $v = 5$, so $c = 5$.
$v = t^2 - 6t + 5 = (t-1)(t-5)$
$s = \dfrac{t^3}{3} - 3t^2 + 5t + c_2$. When $t = 0$, $s = 0$, so $c_2 = 0$.
Minimum velocity: occurs when $a = 0$, so $t = 3$. Then $v = 9 - 18 + 5 = -4$ m/s. So the minimum velocity is $-4$ m/s, but the maximum speed in that interval is 4 m/s. Write both if the question says "speed": speed is the magnitude of velocity.
Five habits that protect your marks
- Sketch the velocity graph. For a quadratic $v$, a quick sketch tells you instantly where $v$ is negative and where direction changes.
- Always split total distance at $v = 0$. No exceptions.
- Include units. m, m/s, m/s². Free marks lost every year.
- Reject invalid times. If you get $t = -2$, state "$t \ge 0$, so $t = -2$ is rejected."
- Answer the actual question. If it asks for the acceleration when the particle returns to O, you need $s = 0$ first, then substitute into $a$, not into $v$.
When it gets harder
Higher-mark parts often combine kinematics with other A Math skills: velocity expressed with trigonometric functions such as $v = 4\cos 2t$, exponential decay models like $v = 10e^{-0.5t}$, or definite integrals for distance over an interval, $\int_a^b |v| , dt$. The calculus is the same, but you now need your differentiation and integration of trigonometric and exponential functions to be automatic. Drill those separately so kinematics questions become pure interpretation.
Kinematics rewards students who are systematic. Write down what you are given, decide whether you are going up or down the chain, then interpret carefully.
If you would like a tutor who sat these papers recently to walk you through a full past-paper kinematics question live, start a free trial class with us on WhatsApp: https://wa.link/dsgbkf