SPM Paper 2 Section C Priority (10 Marks)

Displacement, Velocity, Acceleration & Total Distance

Kinematics is the physical manifestation of calculus (differentiation and integration). Master the calculus loop connecting displacement $s(t)$, velocity $v(t)$, and acceleration $a(t)$, deciphering physical conditions (instantaneous rest, passing origin $O$), and calculating total distance without losing marks on turning points.

1. The Kinematic Calculus Cycle

Displacement (Sesaran)
$s(t)$

Distance & direction from origin $O$. At origin $O \implies s = 0$.

Velocity (Halaju)
$v(t) = \frac{ds}{dt}$

Rate of change of $s$. Instantaneous rest / reverse $\implies v = 0$.

Acceleration (Pecutan)
$a(t) = \frac{dv}{dt} = \frac{d^2s}{dt^2}$

Rate of change of $v$. Maximum velocity $\implies a = 0$.

Physical Condition Mathematical Meaning Marking Scheme Implication
Initial position / Initial velocity Substitute $t = 0$ Yields constant of integration $c$
Particle stops momentarily / Reverses direction Set $v = 0$ Solves for turning time $t$
Particle passes through fixed origin $O$ Set $s = 0$ Solves for return time $t$
Maximum or Minimum velocity Set $a = 0$ Solve for $t$, then substitute back into $v(t)$
Uniform velocity (Halaju seragam) $a = 0$ Acceleration is zero

2. SPM Examiner Pitfalls in Kinematics

Trap 1: Confusing Total Distance with Displacement:

Students often simply compute $s(t_2) - s(t_1)$. That ONLY gives net displacement! If the particle reverses direction ($v = 0$ at $t = t_{\text{turn}}$ between $t_1$ and $t_2$), you MUST calculate the distance traveled in each segment: $\text{Total Distance} = |s(t_{\text{turn}}) - s(t_1)| + |s(t_2) - s(t_{\text{turn}})|$. Missing this loses 3 marks!

Trap 2: Forgetting the Constant of Integration $+ c$:

When integrating $a(t)$ to find $v(t)$, never assume $c = 0$ unless initial velocity is zero! Check if the particle passes with an initial velocity $u$ or initial displacement $s_0 \neq 0$.

3. Full 10-Mark SPM Paper 2 Section C Model Worked Example

SPM Paper 2 Section C Full Blueprint Model [10 Marks]
A particle moves along a straight line and passes through a fixed point $O$ with an initial velocity of $12\text{ ms}^{-1}$. Its acceleration, $a\text{ ms}^{-2}$, $t$ seconds after passing through $O$ is given by $a = 6 - 4t$.

(a) Find the maximum velocity of the particle. [3 marks]

(b) Find the time $t$, in seconds, when the particle stops momentarily. [2 marks]

(c) Calculate the displacement, in $\text{m}$, of the particle when it stops momentarily. [2 marks]

(d) Calculate the total distance, in $\text{m}$, travelled by the particle in the first $5$ seconds after passing through $O$. [3 marks]

Step-by-Step Marking Scheme:
(a) $v = \int a\,dt = \int (6 - 4t)\,dt = 6t - 2t^2 + c$. [1m: K1: Integrate $a(t)$]
At $t = 0, v = 12 \implies c = 12 \implies v(t) = 12 + 6t - 2t^2$. Max velocity occurs when $a = 0 \implies 6 - 4t = 0 \implies t = 1.5\text{ s}$. [1m: K1: Find $t$ for $a = 0$]
$v_{\max} = 12 + 6(1.5) - 2(1.5)^2 = 12 + 9 - 4.5 = 16.5\text{ ms}^{-1}$. [1m: N1: Correct maximum velocity]
(b) Stops momentarily when $v = 0 \implies -2t^2 + 6t + 12 = 0 \implies t^2 - 3t - 6 = 0$. [1m: K1: Equate $v(t) = 0$]
$t = \frac{3 \pm \sqrt{(-3)^2 - 4(1)(-6)}}{2} = \frac{3 \pm \sqrt{33}}{2} = \frac{3 + 5.7446}{2} = 4.37\text{ s}$ (reject $t < 0$). [1m: N1: $t = 4.37\text{ s}$]
(c) $s = \int v\,dt = \int (12 + 6t - 2t^2)\,dt = 12t + 3t^2 - \frac{2}{3}t^3 + c_2$. Passes $O$ at $t = 0 \implies c_2 = 0$. [1m: K1: Integrate $v(t)$]
At $t = 4.37$: $s(4.37) = 12(4.37) + 3(4.37)^2 - \frac{2}{3}(4.37)^3 = 52.44 + 57.29 - 55.60 = 54.13\text{ m}$. [1m: N1: Displacement $= 54.13\text{ m}$]
(d) Turning point occurs at $t = 4.37\text{ s}$, which lies between $0 \le t \le 5$. [1m: P1: Identify turning point in interval]
Displacements: $s(0) = 0$, $s(4.37) = 54.13\text{ m}$. At $t = 5$: $s(5) = 12(5) + 3(5)^2 - \frac{2}{3}(5)^3 = 60 + 75 - 83.33 = 51.67\text{ m}$. [1m: K1: Calculate $s(5)$]
$\text{Total Distance} = |54.13 - 0| + |51.67 - 54.13| = 54.13 + 2.46 = 56.59\text{ m}$. [1m: N1: Correct total distance]
Open Form 5 Chapter 8 Section C Worksheet