MRSM Trial 2025 Add Maths Paper 2, Question 15Kinematics of Linear Motion

Solved on video in English · Form 5, Kinematics of Linear Motion · 8:34 video

The question

(a)

Solution by graph sketching is not accepted. A particle starts moving from a fixed point O along a straight line. The acceleration function is represented by a = h(4 − t²), where h is a constant. The velocity of the particle when t = 3 is 12 ms⁻¹. [Assume motion to the right is positive.]

(a) Express the velocity function of the particle in terms of t.

(b)

A particle moves from a fixed point O with velocity v = 16t − (4/3)t³ (acceleration a = 4(4 − t²), motion to the right positive).

(b) Calculate the displacement when the velocity is maximum.

(c)

A particle moves from a fixed point O with velocity v = 16t − (4/3)t³ and displacement s = 8t² − t⁴/3 (motion to the right positive).

(c) Find the total distance travelled by the particle in the 4th second.

MRSM Trial 2025 Add Maths Paper 2, Question 15: the question as drawn in Eduly's video8:34

Part (a)

Step 1

Velocity is the integral of acceleration: v = ∫h(4 − t²) dt = h(4t − t³/3) + c.

Why this step

Acceleration is how fast velocity changes, so we go backwards by integrating. Always add + c, which stays unknown until we use a condition.

3 more steps, each with the reason for it, in the full solution

Final answer

v = 16t − (4/3)t³

Where students lose marks

Watch out: never forget + c. 'Starts from O' means v = 0 when t = 0, so c = 0, then find h.

Part (b)

3 more steps, each with the reason for it, in the full solution

Final answer

s = 80/3 m (≈ 26.67 m)

Where students lose marks

Watch out: maximum velocity is where a = 0. Solve for t carefully (t > 0), then substitute that t into s.

Part (c)

3 more steps, each with the reason for it, in the full solution

Final answer

25/3 m (≈ 8.33 m)

Where students lose marks

Watch out: the 4th second is t = 3 to 4. If v = 0 inside it, s(4) − s(3) is only net displacement.

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