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

Solved on video in English · Form 5, Kinematics of Linear Motion · 7:31 video

The question

(a)

(Solution by graph sketching is not accepted.) A particle moves along a straight line such that its velocity v ms⁻¹ is given by v = −t² + 2t + 8, where t is the time in seconds after passing through the fixed point O.

(a) Calculate the initial velocity, in ms⁻¹, of the particle.

(b)

A particle moves along a straight line such that its velocity v ms⁻¹ is given by v = −t² + 2t + 8, where t is the time in seconds after passing through the fixed point O.

(b) Calculate the maximum velocity, in ms⁻¹, of the particle.

(c)

(c) Calculate the acceleration, in ms⁻², when the particle reverses its direction.

(d)

(d) Calculate the total distance, in m, travelled by the particle during the first 5 seconds after passing through O.

Kuala Lumpur Trial 2025 Add Maths Paper 2, Question 15: the question as drawn in Eduly's video7:31

Part (a)

Step 1

The initial velocity is the velocity at t = 0.

Why this step

'Initial' means the very start, so t = 0. We just substitute; differentiating would give acceleration, not velocity.

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

Final answer

8 ms⁻¹

Where students lose marks

Watch out: for initial velocity just substitute t = 0. Do not differentiate first, and do not drop the +8.

Part (b)

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

Final answer

9 ms⁻¹

Where students lose marks

Watch out: t = 1 is only the time. Substitute it back into v, and set dv/dt = 0, not v = 0.

Part (c)

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

Final answer

−6 ms⁻²

Where students lose marks

Watch out: reversal means v = 0, not dv/dt = 0. Reject t = −2 because time can't be negative.

Part (d)

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

Final answer

30 m

Where students lose marks

Watch out: split at t = 4 and add the magnitudes. Integrating 0 to 5 gives net displacement 70/3, not distance.

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