Selangor Trial 2025 Add Maths Paper 2, Question 12Kinematics of Linear Motion

Solved on video in English · Form 5, Kinematics of Linear Motion · 6:38 video

The question

(a)

A particle moves along a straight line and passes through a fixed point O. The velocity of the particle, v m s⁻¹, is given by v = 3t² − 2t − 8, where t is the time, in seconds, after passing through O. [Assume motion to the right is positive.] (a) Find the acceleration, in m s⁻², when the particle stops instantaneously.

(b)(i)

(b) Find the distance, in m, travelled by the particle (i) from the maximum velocity until the particle stops instantaneously.

(b)(ii)

(b)(ii) Find the distance, in m, travelled by the particle in the first 7 seconds.

Selangor Trial 2025 Add Maths Paper 2, Question 12: the question as drawn in Eduly's video6:38

Part (a)

Step 1

The particle stops instantaneously when v = 0: 3t² − 2t − 8 = 0.

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

Final answer

a = 10 m s⁻²

Where students lose marks

Watch out: t = −4/3 is not allowed because time can't be negative, and acceleration comes from differentiating v, not integrating it.

Part (b)(i)

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

Final answer

250/27 m (9 7/27 ≈ 9.259 m)

Where students lose marks

Watch out: get t = 1/3 from dv/dt = 0, then take the absolute value of s(2) − s(1/3) and check fraction signs.

Part (b)(ii)

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

Final answer

262 m

Where students lose marks

Watch out: don't just use |s(7) − s(0)| = 238. Split at t = 2 where the direction changes, then add both distances.

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