SPM 2024 Add Maths Paper 2, Question 14Kinematics of Linear Motion

Solved on video in English and Bahasa Melayu · 4 marks · Form 5, Kinematics of Linear Motion · 6:45 video

The question

(a)

Solution by graph sketching is not accepted.

Particle P moves along a straight line such that fixed points A and B lie on the straight line. Its velocity, in m s⁻¹, is given by v = 3t² + 4t − 15, such that t is the time, in seconds, after passing through point A towards point B.

[Assume motion to the right is positive.]

(a) Determine whether point A is to the left or to the right of point B. Justify your answer.

Determine whether point A is to the left or to the right of point B. Justify your answer.

(b)

(b) Find the acceleration, in m s⁻², of the particle at t = 1.3 s.

Find the acceleration, in m s⁻², of the particle at t = 1.3 s.

(c)

(c) If the distance between points A and B is 15 m, determine whether the particle will pass point B or not. Give a reason for your answer. [4 marks]

If the distance between points A and B is 15 m, determine whether the particle will pass point B or not. Give a reason for your answer. [4 marks]

[4 marks]

(d)

(d) Particle Q moves along the same straight line such that its velocity, v m s⁻¹, is given by v = 2t² − 7t + 6, such that t is the time, in seconds, after passing through point A.

Determine the range of time, in seconds, when both particles move to the left.

Particle Q moves along the same straight line such that its velocity, v m s⁻¹, is given by v = 2t² − 7t + 6, such that t is the time, in seconds, after passing through point A.

Determine the range of time, in seconds, when both particles move to the left.

Part (a)

Step 1

Substitute t = 0 (at point A) into the velocity function: v = 3(0)² + 4(0) − 15 = −15 m s⁻¹.

Why this step

t is counted from passing A, so t = 0 is exactly at A. Its velocity there shows which way P sets off towards B.

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

Final answer

A is to the right of B

Where students lose marks

Watch out: v < 0 means the particle heads left towards B, so B is left of A, not the other way round.

Part (b)

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

Final answer

a = 11.8 m s⁻²

Where students lose marks

Watch out: differentiate carefully, 3t² becomes 6t (not 5t), then substitute t = 1.3.

Part (c)

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

Final answer

No, P does not pass B (400/27 m < 15 m)

Where students lose marks

Watch out: finding −400/27 m is not enough. Compare its size with 15 m and state clearly that P does not reach B.

Part (d)

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

Final answer

3/2 < t < 5/3

Where students lose marks

Watch out: 'both' means the overlap of the two time ranges. Draw a number line and keep only the part shaded by both.

Watch every step of this question solved on video (6:45), with the reason behind each step and a box that checks your own answer.

No card. Every past-year question on video, in English and BM.

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