SPM 2019 Add Maths Paper 2, Question 12Kinematics of Linear Motion
Solved on video in English and Bahasa Melayu · 10 marks · Form 5, Kinematics of Linear Motion · 7:28 video
The question
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 = t³ - 4t² + 3t, where t is time, in seconds, after passing through a fixed point O. Find (a) the initial acceleration, in ms⁻², of the particle, (b) the time interval, in seconds, when the acceleration of the particle is less than 6 ms⁻², (c) the time, in seconds, when the particle stops instantaneously, (d) the total distance, in m, travelled by the particle until the particle returned to the fixed point O for the second time.
[10 marks]
7:28Step 1
a = dv/dt = 3t² - 8t + 3. At t=0: a = 3 m/s² (initial acceleration).
8 more steps, each with the reason for it, in the full solution
Final answer
(a) 3 m s⁻²; (b) 0 ≤ t < 3; (c) t = 1 s and t = 3 s; (d) 16/3 m ≈ 5.33 m
Where students lose marks
Watch out: add each forward and backward leg for total distance, not net displacement s(3.721) − s(0). In (b), reject t = −1/3.
Watch every step of this question solved on video (7:28), 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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