SPM 2023 Add Maths Paper 2, Question 13Kinematics of Linear Motion

Solved on video in English and Bahasa Melayu · Form 5, Kinematics of Linear Motion · 8:13 video

The question

(a)(i)

(Solutions by graph sketching are not accepted.) A particle moves along a straight line from a fixed point O. Diagram 7 shows the velocity-time graph of the particle's motion. [Assume motion to the right is positive.]

(a)(i) State the time, in seconds, when the particle stops instantaneously.

(a)(ii)

A particle moves along a straight line from O; Diagram 7 shows its velocity-time graph (motion to the right is positive).

(a)(ii) State the range of times, in seconds, when the particle moves towards the left.

(b)(i)

The velocity-time graph (Diagram 7) passes through the origin and (6, 12), and cuts the t-axis at t = 5. It is given that the velocity-time graph is a quadratic function.

(b)(i) Show that v = 2t² − 10t.

(b)(ii)

The velocity is v = 2t² − 10t.

(b)(ii) Find the acceleration, in m s^−2, when t = 2 seconds.

(b)(iii)

The velocity is v = 2t² − 10t.

(b)(iii) Find the distance, in m, travelled by the particle from p seconds until the 6th second, where p is the time when the maximum velocity is achieved before it changes direction.

SPM 2023 Add Maths Paper 2, Question 13: the question as drawn in Eduly's video8:13

Part (a)(i)

Step 1

The particle stops instantaneously when v = 0 (the graph cuts the t-axis).

1 more step, with the reason for it, in the full solution

Final answer

t = 5 s

Part (a)(ii)

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

Final answer

0 < t < 5

Where students lose marks

Watch out: use strict inequalities. Write 0 < t < 5, not 0 ≤ t ≤ 5, because v = 0 at both ends.

Part (b)(i)

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

Final answer

Shown

Where students lose marks

Watch out: don't just check t = 6 in the given v. Build v from roots 0 and 5, then use (6, 12).

Part (b)(ii)

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

Final answer

a = −2 m s⁻²

Where students lose marks

Watch out: differentiate every term of v. Stopping at 4t gives a = 8 instead of a = −2.

Part (b)(iii)

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

Final answer

p = 5/2 s; distance = 26.5 m

Where students lose marks

Watch out: the particle turns at t = 5. Split there and add the two distances; one net displacement (15.17 m) is wrong.

Watch every step of this question solved on video (8:13), with the reason behind each step and a box that checks your own answer.

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