SPM Mathematics Trial 2021 (Kelantan), Paper 2 (Question 13)

Question 13:
Table 2 shows the speed and time of a particle in 10 seconds.

Speed (ms-1)	0	15	15	30
Time (s)	0	2	6	10

(a) Based on Table 2, draw the speed-time graph in figure in the answer space. [2 marks]

(b) Based on the graph drawn in Figure in the answer space, state:
(i) Uniform speed, in ms^-1, of the particle. [1 mark]
(ii) The duration of time, in second, the particle moves in uniform speed. [1 mark]

(c) Calculate the rate of change of speed, in ms^-2, for the last 4 seconds. [2 marks]

(d) Calculate the average speed, in ms^-1 of the particle in period of 10 seconds. [3 marks]

Answer:
(a)


Solution:
(a)


(b)(i)
Uniform speed of the particle is 15 ms^-1.

(b)(ii)
Duration of time the particle moves in uniform speed = 6 – 2 = 4 seconds 

(c)

$\begin{array}{l}\text{Rate of change of speed of the particle for the last 4 seconds}\\ =\text{ Gradient of the graph for the last 4 seconds}\\ =\frac{y_2-y_1}{x_2-x_1}\\ =\frac{30-15}{10-6}\\ =\frac{15}{4}{\text{ ms}}^{\text{-2}}\text{ or }3.75{\text{ ms}}^{\text{-2}}\end{array}$


(d)

$\begin{array}{l}\text{Average speed}=\frac{\text{Total distance}}{\text{Total time taken}}\\ \\ \text{For a speed-time graph:}\\ \text{Area under the graph = Distance travelled}\\ \\ \text{Thus, the average speed}\\ =\frac{\text{Area of triangle + area of rectangle + area of trapezium}}{10}\\ \\ =\frac{\left[\frac{1}{2}\left(2\right)\left(15\right)\right]+\left(6-2\right)\left(15\right)+\frac{1}{2}\left(15+30\right)\left(10-6\right)}{10}\\ \\ =\frac{15+60+90}{10}\\ =16.5{\text{ ms}}^{\text{-1}}\end{array}$