6.3 SPM Practice (Long Questions)

Question 9:
Diagram below shows the speed-time graph for the movement of an object for a period of 34 seconds.


(a) State the duration of time, in seconds, for which the object moves with uniform speed.
(b) Calculate the rate of change of speed, in ms^-2, of the object for the first 8 seconds.
(c) Calculate the value of u, if the average of speed of the object for the last 26 seconds is 6 ms^-1.

Solution:
(a) Duration of time = 26s – 20s = 6s

$\begin{array}{l}\text{(b)}\\ \text{Rate of change of speed for the first 8 seconds}\\ =\frac{10-6}{0-8}\\ =-\frac{4}{8}\\ =-\frac{1}{2}{\text{ ms}}^{-2}\end{array}$

$\begin{array}{l}\text{(c)}\\ \text{Speed}=\frac{\text{Distance}}{\text{Time}}\\ \frac{\left(\frac{1}{2}\times 12\times \left(6+u\right)\right)+\left(6\times u\right)+\left(\frac{1}{2}\times 8\times u\right)}{26}=6\\ 36+6u+6u+4u=156\\ 16u=120\\ u=7.5{\text{ ms}}^{-1}\end{array}$

Question 10 (6 marks):
Diagram 11 shows the speed-time graph for the movement of a particle for a period of t seconds.

Diagram 11

(a) State the uniform speed, in ms^-1, of the particle.

(b) Calculate the rate of change of speed, in ms^-2, of the particle in the first 4 seconds.

(c) Calculate the value of t, if the distance travelled in the first 4 seconds is half of the distance travelled from the 6^th second to the t^th second.


Solution:
(a)
 Uniform speed of the particle = 12 ms^-1

(b)
$\begin{array}{l}\text{Rate of change of speed of particle}\\ =\frac{12}{4}\\ =3{\text{ ms}}^{-2}\end{array}$

(c)
$\begin{array}{l}\text{Distance travelled in the first 4 seconds }\\ =\frac{1}{2}\left(\begin{array}{l}\text{Distance travelled from the }\\ {\text{6}}^{\text{th}}\text{ second to }{t}^{\text{th}}\text{ second}\end{array}\right)\\ \frac{1}{2}\times 4\times 12=\frac{1}{2}\left[\frac{1}{2}\left(12+20\right)\left(t-6\right)\right]\\ 24=\frac{1}{2}\left[16\left(t-6\right)\right]\\ 24=8\left(t-6\right)\\ 24=8t-48\\ 3=t-6\\ t=9\end{array}$