Question 12 (5 marks):
Diagram 7 shows a distance-time graph for the journey of a car from Kuala Lumpur to Ipoh.


(a) State the duration of time, in minutes, the driver stopped and rest at Tapah.
(b) Calculate the speed, in kmh^-1, of the car from Kuala Lumpur to Tapah.
(c) Calculate the average speed, in kmh^-1, of the car for whole journey.


Solution:
(a)
1400 – 1345 = 15 minutes
The driver stopped and rest at Tapah for 15 minutes.


(b)
$\begin{array}{l}\text{Speed}=\frac{\text{Distance}}{\text{Time}}\\ \text{Speed}=\frac{152\text{ km}}{1\frac{3}{4}\text{ h}}\leftarrow \boxed{\begin{array}{l}1345-1200\\ =1\text{ h }45\mathrm{mins}\end{array}}\\ \text{Speed}=86.86{\text{ kmh}}^{-1}\\ \\ \text{Speed of the car from Kuala Lumpur}\\ \text{to Tapah}=86.86{\text{ kmh}}^{-1}\end{array}$


(c)
$\begin{array}{l}\text{Average speed}=\frac{\text{Total distance}}{\text{Time taken}}\\ =\frac{205\text{ km}}{3\frac{1}{3}\text{ h}}\leftarrow \boxed{\begin{array}{l}1520-1200\\ =3\text{ h 20 mins}\end{array}}\\ \text{Average speed}=61.5{\text{ kmh}}^{-1}\\ \\ \text{Average speed of the car for the}\\ \text{whole journey}=61.5{\text{ kmh}}^{-1}\end{array}$

Question 11 (6 marks):
Diagram 6 shows the distance-time graph of Ursula, Janet and Maria in a 100 m race.

Diagram 6



(a) Who won the race?

(b) During the race, Ursula slipped and fell over. After that, she continued her run.
State the duration, in seconds, before Ursula continued her run.

(c) During the race, Janet was injured and she stopped running.
State Janet's distance, in m, from the finishing line when she stopped running.

(d) Calculate the average speed, in ms^-1, of Ursula.


Solution:
(a)
Maria won the race: 100 km in 16 seconds.

(b)
Duration before Ursula continued her run from slipping and falling over
= 18 seconds – 9 seconds
= 9 seconds

(c)
Janet's distance from the finishing line when she stopped running
= 100 m – 70 m
= 30 m

(d)
$\begin{array}{l}\text{Average speed}=\frac{\text{Distance}}{\text{Time}}\\ =\frac{100\text{ m}}{20\text{ s}}\\ =5{\text{ ms}}^{-1}\\ \\ \text{Average speed of Ursula}=5{\text{ ms}}^{-1}\end{array}$

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

Question 7:




Solution:
(a)
Distance travel by the object for the first 12 seconds

Question 8:



Solution:
(a)
$\begin{array}{l}\text{Acceleration from }t=6s\text{ to }t=12s\\ =\text{ Gradient of speed-time graph}\\ =\frac{6-6}{12-6}\\ =0{\text{ ms}}^{-2}\leftarrow \boxed{\begin{array}{l}\text{The moving object is moving}\\ \text{ at a uniform speed}\end{array}}\end{array}$

(c)