**Question 4**:

The diagram above shows the speed-time graph of a moving particle for 10 seconds. From the graph above, find

**(a)**the total distance travel by the particle for the whole journey.

**(b)**the average speed for the whole journey.

*Solution:***(a)**

Total distance travelled

= Area under the speed-time graph

= Area of triangle

= ½ × 15 × 10

= **75 m**

**(b)**

Average speed for the whole journey

**$\begin{array}{l}=\frac{\text{Total distance travelled}}{\text{Total time taken}}\\ =\frac{75}{10}\\ =7.5\text{}m{s}^{-1}\end{array}$**

**Question 5**:

The diagram above shows the speed-time graph of a moving particle for 12 seconds. Find

**(a)**the length of the time, in

*s*that the particle move with uniform speed.

**(b)**

**the distance travel by the particle when it move with constant speed.**

**(c)**the distance travel by the particle when the rate of the speed change is negative.

Solution:Solution:

**(a)**

Length of the time the particle move with uniform speed

= 10 – 6

=

**4***s*

**(b)**

Distance travel by the particle when it move with constant speed

= Area under the speed-time graph

= Area of rectangle

= 4 × 10

=

**40 m****(c)**

Distance travel by the particle when the rate of the speed change is negative

= Area under the speed-time graph for the first 6 s

= Area of trapezium

= ½ (10 + 25)(6)

= **105 m**

**Question 6**:

Diagram below shows the speed-time graph for the movement of an object for a period of 40 seconds.

(a) State the duration of time, in s, for which the object moves with uniform speed.

(b) Calculate the rate of change of speed, in ms

^{-2}, of the object for the last 12 seconds.

(c) Calculate the value of

*v*, if the total distance travelled for the period of 40 seconds is 500 m.

**(a) Duration of time the object moves with uniform speed = 28s – 10s = 18s**

*Solution:*

$\begin{array}{l}\text{(b)}\\ \text{Rateofchangeofspeed}\\ =-\frac{15}{12}{\text{ms}}^{-2}\\ =-1.25{\text{ms}}^{-2}\\ \\ \text{(c)}\\ \text{AreaoftrapeziumI+AreaoftrapeziumII}=500\\ \frac{1}{2}\left(v+15\right)10+\frac{1}{2}\left(18+30\right)15=500\\ \text{}5v+75+360=500\\ \text{}5v=65\\ \text{}v=13\end{array}$