**Question 7**:

The diagram above shows the speed-time graph of a moving object for 15 seconds. Find

**(a)**

**the speed of the object at**

*t*= 9s.

**(b)**the distance travel by the object for the first 12 seconds.

*Solution:***(a)**

The speed of the object at

*t*= 9s is**6 ms**^{-1}

(b)

(b)

= Area under the speed-time graph

= Area of triangle

= ½ × 12 × 8

=

**48 m**

**Question 8**:

The diagram above shows the speed-time graph of a moving object for 15 seconds.

**(a)**Find the time interval when the object moves with constant speed.

**(b)**Find the acceleration from

*t*= 6 s to

*t*= 12 s.

**(c)**State the time when the object is stationary.

Solution:Solution:

**(a)**

Time interval when the object moves with constant speed

= 12 – 6

=

**6 s**

**(b)**

(c)

(c)

Time when the object is stationary is at

**15s**.

**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}.

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

*Solution:*

$\begin{array}{l}\text{(b)}\\ \text{Rateofchangeofspeedforthefirst8seconds}\\ =\frac{10-6}{0-8}\\ =-\frac{4}{8}\\ =-\frac{1}{2}{\text{ms}}^{-2}\end{array}$

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