**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

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

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

**Diagram shows the speed-time graph for the movement of two particles,**

**Question**10:*J*and

*K*, for a period of

*t*s. The graph

*ABCD*represents the movement of

*J*and the graph

*AE*represents the movement of

*K*. Both particles start at the same point and move along the same route.

(a) State the uniform speed, in

*ms*^{-1}, of particle*J*.(b) Calculate the rate of change of speed, in

*ms*^{-2}, of particle*J*for the first 13 s.(c) At

*t*s, the difference between the distance travelled by*J*and*K*is 169*m*. Calculate the value of*t*.[ 6 marks ]

*Solution:***(a)**

Uniform speed of particle

*J*=**26***ms*^{-1}

(b)

(b)

Rate of change of speed of particle

$=\frac{26-0}{13-0}=2m{s}^{-2}$
*J*for the first 13 s**(c)**

Given at

*t*s, the difference between the distance travelled by*J*and*K*is 169(distance travelled by particle

*J*) – (distance travelled by particle*K*) = 169[ ½ (

*t*– 13 +*t*) × 26 ] – [ ½ (26) (*t*)] = 169[ 13 (2

*t*– 13) ] – 13*t*= 169( 26

*t*– 169 – 13*t*) = 16913

*t*= 338

*t*= 26