**Question 1**:

The diagram above shows the distance-time graph of a moving particle for 5 seconds. Find

**(a)**the distance travel by the particle from the time 2 second to 5 second.

**(b)**

**the speed of the particle for the first 2 seconds.**

Solution:Solution:

**(a)**

Distance travel by the particle from the time 2 second to 5 second

= 20 – 15

=

**5 m**

(b)

(b)

The speed of the particle for the first 2 seconds

= Gradient

$\begin{array}{l}=\frac{15-0}{2-0}\\ =7.5{\text{ms}}^{-1}\end{array}$**Question 2**:

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

**(a)**

**the value of v, if the average speed of the car for the first 6 seconds is 2 ms**

^{-1}.

**(b)**

**average speed of the car for the first 8 seconds.**

*Solution:***(a)**

$\begin{array}{l}{\text{Averagespeedofthecarforthefirst6secondsis2ms}}^{\text{-1}}\\ \frac{\text{Totaldistancetravelled}}{\text{Totaltimetaken}}=2\\ \frac{v}{6}=2\\ v=12\end{array}$

**(b)**

$\begin{array}{l}\text{Averagespeedofthecarforthefirst8seconds}\\ =\frac{15}{8}\\ =1.875\text{}m{s}^{-1}\end{array}$

**Question 3**:

Diagram below shows the distance-time graph for the journey of a train from one town to another for a period of 90 minutes.

(a) State the duration of time, in minutes, during which the train is stationery.

(b) Calculate the speed, in km h

^{-1}, of the train in the first 40 minutes.

(c) Find the distance, in km, travelled by the train for the last 25 minutes.

**(a) Duration the train is stationery = 65 – 40 = 25 minutes**

*Solution:*

$\begin{array}{l}\text{(b)}\\ \text{Speedofthetraininthefirst40minutes}\\ =\frac{150-90\text{km}}{40\text{minutes}}\\ =\frac{60\text{km}}{\frac{40}{60}\text{h}}\\ =90\text{km/h}\end{array}$

(c) 90 – 0 = 90 km

Want more activities in this chapter

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