In the velocity-time graph shown below, the ratio of the distance travelled by the object in the last 2 s and the distance travelled in 7 s is :

1. $\dfrac{1}{2}$

2. $\dfrac{1}{4}$

3. $\dfrac{1}{3}$

4. $\dfrac{2}{3}$

$\dfrac{1}{4}$

Reason — Distance covered in last 2 s = area of triangle = $\dfrac{1}{2}$ x 2 x 10 = 10 m

Distance covered in last 7 s = area of trapezium = $\dfrac{1}{2}$ x 10 x (2 + 6) = 40 m

Hence, ratio of of the distance travelled by the object in the last 2 s and the distance travelled in 7 s is = $\dfrac{10}{40}$ = $\dfrac{1}{4}$