Question

Figure below shows a velocity-time graph for a car starting from rest. The graph has three parts AB, BC and CD. Compare the distance travelled in part BC with the distance travelled in part AB.

Answer 1

The area enclosed by the velocity-time graph gives the distance travelled in parts BC and AB. Distance travelled in part BC
= Area of the rectangle
$=$ Length $\times$ breadth
$=(2t-t)\times v_0$
$=v_{0}t$
Distance travelled in part AB
$=$ Area of the first triangle
$=\frac{1}{2}\times \text{base}\times \text{height}$

$=\frac{1}{2}\times t\times v_0$

$=\frac{1}{2}{v}_{0}t$

Hence, distance travelled in part BC is two times of the distance travelled in part AB.