Question

Fig. 2.42 given below shows a velocity-time graph for a car starting from rest.The graph has three parts AB,BC and CD.

(i) State how is the distance travelled in any part is determined from this graph.

(ii) Compare the distance travelled in part BC with the distance travelled in part AB.

(iii) Which part of graph -shows motion with uniform (a) velocity (b) acceleration (c) retardation ?

(iv) (a) Is the magnitude of acceleration higher or lower than that of retardation ? Give a reason.(b) Compare the magnitude of acceleration and retardation.

Answer 1

(i) Distance travelled in any part of the graph can be determined by finding the area enclosed by the graph in that part with the time axis.

(ii) Distance travelled in part BC = Area of the rectangle tBC2t = base × height.
= (2t – t) × v_o
= v_ot

Distance travelled in part AB = Area of the triangle ABt
= (1/2) × base × height
= (1/2) × t × v_o
= (1/2) v_o t
Therefore, distance travelled in part BC:distance travelled in part AB :: 2:1.

(iii) (a) BC shows motion with uniform velocity.
(b) AB shows motion with uniform acceleration.
(c) CD shows motion with uniform retardation.

(iv) (a) The magnitude of acceleration is lower as the slope of line AB is less than that of line CD.
(b) Slope of line AB = $\frac{v_o}{t}$
Slope of line CD = $\frac{v_o}{0.5t}$
$\frac{SlopeoflineAB}{SlopeoflineCD}=\frac{\left(vo/t\right)}{\frac{(vo}{0.5t)}}$$\frac{SlopeoflineAB}{SlopeoflineCD}$ ::$\frac{1}{2}$