Question

Acceleration $(a)$ -displacement( $s$ ) graph of a particle moving in a straight line is shown in adjacent figure. The initial velocity of the particle is zero. The $v-s$ graph of the particle would be

• A.
• B.
• C.
• D.

Answer 1

Answer: (D)

Acceleration, $a=\frac{d v}{d t}=\frac{d v}{d s} \times \frac{d s}{d t}=v \frac{d v}{d s}$
From given graph $a=k s$
So $\int\limits_{0}^{v} v d v=k \int\limits_{0}^{s} s d s$
or $ \frac{v^{2}}{2}=\frac{1}{2} k s^{2}$
or $ v=\sqrt{k} s $ i.e. $ v \propto s$
So $v-s$ graph is a straight line.