Question

Corresponding to given $\text{v-s graph}$ of a particle moving in a straight line, Which of the following is the correct $\text{a-s graph}$$($$s$ is displacement, $v$ is velocity and $a$ is acceleration $)$.

• A.
• B.
• C.
• D.

Answer 1

The correct option is D

The given $\text{v-s graph}$ is a straight line with positive slope $($say $m)$ and positive $\text{y-intercept}$ $($say $c)$.

Therefore, we can write $v-s$ equation as,
$v=ms+c.....\left(1\right)$

on differentiating above equation, we get,

$\frac{dv}{ds}=m$

Multiplying both sides by $v$,

$\Rightarrow v\frac{dv}{ds}=vm$

$\Rightarrow a=(ms+c)m$, ( from $1$ and $v\frac{dv}{ds}=a$)

Rearranging the above equation,

$\Rightarrow a=m^{2}s+mc$

Thus we can conclude that, $\text{a-s}$ equation is a linear equation representing a straight line with positive slope $(=m^2)$ and positive $\text{y-intercept}$ $(=mc$).

The $\text{a-s graph}$ is as shown below.

Hence, option $\left(d\right)$ is correct option.