Question

The acceleration versus velocity graph of a particle moving in a straight line starting from rest is as shown in the figure.

The corresponding velocity-time graph may be

• A.
• B.
• C.
• D.

Answer 1

The correct option is D

From the graph, $a=-kv+c$ ($k$, $c$ >0)
$\Rightarrow \frac{dv}{dt}=-kv+c$
$\Rightarrow \int \frac{dv}{-kv+c}=\int dt$
$\Rightarrow -\frac{1}{k}ln(-kv+c)=t+c^{\prime }$
On rearranging,
$v=\frac{c}{k}-\frac{e^{-k(t+c^{\prime })}}{k}$
$v=-K^{\prime }{e}^{-kt}+C^{\prime }$
where $K^{\prime }=\frac{e^{-k{c}^{\prime }}}{k},C^{\prime }=\frac{c}{k}$
Only (d) option may be correct if $K^{\prime }=C^{\prime }$