Question

Acceleration versus velocity graph of a particle moving in a straight line starting from rest is as shown in figure. The corresponding velocity-time graph would be:

• A.
• B.
• C.
• D.

Answer 1

The correct option is D

We, know that $$a=\frac{dv}{dt}...\left(1\right)$$ From the given $a-v$ graph, we have,

$a=-mv+C...\left(2\right)$
$m$ & $C$ are positive constants.

So, from equation $\left(1\right)$ and $\left(2\right)$, we have
$-mv+C=\frac{dv}{dt}$

Integrating on both sides we get,

${\int }_0^{t}dt={\int }_0^v\frac{dv}{C-mv}$

$\Rightarrow -\frac{1}{m}[\ln |C-mv|{]}_0^v=t$

$\Rightarrow \ln \frac{C-mv}{C}=-mt$

$\Rightarrow \frac{C-mv}{C}=\exp (-mt)$

$\Rightarrow C-mv=C\exp (-mt)$

$\Rightarrow mv=C(1-\exp (-mt)$

$\therefore v=\frac{C(1-\exp (-mt))}{m}$

Hence, the velocity varies exponentially with time.
When, $t=\infty$, $v=v_{max}=\frac{C}{m}$ and the curve gets flatten. This is possible only for option (d).


Hence, option (d) is the correct alternative.