Question

A particle of mass $m$ is at rest at the origin at time $t=0$. It is subjected to a force $F\left(t\right)=F_0{e}^{-bt}$ in the x-direction .Its speed $v\left(t\right)$ is depicted by which of the following curves ?

• A.
• B.
• C.
• D.

Answer 1

The correct option is B

We will try to solve this question by using some common sense.
$F=F_0{e}^{-bt}$ will always be a positive quantity as exponential function always remains positve.
It means that the acceleration will always remain in positive x- direction.

If the acceleration is in +ve x- direction and the body is starting from rest then the velocity will also be in +ve x-direction.
The velocity will keep increasing as there is no reason for the velocity to change its direction . So, the velocity will keep increasing in +ve x-direction.

From this we can easily eliminate options (a) and (d)
Now the only difference that lies between option (c) and (d) is its peak value on the graph.
We can compare them by dimensional analysis
$F=F_0{e}^{-bt}$ Here $F$ and $F_0$ both are forces which means $e^{-bt}$ is an dimensionless quantity.
$⟹b=\frac{1}{t}$
and its unit is $se{c}^{-1}$
This means that $\frac{F_{0}b}{m}$ is dimensionally incorrect.
$\therefore$ Option B is correct