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. It's speed $v\left(t\right)$ is depicted by which of the following curves?

• A.
• B.
• C.
• D.

Answer 1

The correct option is C

Given, $F=F_0{e}^{-bt}$
$\Rightarrow ma=F_0{e}^{-bt}$
$\Rightarrow m\frac{dv}{dt}=F_0{e}^{-bt}$
$\Rightarrow {\int }_0^{v}dv=\frac{F_0}{m}{\int }_0^t{e}^{-bt}dt$
($\because v=0$ at $t=0$)
$\Rightarrow v=-\frac{F_0}{mb}[e^{-bt}{]}_0^t=\frac{-F_0}{mb}[e^{-bt}-1]$
or, $v=\frac{F_0}{mb}[1-e^{-bt}]$
Now from the boundary conditions, we have
At $t=0,v=0$,
and as $t\to \infty ,v=\frac{F_0}{mb}$
Option C is the matching one.