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

Given that,

$F\left(t\right)=F_0{e}^{-bt}$
Using newton's second law of motion,
$F\left(t\right)=m\frac{dv}{dt}$

$\Rightarrow m\frac{dv}{dt}=F_0{e}^{-bt}$

$\Rightarrow \frac{dv}{dt}=\frac{F_0}{m}{e}^{-bt}$

Rearranging and integrating the above equation,

${\int }_0^{v}dv=\frac{F_0}{m}{\int }_0^t{e}^{-bt}$

$\Rightarrow v=\frac{F_0}{m}{\left[\frac{e^{-bt}}{-b}\right]}_0^t=\frac{F_0}{mb}[-(e^{-bt}-e^{-0})]$

$\Rightarrow v=\frac{F_0}{mb}[1-e^{-bt}]$

At $t=0$,

$v=\frac{F_0}{mb}[1-e^{-b\times 0}]=\frac{F_0}{mb}[1-1]=0$

At, $t=\infty$

$v=\frac{F_0}{mb}[1-e^{-b\times \infty }]=\frac{F_0}{mb}[1-0]=\frac{F_0}{mb}$

On plotting the given data we get,


Hence, $\left(B\right)$ is the correct answer.