Question

A particle, initially at rest is subject to two forces; one is constant, the other is a retarding force proportional to the particle velocity. In the subsequent motion of the particle,

• A. the acceleration will increase from zero to a constant value
• B. the acceleration will remain constant with respect to time
• C. the velocity will increase from zero to a maximum and then decrease
• D. the velocity will increase from zero to a constant value

Answer 1

The correct option is D the velocity will increase from zero to a constant value

$F_1=\text{constant}$ and $F_2\propto v$
$\Rightarrow F_2=-kv$
also $(\overrightarrow{F_1}+\overrightarrow{F_2})=m\overrightarrow{a}$ $\therefore F_1-kv=\frac{mdv}{dt}$
$\therefore a=\left(\frac{F_1-kv}{m}\right)$ decrease from its initial value.
Now, $\int \frac{dv}{F_1-kv}=\int \frac{dt}{m}$
$\Rightarrow \frac{ln(F_1-kv)}{-k}=\frac{1}{m}t+c$
$\because$ at $t=0,v=0\therefore \frac{-ln{F}_1}{k}=c$
$\therefore \frac{ln(F_1-kv)}{-k}=\frac{t}{m}-\frac{ln{F}_1}{k}$
$\therefore -\frac{1}{k}\left[ln\right(F_1-kv)-ln{F}_1]=\frac{t}{m}$
i.e., $-\frac{1}{k}\left(ln\left(\frac{F_1-kv}{F_1}\right)\right)=\frac{t}{m}$
$\Rightarrow ln\left(\frac{F_1-kv}{F_1}\right)=-\frac{kt}{m}\therefore \frac{F_1-kv}{F_1}=e^{-kt/m}$
$1-\frac{kv}{F_1}=e^{-kt/m}$
$1-e^{-kt/m}=\frac{kv}{F_1}$
$v=\frac{F_1}{k}(1-e^{-kt/m})$
i.e. velocity will increase from zero to a constant value.