Question

If velocity $v$ of a particle as a function of position $x$ is given as $v=(\alpha -\beta x)$ where at $t=0,x=0$. Then,

• A. Maximum displacement of particle is $\frac{\alpha }{\beta }$.
• B. Maximum displacement of particle is $\infty$.
• C. It takes $\infty$ time for particle to stop.
• D. Time taken to reduce to half of initial velocity is $\frac{ln\left(2\right)}{\beta }$.

Answer 1

Answer: (A), (C), (D)

Time taken to reduce to half of initial velocity is $\frac{ln\left(2\right)}{\beta }$.

$v=\alpha -\beta x\Rightarrow \frac{dx}{dt}=\alpha -\beta x\Rightarrow {\int }_0^x\frac{dx}{\alpha -\beta x}={\int }_0^{t}dt$
$x=\frac{\alpha }{\beta }(1-e^{-\beta t})$.
$\therefore$ The maximum value of $x$ is $\frac{\alpha }{\beta }$, and it happens at $t=\infty$. Since, displacement is maximum at $t=\infty$, its velocity should be zero.
$v=\frac{dx}{dt}=-\alpha e^{-\beta t}{v}_0=-\alpha$
Say at $t=t_{1/2},v=v_0/2$
$\Rightarrow \frac{v_0}{2}=-\alpha e^{-\beta t_{1/2}}\Rightarrow \frac{v_0}{2}=-v_0{e}^{-\beta t_{1/2}}\Rightarrow t_{1/2}=\frac{ln2}{\beta }$