Question

The position $x$ of a particle varies with time $t$ as $x=a{e}^{-\alpha t}+b{e}^{\beta t}$, where $a,b,\alpha$ and $\beta$ are positive constants. The velocity of the particle will

• A. go on decreasing with time.
• B. be independent of $\alpha$ and $\beta$ .
• C. drop to zero when $\alpha =\beta$.
• D. go on increasing with time.

Answer 1

The correct option is D go on increasing with time.

Differentiate the displacement $x=a{e}^{-\alpha t}+b{e}^{\beta t}$, we get
$v=\frac{dx}{dt}=-a\alpha e^{-\alpha t}+b\beta e^{\beta t}$
Differentiate the velocity to get the acceleration
$\frac{dv}{dt}=a{\alpha }^2{e}^{-\alpha t}+b{\beta }^2{e}^{\beta t}$
$\Rightarrow \frac{dv}{dt}=\frac{a{\alpha }^2}{e^{\alpha t}}+b{\beta }^2{e}^{\beta t}$
From the equation above , it is easy to see that $\frac{dv}{dt}>0$ for all $t\ge 0$ as the two positive terms are added. Thus , velocity $v$ will keep on increasing with time.