Question

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

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

Answer 1

The correct option is D go on increasing with time

It is given that
$x=a{e}^{-\alpha t}+b{e}^{\beta t}v=\frac{dx}{dt}=-a\alpha e^{-\alpha t}+b\beta e^{\beta t}a=\frac{dv}{dt}=a{\alpha }^2{e}^{-\alpha t}+b{\beta }^2{e}^{\beta t}$

Clearly acceleration is positive.
Thus, velocity will increase with time.