Question

The displacement $x$ 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. 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

Answer: (D)

go on increasing with time

Given, $x=a{e}^{-\alpha t}+b{e}^{\beta t}V=\frac{dx}{dt}=-a\alpha e^{-\alpha t}+b\beta e^{\beta t}\Rightarrow V=\frac{-a\alpha }{e^{\alpha t}}+b\beta e^{\beta t}$
Thus, velocity will go on increasing