Question

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

• A. Decrease with time
• B. Be independent of $\alpha$ and $\beta$
• C. Drop to zero when $\alpha =\beta$
• D. Increase with time

Answer 1

Answer: (D)

Increase with time

Given,
$x={ae}^{-\alpha t}+{be}^{\beta t}$
We know that, velocity, $v=\frac{dx}{dt}$
$\frac{d}{dt}({ae}^{-\alpha t}+{be}^{\beta t})$
$=-a\alpha e^{-\alpha t}+b\beta e^{\beta t}$
$=A+B$
where, $A=-a\alpha e^{-\alpha t},B=b\beta e^{\beta t}$
The value of term $A=-a\alpha e^{-\alpha t}$ increases and of term $B=b\beta e^{-\beta t}$ increases with time. As a result, velocity goes on increasing with time.