Question

The displacement of a particle varies with time t, $x=a{e}^{-at}+b{e}^{\beta t}$ 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

$x=a{e}^{-\alpha t}+b{e}^{\beta t}$
Velocity v = $\frac{dx}{dt}=\frac{d}{dt}\left(a{e}^{-\alpha t+b{e}^{\beta t}}\right)$
$=a{e}^{-\alpha t}(-\alpha )+b{e}^{\beta t}\left(\beta \right)=-a\alpha e^{-\alpha t}+b\beta e^{\beta t}$

From the above equation itself we can tell that velocity will increase with time since the negative component will decrease with increase in time and positive component will increase in the equation given equation.

Let us find acceleration also.
Acceleration = $-a\alpha e^{-\alpha t}(-\alpha )+b\beta e^{\beta t}.\beta$
$=a{\alpha }^2{e}^{-\alpha t}+b{\beta }^2{e}^{\beta t}$
Acceleration is positive so velocity goes on increasing with time.