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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
Be independent of $\alpha$ and $\beta$
B
Go on increasing with time
C
Drop to zero when $\alpha = \beta$
D
Go on decreasing with time
Detailed Solution
$x = ae^{-\alpha t} + be^{\beta t}$
Velocity: $v = \dfrac{dx}{dt} = -a\alpha e^{-\alpha t} + b\beta e^{\beta t}$
It depends on $\alpha$ and $\beta$.
Acceleration: $\dfrac{dv}{dt} = a\alpha^2 e^{-\alpha t} + b\beta^2 e^{\beta t}$
All the constants are positive and exponentials are always positive, so $\dfrac{dv}{dt} \gt 0$ always.
Hence $v$ is an increasing function of $t$: the velocity goes on increasing with time.
Setting $\alpha = \beta$ does not make $v$ zero in general.
Velocity: $v = \dfrac{dx}{dt} = -a\alpha e^{-\alpha t} + b\beta e^{\beta t}$
It depends on $\alpha$ and $\beta$.
Acceleration: $\dfrac{dv}{dt} = a\alpha^2 e^{-\alpha t} + b\beta^2 e^{\beta t}$
All the constants are positive and exponentials are always positive, so $\dfrac{dv}{dt} \gt 0$ always.
Hence $v$ is an increasing function of $t$: the velocity goes on increasing with time.
Setting $\alpha = \beta$ does not make $v$ zero in general.