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(D) Given displacement $x = a e^{-\alpha t} + b e^{\beta t}$.Velocity $v$ is the rate of change of displacement with respect to time:$v = \frac{dx}{dt

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Go on increasing with time

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(D) Given displacement $x = a e^{-\alpha t} + b e^{\beta t}$.Velocity $v$ is the rate of change of displacement with respect to time:$v = \frac{dx}{dt} = \frac{d}{dt}(a e^{-\alpha t} + b e^{\beta t})$$v = -a\alpha e^{-\alpha t} + b\beta e^{\beta t}$.To determine how velocity changes with time,we find the acceleration $a_{acc}$:$a_{acc} = \frac{dv}{dt} = \frac{d}{dt}(-a\alpha e^{-\alpha t} + b\beta e^{\beta t})$$a_{acc} = a\alpha^2 e^{-\alpha t} + b\beta^2 e^{\beta t}$.Since $a, b, \alpha, \beta$ are positive constants and the exponential functions $e^{-\alpha t}$ and $e^{\beta t}$ are always positive for any time $t$,the acceleration $a_{acc}$ is always positive.Because the acceleration is positive,the velocity $v$ will go on increasing with time.

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

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