The displacement $x$ of a particle varies with time $t$ as $x = ae^{-\alpha t} + be^{\beta t}$, where $a, b, \

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:

Accepted Answer

1.  **Find Velocity ($v$):** Velocity is the rate of change of displacement ($v = \frac{dx}{dt}$) .
    Given $x = ae^{-\alpha t} + be^{\beta t}$.
    Differentiating with respect to $t$:
    $v = \frac{d}{dt}(ae^{-\alpha t}) + \frac{d}{dt}(be^{\beta t})$
    $v = -a\alpha e^{-\alpha t} + b\beta e^{\beta t}$.
2.  **Analyze the Change in Velocity:** To determine if the velocity is increasing or decreasing, we look at the acceleration ($a_{acc} = \frac{dv}{dt}$) .
    Differentiating $v$ with respect to $t$:
    $a_{acc} = \frac{d}{dt}(-a\alpha e^{-\alpha t}) + \frac{d}{dt}(b\beta e^{\beta t})$
    $a_{acc} = -a\alpha(-\alpha)e^{-\alpha t} + b\beta(\beta)e^{\beta t}$
    $a_{acc} = a\alpha^2 e^{-\alpha t} + b\beta^2 e^{\beta t}$.
3.  **Conclusion:** Since $a, b, \alpha, \beta$ are positive constants and the exponential function $e^{x}$ is always positive for any real $x$, both terms in the acceleration equation are positive. Therefore, the acceleration $a_{acc}$ is always positive ($a_{acc} > 0$). A positive acceleration implies that the velocity is constantly increasing with time.

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