Solved: PHYSICS Question for NEET

NEET PHYSICSMedium

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:

Go on decreasing with time

Be independent of $\alpha$ and $\beta$

Drop to zero when $\alpha = \beta$

Go on increasing with time

Step-by-Step Solution

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 $t$ : $v = \frac{d}{dt}(ae^{-\alpha t}) + \frac{d}{dt}(be^{\beta t})$ $v = -a\alpha e^{-\alpha t} + b\beta e^{\beta t}$ .
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 $t$ : $a_{acc} = \frac{d}{dt}(-a\alpha e^{-\alpha t}) + \frac{d}{dt}(b\beta e^{\beta t})$ $a_{acc} = -a\alpha(-\alpha)e^{-\alpha t} + b\beta(\beta)e^{\beta t}$ $a_{acc} = a\alpha^2 e^{-\alpha t} + b\beta^2 e^{\beta t}$ .
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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