The displacement x of a particle varies with time t as x = ae -α t + be β t, where a, b, α and β are positive constants. The velocity of the particle will (a) be independent of β (b) drop to zero when α = β (c) go on decreasing with time (d) go on increasing with time.

Answer: (d). v = -aα e-α t + bβ eβ t dv/dt = aα² e-α t + bβ² eβ t > 0 for all t Acceleration is always positive, so v keeps increasing. So the answer is (d).

The displacement x of a particle varies with time t as x = ae -α t +… | NEET 2005 (AIPMT)

Q44NEET 2005 · AIPMTMedium

Velocity, Acceleration & Variable Motion

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

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Answer: (d)

$v = -a\alpha e^{-\alpha t} + b\beta e^{\beta t}$
$\dfrac{dv}{dt} = a\alpha^2 e^{-\alpha t} + b\beta^2 e^{\beta t} > 0$ for all $t$
Acceleration is always positive, so $v$ keeps increasing.
So the answer is (d).

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