The displacement x of a particle varies with time according to the relation $x = \frac{a}{b} (1 - e^{- bt})$ . Then

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The velocity and acceleration of the particle at t = 0 are a and –ab respectively

Both 1 and 2

At t = 1/b, the displacement of the particle is (3/5) (a/b)

At t = 1/b, the displacement of the particle in nearly (1/3) (a/b)

answer is B.

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Detailed Solution

Velocity $= \frac{dx}{dt} = \frac{a}{b} [1 - e^{- bt} (- b)]$
$v = \frac{a}{b} [{be}^{- bt}]$ ; At t = 0; v = a
Acceleration $= \frac{dv}{dt} = \frac{a}{b} [{be}^{- bt} (- b)]$
At t = 0; Acceleration = – ab

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