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Question: The motion of a particle is described \(byx = x_{0}(1 - e^{- kt});t \geq 0,x_{0} > 0,k > 0.\) With ...

The motion of a particle is described

byx=x0(1−e−kt);t≥0,x0>0,k>0.byx = x_{0}(1 - e^{- kt});t \geq 0,x_{0} > 0,k > 0. With what velocity does the particle start?

A

x0k\frac{x_{0}}{k}

B

x0kx_{0}k

C

kx0\frac{k}{x_{0}}

D

2x0k2x_{0}k

Answer

x0kx_{0}k

Explanation

Solution

Given : x =x0(1−e−kt)x_{0}(1 - e^{- kt})

Velocity, v=dxdt=x0ke−ktv = \frac{dx}{dt} = x_{0}ke^{- kt}

At t = 0, v=x0kv = x_{0}k