Question

The motion of a particle is described by $x=x_0(1-e^{-kt})$; $t\ge 0$; $x_0>0$, k>0. With what velocity does the particle start?

• A. $\frac{x_0}{k}$
• B. $x_{0}k$
• C. $\frac{k}{x_0}$
• D. $2x_{0}k$

Answer 1

The correct option is B $x_{0}k$

Given:

$x=x_0(1-e^{-kt})$

$\frac{dx}{dt}=x_0\left(K{e}^{-Kt}\right)$

initial velocity is at $t=0$

$v=\frac{dx}{dt}{\mid }_{t=0}=x_{0}K{e}^{-K0}$

$v=K{x}_0$