Question

The rectilinear motion of a particular is defined by the displacement time equation

x = $x_0[2e^{-kt}-e-2kt]$

Where $x_0$ is the initial displacement, K is a constant. What will be the maximum velocity of the particle?

• A. $-\frac{3}{2}k{x}_0$
• B. $-\frac{1}{2}k{x}_0$
• C. $\frac{1}{2}k{x}_0$
• D. $\frac{3}{2}k{x}_0$

Answer 1

The correct option is B $-\frac{1}{2}k{x}_0$

x = $x_0[2e^{-kt}-e-2kt]$

$v=\frac{dx}{dt}=-2k{x}_0[e^{-kt}-e^{-2kt}]$

$\frac{dv}{dt}=-2k{x}_0[-k{e}^{-kt}+2k{e}^{-2kt}]$

$-k{e}^{-kt}+2k{e}^{-2kt}=0$

$2e^{-kt}=1$

$-kt=In\frac{1}{2}$

$t=-\frac{1}{k}ln\frac{1}{2}=\left(\frac{ln2}{k}\right)$

$v=-2k{x}_0[e^{-kx\frac{ln2}{k}}-e^{-2kx\frac{ln2}{k}}]$

$=-2k{x}_0[e^{-ln2}-e^{-2ln2}]=-2k{x}_0[2^{-1}-e^{ln{2}^2}]$

$=-2k{x}_0[\frac{1}{2}-\frac{1}{4}]$

$=-2k{x}_0\left[\frac{2-1}{4}\right]=-2k{x}_0\left[\frac{1}{4}\right]=-\frac{k{x}_0}{2}$

$=-\frac{1}{2}k{x}_0$