Question

A particle of mass $\frac{10}{7}Kg$ is moving in the positive direction of $x$. Its initial position $x=0$ & initial velocity is $1m/s$. The velocity at $x=10$ is (in $m/s$ )

Answer 1

Answer: 4.00

$P=m{V}^2\frac{dV}{dx}$
$P=2+\frac{2}{10}\times x$
$P=2+\frac{x}{5}$
$P=mV\frac{dV}{dx}\times V$
$=\frac{10}{7}\times V^2\frac{dV}{dx}$
$\frac{10}{7}\underset{1}{\overset{v}{\int }}{V}^{2}dV=\underset{x=0}{\overset{x=10}{\int }}\left(2+\frac{x}{5}\right)dx$
$\frac{10}{7}\left[\frac{V^3-1}{3}\right]=[2x{]}_0^{10}+\frac{1}{5}{\left[\frac{x^2}{2}\right]}_0^{10}$
$\frac{10}{21}\left[V^3-1\right]=20+\frac{1}{10}\times 100$
$V^3-1=30\times \frac{21}{10}$
$V^3=64$
$V=4m/\sec .$