Question

A particle moves in a straight path such that its velocity is always $4$ times its acceleration. if its velocity at time $t$=0 is $2m/sec$, its velocity at $T=2sec,$ is

• A. $\sqrt{2e}m/\sec$
• B. $2\sqrt{e}m/\sec$
• C. $\frac{\sqrt{e}}{2}m/\sec$
• D. $\frac{2}{\sqrt{e}}m/\sec$

Answer 1

Answer: (B)

$2\sqrt{e}m/\sec$

Given,

$4\frac{dv}{dt}=v$

$4ln|v|=t+C$

Also given $v=2m/sec$ at $t=0$

or, $C=4ln|2|$

$4ln|\frac{v}{2}|=t$

At $t=2sec$

$ln|\frac{v}{2}|=\frac{1}{2}$

$v=2e^{0.5}=2\sqrt{e}$