Question

A train starting from rest is accelerated and the instantaneous acceleration is given by $\frac{10}{v+1}m/s^2$. Where $v$ is the velocity in $m/s$. Find the distance in which the train attains a velocity of $54km/hr$.

• A. $123.75m$
• B. $112.5m$
• C. $11.25m$
• D. $77.25m$

Answer 1

The correct option is A $123.75m$

Given $v=54km/hr=15m/s$,

$a=\frac{10}{v+1}$ $\Rightarrow \frac{dv}{dt}=\frac{10}{v+1}$

$\Rightarrow \frac{dv}{dx}\frac{dx}{dt}=\frac{10}{v+1}$

$\Rightarrow v\frac{dv}{dx}=\frac{10}{v+1}$

$\Rightarrow {\int }_0^{15}v(v+1)dv={\int }_0^{x}10dx$

$\Rightarrow (\frac{v^3}{3}+\frac{v^2}{2}){|}_0^{15}=10x$ $\Rightarrow x=123.75$ $m$