If the engine of a long train moving with constant acceleration crosses a tree with velocity ' $u$ ' and the…
If the engine of a long train moving with constant acceleration crosses a tree with velocity ' $u$ ' and the last compartment of the train crosses the same tree with velocity ' $v$ ', then the velocity with which the middle compartment crosses the same tree is
$\frac{(v+u)}{2}$
$\frac{2 u v}{(u+v)}$
$\sqrt{\frac{\left(v^2+u^2\right)}{2}}$
$\sqrt{2\left(u^2+v^2\right)}$
Solution
Using equation of motion,
$\mathrm{v}^2-\mathrm{u}^2=2 \mathrm{as}$
let the length of train is L,
on rearranging $v^2-u^2=2 a L$
$a=\frac{v^2-u^2}{2 L}$ ....(1)
The velocity with which middle compartment crosses pain, is given by,
$\begin{aligned} & v_f^2-u^2=2 a S_f \\ & v_f^2=u^2+2 a\left(\frac{L}{2}\right)=u^2+a L\end{aligned}$
Substitute "a" from equation (1)
$v_f^2=u^2+\left(\frac{v^2-u^2}{2 L}\right) L$
$\therefore \mathrm{v}_{\mathrm{f}}=\sqrt{\frac{\mathrm{v}^2+\mathrm{u}^2}{2}}$