Question

A train moving with constant acceleration takes $t$ seconds to pass a certain fixed point and the front and back end of the train passes the fixed point with velocities $u$ and $v$ respectively. Show that the length of the train is $\frac{1}{2}(u+v)t$.

Answer 1

Since, $v=u+at$
$\Rightarrow a=\frac{v-u}{t}$

and $v^2=u^2+2as$

$\Rightarrow s=\frac{v^2-u^2}{2a}$

$=\frac{(v+u)(v-u)}{2a}$

$=\frac{at(v+u)}{2a}$

$=\frac{u+v}{2}\cdot t=\frac{1}{2}(u+v)t$