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Q. A particle moves with constant acceleration along a straight line. If $v_{1}$ , $v_{2}$ and $v_{3}$ are the average velocities in the three successive intervals $t_{1}$ , $t_{2}$ and $t_{3}$ of time, then the correct relation is

NTA AbhyasNTA Abhyas 2020Motion in a Straight Line

Solution:

If $v_{1}' v_{2}'$ and $ v_{3}'$ be the velocities at the ends of time intervals $t_{1}, \, t_{1}+t_{2} $ and $ t_{1}+t_{2}+t_{3}$ respectively, then $v_{1}'=u+at_{1}, \, v_{2}'=u+a\left(t_{1} + t_{2}\right), \, v_{3}'=u+a\left(t_{1} + t_{2} + t_{3}\right)$
$v_{1}=\frac{u + v_{1} '}{2}=\frac{u + u + a t_{1}}{2}=u+\frac{1}{2} \, at_{1}$
$v_{2}=\frac{v_{1}' \, + \, v_{2}'}{2}=\frac{\left(u + a t_{1}\right) + \left[u + a \left(t_{1} + t_{2}\right)\right]}{2}$
$=u+at_{1}+\frac{1}{2}at_{2}$
$v_{3}=\frac{u_{2}' + v_{3} '}{2}=\frac{u + a t_{1} + a t_{2} + u + a \left(t_{1} + t_{2} + t_{3}\right)}{2}$
Or $v_{3}=u+at_{1}+at_{2}+\frac{1}{2} \, at_{3}$
$v_{1}-v_{2}=u+\frac{1}{2} \, at_{1}-u-at_{1}-\frac{1}{2} \, at_{2}=-\frac{1}{2}a\left(t_{1} + t_{2}\right)$
$v_{2}-v_{3}=u+at_{1}+\frac{1}{2} \, at_{2}-u-at_{1}-at_{2}-\frac{1}{2}at_{3}$
$= \, -\frac{1}{2} \, at_{2}-\frac{1}{2} \, at_{3}=-\frac{1}{2} \, a \, \left(t_{2} + t_{3}\right)$
$\frac{v_{1} - \, v_{2}}{v_{2} - v_{3}}=\frac{t_{1} + t_{2}}{t_{2} + t_{3}}$