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A point moves...

Question

A point moves with uniform acceleration and $v_{1}, v_{2}, v_{3}$ denote the average velocities in three successive intervals of time $t_{1}, t_{2}, t_{3}$ . Then, the relation $\frac{v_{1} - v_{2}}{v_{2} - v_{3}}$ is

\frac{t_{1} - t_{2}}{t_{2} + t_{3}}

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\frac{t_{1} + t_{2}}{t_{2} + t_{3}}

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\frac{t_{1} - t_{2}}{t_{1} + t_{3}}

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\frac{t_{1} - t_{2}}{t_{2} - t_{3}}

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Solution

The correct option is B $\frac{t_{1} + t_{2}}{t_{2} + t_{3}}$
Let $u$ be the initial velocity.
$∴ v_{1}^{'} = u + a t_{1}, v_{2}^{'} = u + a (t_{1} + t_{2})$ and $v_{3}^{'} = u + a (t_{1} + t_{2} + t_{3})$
Now $v_{1} = \frac{u + v_{1}^{'}}{2} = \frac{u + (u + a t_{1})}{2} = u + \frac{1}{2} a t_{1}$
$v_{2} = \frac{v_{1}^{'} + v_{2}^{'}}{2} = u + a t_{1} + \frac{1}{2} a t_{2}$
$v_{3} = \frac{v_{2}^{'} + v_{3}^{'}}{2} = u + a t_{1} + a t_{2} + \frac{1}{2} a t_{3}$
So $, v_{1} - v_{2} = - \frac{1}{2} a (t_{1} + t_{2})$ and $v_{2} - v_{3} = - \frac{1}{2} a (t_{2} + t_{3})$
$∴ (v_{1} - v_{2}) : (v_{2} - v_{3}) = (t_{1} + t_{2}) : (t_{2} + t_{3})$

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