A particle is travelling with uniform acceleration of magnitude a- During successive time intervals of - t 1- - t 2 and - t 3- its average velocities are v 1- v 2 and v 3 respectively- Then-A- v 2- v 1 - t 2- t 1- v 3- v 2 - t 3- t 2- v 2- v 1 - t 2- t 1- v 3- v 2 - t 3- t 2- v 2- v 1 - t 2- t 1- v 3- v 2 - t 3- t 2D- v 2- v 1 - t 2- t 1- v 3- v 2 - t 3- t 2

The correct option is A (v_2 v_1)/(Δ t_2 Δ t_1) = nbsp;(v_3 v_2)/(Δ t_3 Δ t_2)Let u be the initial velocity Let t_1 = Δ t_1, t_2 = Δ t_1+Δ t_2 amp; t_ ...

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Standard XII

Physics

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A particle is...

Question

A particle is travelling with uniform acceleration of magnitude a. During successive time intervals of $Δ t_{1}, Δ t_{2}$ and $Δ t_{3}$ , its average velocities are $v_{1}, v_{2}$ and $v_{3}$ respectively. Then,

(v_{2} - v_{1}) / (Δ t_{2} - Δ t_{1}) = (v_{3} - v_{2}) / (Δ t_{3} - Δ t_{2})

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(v_{2} - v_{1}) / (Δ t_{2} + Δ t_{1}) = (v_{3} - v_{2}) / (Δ t_{3} + Δ t_{2})

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(v_{2} + v_{1}) / (Δ t_{2} - Δ t_{1}) = (v_{3} + v_{2}) / (Δ t_{3} + Δ t_{2})

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(v_{2} + v_{1}) / (Δ t_{2} - Δ t_{1}) = (v_{3} + v_{2}) / (Δ t_{3} - Δ t_{2})

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Solution

The correct option is A $(v_{2} - v_{1}) / (Δ t_{2} - Δ t_{1}) = (v_{3} - v_{2}) / (Δ t_{3} - Δ t_{2})$
Let $u$ be the initial velocity
Let $t_{1} = Δ t_{1}, t_{2} = Δ t_{1} + Δ t_{2}$ & $t_{3} = Δ t_{1} + Δ t_{2} + Δ t_{3}$

Since the motion is uniformly accelerated,
$v_{1} = \frac{u + u + a t_{1}}{2} = u + a (\frac{t_{1}}{2}) = u + a (\frac{Δ t_{1}}{2})$
$v_{2} = \frac{u + a t_{1} + u + a t_{2}}{2} = u + a (\frac{t_{1} + t_{2}}{2}) = u + a (Δ t_{1} + \frac{Δ t_{2}}{2})$
$v_{3} = \frac{u + a t_{2} + u + a t_{3}}{2} = u + a (\frac{t_{2} + t_{3}}{2}) = u + a (Δ t_{1} + Δ t_{2} + \frac{t_{3}}{2})$

$v_{2} - v_{1} = a (\frac{Δ t_{1} + Δ t_{2}}{2})$
$v_{3} - v_{2} = a (\frac{Δ t_{2} + Δ t_{3}}{2})$

$∴ \frac{v_{2} - v_{1}}{Δ t_{1} + Δ t_{2}} = \frac{v_{3} - v_{2}}{Δ t_{2} + Δ t_{3}}$

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