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1st Equation of Motion

A point moves...

Question

A point moves with uniform acceleration. Let $v_{1}$ , $v_{2}$ , $v_{3}$ denote the average velocities in three successive intervals of time $t_{1}$ , $t_{2}$ , $t_{3}$ . Correct relation among the following is

(v_{1} - v_{2})

(v_{2} - v_{3})

(t_{1} - t_{2})

(t_{2} - t_{3})

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(v_{1} - v_{2})

(v_{2} - v_{3})

(t_{1} + t_{2})

(t_{2} + t_{3})

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(v_{1} - v_{2})

(v_{2} - v_{3})

(t_{1} - t_{2})

(t_{2} + t_{3})

No worries! We‘ve got your back. Try BYJU‘S free classes today!

(v_{1} - v_{2})

(v_{2} - v_{3})

(t_{1} + t_{2})

(t_{2} - t_{3})

No worries! We‘ve got your back. Try BYJU‘S free classes today!

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Solution

The correct option is A $(v_{1} - v_{2})$ : $(v_{2} - v_{3})$ = $(t_{1} - t_{2})$ : $(t_{2} - t_{3})$
$v_{a v g} = \frac{u t + \frac{1}{2} a t^{2}}{t} = u + \frac{1}{2} a t$

$V_{1} = U + \frac{a t_{1}}{2}$

$V_{2} = U_{1} + \frac{a t_{2}}{2}$

$V_{3} = U_{2} + \frac{a t_{3}}{2}$

so we can find,

$(V_{1} - V_{2}) : (V_{2} - V_{3}) = (t_{1} - t_{2}) : (t_{2} - t_{3})$

Alternatively,
Since acceleration is constant, acceleration in the interval $t_{1} - t_{2}$ will be same as acceleration in the interval $t_{2} - t_{3}$
Hence,
$(V_{1} - V_{2}) : (V_{2} - V_{3}) = (t_{1} - t_{2}) : (t_{2} - t_{3})$

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