A point moves with uniform acceleration and its initial speed and final speed are $ 1 \mathrm{~m} \mathrm{~s}^{-1} $ and $ 2 \mathrm{~m} \mathrm{~s}^{-1} $ respectively then, the space average of velocity over the distance moved is. (in $ \mathrm{m} \mathrm{s}^{-1} $ ) :-

Question

A point moves with uniform acceleration and its initial speed and final speed are $ 1 \mathrm{~m} \mathrm{~s}^{-1} $ and $ 2 \mathrm{~m} \mathrm{~s}^{-1} $ respectively then, the space average of velocity over the distance moved is. (in $ \mathrm{m} \mathrm{s}^{-1} $ ) :-, $ \frac{14}{9} \mathrm{~m} \mathrm{~s}^{-1} $, $ \frac{3}{2} \mathrm{~ms}^{-1} $, $ \frac{5}{2} \mathrm{~ms}^{-1} $, None of these

MARKS App · Accepted Answer

v > space = ∫ v d x ∫ d x But a = v dv dx So vdx = v 2 a dv and d x = v a dv So space = ∫ v 2 a dv ∫ v a dv = ∫ 1 2 v 2 d v ∫ 1 2 vdv = v 3 3 1 2 v 2 2 1 2 = 14 9 m s - 1

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