The average velocity of a body moving with uniform acceleration after travelling a distance of 3.06 m is 0.34 ms^-1. If the change in velocity of the body is 0.18 ms^-1 during this time, its uniform acceleration is

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0.01 ms-²

0.02 ms-²

0.03 ms-²

0.04 ms-²

answer is B.

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Detailed Solution

$Given v_{a v} = \frac{v + u}{2} = 0.34 and v - u = 0.18 S o l v i n g t h e s e t w o e q u a t i o n s, w e g e t u = 0.25 {m s}^{- 1}, v = 0.43 {m s}^{- 1} Given s = 3.06 m Now use v^{2} - u^{2} = las to find a = 0.02 {m s}^{- 2}$

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