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

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${0.04 ms}^{- 2}$

$0.03 {ms}^{2}$

${0.02 ms}^{- 2}$

${0.01 ms}^{- 2}$

answer is B.

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Detailed Solution

$s = 3.06m u = 0.34mts$

$t = \frac{s}{v} = \frac{3.06}{0.34}$

$t = 9 s$

$a = \frac{v - u}{t} = \frac{0.18}{9} = 0.0 {2m/s}^{2}$

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