The average velocity of a body moving with uniform acceleration 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

Options

A. `0.01 m s^(-2)`
B. `0.02 m s^(-2)`
C. `0.013 m s^(-2)`
D. `0.04 m s^(-2)`

Answer

B

Explanation

Given `v_(av) =(v+u)/(2) =0.34` and `v-u =0.18 `
Solving these two equations, we get
`u = 0.25 m s^(-1)`, `v=0.43 m s^(-1)`. Given `s=3. 06 m`
None use `v^(2)-u^(2) =2` as to find `a=0.02 m s^(-2)`.