A particle moves towards east with velocity `5m//s`. After `10 seconds` its direction changes towards north with same Velocity. The average acceleration of the particle is

Options

A. Zero
B. `(1)/(sqrt(2)) m//s^(2) N -W`
C. `(1)/(sqrt(2)) m//s^(2) N - E`
D. `(1)/(sqrt(2)) m//s^(2) S -W`

Answer

B

Explanation

`Delta vec(v) = vec(v_(2)) - vec(v_(1)) = vec(v_(2)) + (-vec(v_(1)))`

`Delta v = 5 sqrt(2) m//s`
`theta = 45^(@)`
Change in velocity ` = 5 sqrt(2) m//s , NW`
`vec(a) = (Delta v)/(Delta t) = (5 sqrt(2))/(10) = (1)/(sqrt(2)) m//s^(2) , NW`