A particle is moving eastwards with a velocity of ` 5 ms_(-1)`. In `10 seconds` the velocity changes to `5 ms^(-1)` northwards. The average acceleration in this time 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

`Deltav= 2v sin ((theta)/(2))= 2xx5xxsin 45^(@)= (10)/(sqrt(2))`
`:. a=(Deltav)/(Deltat)= (10sqrt(2))/(10)= 1/(sqrt(2))m//s^(2)`