A particle is moving eastwards with a velocity of ` 5 m//s`. In `10 s` the velocity changes to `5 m//s` nothwards. The average acceleration in this time is

Options

A. Zero
B. `(1)/(sqrt(2)) m//s^(2)` towards north-west
C. `(1)/(2) m//s^(2)` towards north
D. `(1)/(sqrt(2)) m//s^(2)` towards north-east

Answer

B

Explanation

acc. `=(Delta nu)/(t)`
`=(5sqrt(2))/(10)=(1)/(sqrt(2))m//s^(2)`
`vecV_(1)=5m//s` (East)
`vecV_(2)=5m//s` (North)
Change in velocity `=vecV_(t)-vecV_(i)`
`=5hatj-5hati m//s`
`Delta nu=` Magnitude of change is velocity
`=sqrt((5)^(2)+(5)^(2))=5sqrt(2) m//s`