1.

A particle is moving eastwards with a velocity of5 m//s. In 10 s the velocity changes to 5 m//s nothwards. The average acceleration in this time is

Answer»

Zero
`(1)/(sqrt(2)) m//s^(2)` TOWARDS NORTH-west
`(1)/(2) m//s^(2)` towards north
`(1)/(sqrt(2)) m//s^(2)` towards north-east

Solution :
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`


Discussion

No Comment Found

Related InterviewSolutions