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A particle (A) moves due North at 3 kmh^...

A particle (A) moves due North at `3 kmh^(-1)` and another particle (B) moves due West at `4 kmh^(-1)`. The relative velocity of A with respect to B is `(tan 37^(@)=3//4)`

A

`5 kmh^(-1), 37^(@)` North of East

B

`5 kmh^(-1), 37^(@)` East of North

C

`5sqrt(2)kmh^(-1), 53^(@)` East of North

D

`5sqrt(2)kmh^(-1), 53^(@)` North of East

Text Solution

Verified by Experts

The correct Answer is:
B
`v_(AB)=v_(A)-v_(B)`

`upsilon_(AB)=sqrt(upsilon_(AB)=sqrt(upsilon_(A)^(2)+upsilon_(B)^(2))=5 kmh^(-1)`
`alpha = tan^(-1)((upsilon_(A))/(upsilon_(B)))=tan^(-1)((3)/(4))=37^(@)` East of North
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