Question

A man walk for some time T with velocity V due east .then he walks for same time T with velocity V due to north. the average velocity of the man is?

Answer 1

The man walks for some time T with velocity V due east.
So his displacement due east `vec(d_e)=VThate`, where `hate` represents unit displacement vector towards east

Then the man walks for same time T with velocity V due north
So his displacement due east `vec(d_n)=VThatn`, where `hatn` represents unit displacement vector towards north

So net displacement in 2T time `vecd=vec(d_e)+vec(d_n)=VThate+VThatn`

So average velocity `vec(V"avg")=vecd/(2T)=V/2(hate+hatn)`

So magnitude of average velocity

`absvec(V"avg")=sqrt((V/2)^2+(V/2)^2)=V/sqrt2`

The direction of average velocity

`phi=tan^-1((V/2)/(V/2))=45^@` with east towards north