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?

1 Answer
Jun 3, 2017

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