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