Question

A man walks for some time 't' with velocity (v) due
east. Then he walks for same time 't' with velocity
(v) due north. The average velocity of the man is :
(1) 2v
(2)
$\sqrt{2} v$

(3) v
(4)
$v \div \sqrt{2}$
​

Answer 1

Answer:

represents unit displacement vector towards north

So net displacement in 2T time →d=→de+→dn=VTˆe+VTˆn

So average velocity −−−→Vavg=→d2T=V2(ˆe+ˆn)

So magnitude of average velocity

∣∣∣−−−→Vavg∣∣∣=√(V2)2+(V2)2=V√2

The direction of average velocity

ϕ=tan−1(V2V2)=45∘ with east towards north

Answer 2

Step-by-step explanation:

as the direction changed so the man is walking with variable velocity

average velocity=v+v

average velocity=2v ans