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 = A) 2v B)√2v
C)v C)v/√2

Answer 1

Given:

Man walks with velocity v for sometime t due East and again towards North with velocity v for time t .

To find:

Average velocity of man

Concept:

Average velocity is defined as the ratio of the total displacement to the total time taken by the body.

Calculation:

Displacement can be calculated using Pythagoras Theorem :

Let displacement be d :

$d = \sqrt{ {(vt)}^{2} + {(vt)}^{2} } = vt \sqrt{2}$

Now average velocity will be :

$avg. \: v = \dfrac{total \: displacement}{total \: time}$

$= > avg. \: v = \dfrac{vt \sqrt{2} }{t + t}$

$= > avg. \: v = \dfrac{vt \sqrt{2} }{2t}$

$= > avg. \: v = \dfrac{v\sqrt{2} }{2}$

$= > avg. \: v = \dfrac{v}{ \sqrt{2} }$

So final answer is :

Average Velocity is v/√2

Answer 2

Given :

Time taken travelling east = t

Velocity due East = v

Time taken travelling north = t

Velocity due north = v

To Find :

Average velocity of the man =?

Solution :

We know, displacement = velocity ×time

Therefore,

Displacement due East = vt

Displacement due north =vt

And,

$Total \: Displacement = \sqrt{ {(vt)}^{2} + {(vt)}^{2} } \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = \sqrt{2 {v}^{2} {t}^{2} } \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = \sqrt{2} vt$

And, total time taken = t+t =2t.

Therefore,

$Average \: velocity = \frac{total \: displacement}{total \: time } \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = \frac{ \sqrt{2} vt}{2t} \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = \frac{v}{ \sqrt{2} }$

Hence, correct option is (C).