Question

A man walks for some time $'t'$ with velocity $\left( V \right)$ due east. Then he walks for the same time $'t'$ with velocity $\left( V \right)$ due north. The average velocity of the man is
$A.$ $2V$
$B.$ $\sqrt 2 V$
$C.$ $V$
$D.$ $\dfrac{V}{{\sqrt 2 }}$

Answer 1

It is given that the man walks initially with the velocity $V$ in east direction in the diagram below it is indicated by $OA$ .Next the man walks with $V$ in west direction which is indicated by $AB$ . First we will calculate the displacement $S$ and we will calculate the average velocity by using the formula.

Complete step-by-step solution:
According to the given question the figure can be drawn as shown below

From the figure, on applying Pythagoras theorem to the triangle $OAB$
The displacement, $S = OB = \sqrt {{{\left( {OA} \right)}^2} + {{\left( {AB} \right)}^2}}$ …….. $\left( 1 \right)$
And also $OA = AB = Vt$ $\left[ {\because Displacement = velocity \times time} \right]$
Substituting in equation $\left( 1 \right)$ we get
$S = \sqrt {{{\left( {Vt} \right)}^2} + {{\left( {Vt} \right)}^2}}$
Therefore, $S = \sqrt 2 Vt$ ………..$\left( 2 \right)$
Let us consider total time as $T$
And $T = 2t$……….$\left( 3 \right)$
We known that average velocity $\left( {{V_{avg}}} \right)$ $= \dfrac{{Displacement}}{{Total{\text{ }}time}}$ ………..$\left( 4 \right)$
Substituting equation $\left( 2 \right)$ and equation $\left( 3 \right)$ in equation $\left( 4 \right)$ we get
${V_{avg}} = \dfrac{{\sqrt 2 Vt}}{{2t}}$
On simplification
${V_{avg}} = \dfrac{V}{{\sqrt 2 }}$
Hence, option $D$ is correct. Average velocity $\left( {{V_{avg}}} \right)$ $= \dfrac{V}{{\sqrt 2 }}m{s^{ - 1}}$

Note: The average velocity is defined as the total displacement to that of the total time taken. In another world we can define the average velocity as the rate at which an object changes its position from one place to the other place.