Question

Ram moves in east direction at a speed of 6 m/s and shyam moves 30° east of north at a sleed of 6m/s. The magnitude of their relative velocity is.

Answer 1

Velocity of Ram = 6i
Velocity of Shyam = 6Sin30i + 6Cos30j

Relative velocity = 6i - 6Sin30i - 6Cos30j
= 6i - 3i - 3√3j
= 3i - 3√3j

Magnitude of relative velocity = sqrt(3^2 + (3√3)^2)
= 6 m/s

Answer 2

The magnitude of relative velocity is 6m/s (option b).

Given:

Ram moves at the speed=6m/s

Shyam moves 30 degree east of north at the speed=6m/s

To find:

The magnitude of relative velocity.

In the other words,

Velocity of Ram=6i

Velocity of Shyam =$6 \sin 30 i+6 \cos 30 j$

Relative velocity= velocity of Ram –velocity of Shyam

Let us substitute the values of velocity of ram and velocity of shyam in the given expression

$=6 i-(6 \sin 30 i+6 \cos 30 j)$

Multiplying the (–) inside

$=6 i-6 \sin 30 i-6 \cos 30 j$

$\sin 30=\frac{1}{2}$

$\cos 30=\frac{\sqrt{3}}{2}$

By substituting we get,

$=6 i-6\left(\frac{1}{2}\right)-6\left(\frac{\sqrt{3}}{2}\right)$

$=6 i-3 i-3 \sqrt{3} j$

$=3 i-3 \sqrt{3} j$

Hence magnitude of the relative velocity $= \sqrt{\left(3^{2}+3 \sqrt{3}^{2}\right)}$

=6m/s

Therefore, magnitude =6m/s.