Question

Ram moves in east direction at a speed of 6 m/s and Shyam moves 30 degree east of north at a speed of 6 m/s. The magnitude of their relative velocity is:- a) 3m/s b) 6m/s c) 6(3) 1/ 2 m/s d) 6(2) 1/ 2 m/s

Answer 1

Answer:

he magnitude of their relative velocity is 6 m/s. Thus option B is the right answer.

Explanation:

Given  

Ram moves at the speed=6m/s

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

To find the magnitude

In other words  

Velocity of Ram=6i

Velocity of Shyam=6sin30i+6cos30j

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

=6i-(6sin30i+6cos30j)

Multiplying the – inside

=6i-6sin30i-6cos30j

Sin 30=1/2

Cos30=√3/2

By substituting we get

=6i-6(1/2)-6(√3/2)

=6i-3i-3√3j

=3i-3√3j

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

=6m/s

Therefore magnitude =6m/s

Answer 2

Ąńšwər

$he magnitude of their relative velocity is 6 m/s. Thus option B is the right answer.</p><p>Explanation:</p><p>Given  </p><p>Ram moves at the speed=6m/s</p><p>Shyam moves 30 degree east of north at the speed=6m/s</p><p>To find the magnitude</p><p>In other words  </p><p>Velocity of Ram=6i</p><p>Velocity of Shyam=6sin30i+6cos30j</p><p>Relative velocity=velocity of Ram – velocity of Shyam</p><p>Let us substitute the values of velocity of ram and velocity of Shyam in the given expression</p><p>=6i-(6sin30i+6cos30j)</p><p>Multiplying the – inside</p><p>=6i-6sin30i-6cos30j</p><p>Sin 30=1/2</p><p>Cos30=√3/2</p><p>By substituting we get</p><p>=6i-6(1/2)-6(√3/2)</p><p>=6i-3i-3√3j</p><p>=3i-3√3j</p><p>Hence magnitude of the relative velocity= \sqrt{3^{2}+3 \sqrt{3^{2}}}32+332</p><p>=6m/s</p><p>Therefore magnitude =6m/s</p><p>$

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