Question

two particles A and B are moving with velocity 3m/s and 4m/s in mutually perpendicular direction.magnitude of velocity of A w.r.t. B, is how many m/s???​

Answer 1

Given:

Two particles A and B are moving with velocity 3m/s and 4m/s in mutually perpendicular direction.

To find:

Velocity of A w.r.t B ?

Calculation:

The Velocity of Particle A with respect to Particle B can be mathematically represented as the vector difference between velocity vector of A and B.

$\therefore \: \vec{v}_{AB} = \vec{v}_{A} - \vec{v}_{B}$

$\implies \: | \vec{v}_{AB} | = \sqrt{ {| \vec{v}_{A} |}^{2} + { | \vec{v}_{B}|}^{2} + 2 |v_{A}| |v_{B}| \cos( {180}^{ \circ} - {90}^{ \circ} ) }$

$\implies \: | \vec{v}_{AB} | = \sqrt{ {| \vec{v}_{A} |}^{2} + { | \vec{v}_{B}|}^{2} + 2 |v_{A}| |v_{B}| \cos( {90}^{ \circ} ) }$

$\implies \: | \vec{v}_{AB} | = \sqrt{ {| \vec{v}_{A} |}^{2} + { | \vec{v}_{B}|}^{2} + 2 |v_{A}| |v_{B}| (0) }$

$\implies \: | \vec{v}_{AB} | = \sqrt{ {| \vec{v}_{A} |}^{2} + { | \vec{v}_{B}|}^{2} }$

$\implies \: | \vec{v}_{AB} | = \sqrt{ {(3)}^{2} + {(4)}^{2} }$

$\implies \: | \vec{v}_{AB} | = \sqrt{ 9 + 16}$

$\implies \: | \vec{v}_{AB} | = \sqrt{25}$

$\implies \: | \vec{v}_{AB} | = 5 \: m {s}^{ - 1}$

So, velocity of particle A w.r.t particle B is 5 m/s.