Question

two cars A and B are moving with equal speed V is moving along north that while b is moving along east.what is the magnitude of speed of a with respect to b

pls answer my question correctly.i will mark you as brainiest​

Answer 1

Two objects (A and B) are moving with velocity VA and VB . The velocity of A with respect to B is:

VA/B=VA−VB

In your question VA=+V and VB=−V

VA/B=V−(−V)=2V

The relative velocity equation above can be used for any two vectors in any direction.

Answer 2

Given:

Two cars A and B are moving with equal speed V is moving along north that while b is moving along east.

To find:

What is the magnitude of speed of A with respect to B?

Calculation:

Velocity vector of A :

$\rm \therefore \: \vec{v}_{A} = v \: \hat{j}$

Velocity vector of B :

$\rm \therefore \: \vec{v}_{B} = v \: \hat{i}$

Velocity vector of A w.r.t B will be :

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

$\rm\implies\: \vec{v}_{AB} = v \: \hat{j}- v \: \hat{i}$

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

$\rm\implies\: | \vec{v}_{AB} | = \sqrt{ {v}^{2} + {v}^{2} - 2 {v}^{2}(0) }$

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

$\rm\implies\: | \vec{v}_{AB} | = \sqrt{ 2{v}^{2} }$

$\rm\implies\: | \vec{v}_{AB} | = v\sqrt{ 2}$

So, required answer is :

$\boxed{ \bold{\: | \vec{v}_{AB} | = v\sqrt{ 2} }}$