Question

particle A is moving with velocity (6i^+10j^)m/s and particle B is moving with (3i^+6j^)m/s magnitude of velocity ofAw.r.t B is​

Answer 1

Answer:

I think answer is with respect to B we get √136 and with respect to A we get√45, if it is correct please follow me

Answer 2

Given: Particle A is moving with velocity (6i^+10j^)

m/s and particle B is moving with (3i^+6j^)m/s

To find: Magnitude of velocity A w.r.t B

Explanation: Magnitude of velocity is given by the formula:

$|v| = \sqrt{ {(x})^{2} + {(y)}^{2} }$

where x and y are the coefficient of i and j respectively.

Velocity of particle A = (6i^+10j^)

Let magnitude of velocity of particle A be denoted by v1.

${v1}^{2} = {6}^{2} + {10}^{2}$

${v1}^{2} = 36 + 100$

$v1 = \sqrt{136}$

v1 = 11.66 m/s

Velocity of particle B = (3i^+6j^)

Let magnitude of velocity of particle B be denoted by v2.

${v2}^{2} = {3}^{2} + {6}^{2}$

${v2}^{2} = 9 + 36$

$v2= \sqrt{45}$

v2 = 6.70 m/s

Magnitude of velocity of A with respect to B:

= magnitude of velocity of A - magnitude of velocity of B

= v1 - v2

= 11.66- 6.70 m/s

= 4.96 m/s

Therefore, the magnitude of velocity of A with respect to B is 4.96 m/s.