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  • Answer please! A particle A is moving along north with speed 3 m/s and another particle B is moving with velocity 4 m/s at 60owith north, then velocity of B as seen by A is

A particle A is moving along north with speed 3 m/s and another particle B is moving with velocity 4 m/s at 60o with north, then velocity of B as seen by A is 

  • Option 1)

    5 m/s

  • Option 2)

    3.5 m/s

  • Option 3)

    \sqrt{19} m/s

  • Option 4)

    \sqrt{15}m/s

 

Answers (1)

best_answer

As we have learnt,

 

Relative Velocity -

Relative velocity of a body, A with respected body B when the to bodies moving at an angle \Theta.

 

V_{AB}= \sqrt{V_{A}^{2}+V_{B}^{2}+2V_{A}V_{B}\cos \left ( 180-\theta \right )}

        = \sqrt{V_{A}^{2}+V_{B}^{2}-2V_{A}V_{B}\cos \left ( \theta \right )}

 

- wherein

\\*V_{A}= velocity\: of\: A\\* V_{B}= velocity\: of\: B\\* \Theta = angle \: between \: A \: and \: B

 

 |\vec{V}_{BA}| = \sqrt{V^2_A + V^2_B - 2V_A V_B\cdot \cos\theta} = \sqrt{3^2 + 4^2 - 2\times3\times4\times\frac{1}{2}} = \sqrt{19}

 


Option 1)

5 m/s

Option 2)

3.5 m/s

Option 3)

\sqrt{19} m/s

Option 4)

\sqrt{15}m/s

Posted by

Aadil

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