Question

A particle Pis moving along a straight line with velocity of 3 m/s and another particle Q has a velocity of 4 m/s at an angle of 30 to the path P. Find the speed of Q relative to P
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Answer 1

Hey Dear,

◆ Answer -

Vqp = 2.83 m/s

● Explaination -

# Given -

|vA| = 3 m/s

|vB| = 5 m/s

θ = 30°

# Solution -

Relative velocity of q relative to p is calculated as -

Vqp = |vB - vA|

Vqp = √(|vA|² + |vB|² - 2vA.vB.cosθ)

Vqp = √(3² + 5² - 2×3×5×cos30°)

Vqp = √(9 + 25 - 30×0.866)

Vqp = √(9 + 25 - 26)

Vqp = √8

Vqp = 2.83 m/s

Therefore, velocity of q relative to p will be 2.83 m/s.

Thanks dear...