The air transport B is flying with a constant speed of 480 mi/hr in a horizontal

Question

The air transport B is flying with a constant speed of 480 mi/hr in a horizontal arc of 9-mi radius. When B reaches the position shown, aircraft A, flying south-west at a constant speed of 360 mi/hr, crosses the radial line from B to the center of curvature C of its path. Write the vector expression, using the xy axes attached to B, for the velocity of A as measured by an observer in and turning with B

The correct answer is Vrel = -373i - 686j ft/s

Please show me all the correct steps to solve this problem and it will help out greatly

Explanation / Answer

Vb = 480 j

Va = 360 (-1/sqrt(2) i - 1/sqrt(2) j)

Vb wrt a = Vrel = -360/sqrt(2) i - (360/sqrt(2) + 480) j = -254.56 i - 734.56 j mi/hr

Vrel = (-254.56 i - 734.56 j ) mi/hr*1.466( ft/s) /(mi/hr)

Vrel = - 373 i - 1077 j .

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