What is the distance between two ships and the correct solution to calculate it?

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The discussion centers on calculating the distance between two ships, A and B, given their velocities and positions over time. Ship A travels east at 30 km/h, initially spotting Ship B 15 km away at a 30° angle from the north. After 40 minutes, the angle changes to 60° from the north, leading to a calculated distance of 12.14 km between the ships. The proposed solution involves vector calculations using the velocities of both ships, resulting in a velocity for Ship B of 31.82 km/h.

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peripatein
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Hello,

I'd like to verify whether my solution to the problem as described below is indeed correct.

Ship A is traveling east at a constant velocity of 30km/h. It notices ship B 15 km away and 30° from the north. Ship B is traveling at a constant velocity. 20 minutes later ship A notices ship B 45° from the north, and 20 more minutes after that ship B is observed 60° from the north. The distance between ship A and B at that point in time ought to be calculated, including the velocity (magnitude and direction) of ship B.

Proposed solution:

Supposing ship A first spots ship B from O, (0,0):

1) (1/3Va)(OB + 1/3Vb) / |(1/3Va)||(OB + 1/3Vb)| = 1/SQRT(2)

2) (2/3Va)(OB + 2/3Vb) / |(2/3Va)||(OB + 2/3Vb)| = SQRT(3)/2

[Va, Vb, OB are vectors; 1/SQRT(2) = cos 45°; SQRT(3)/2 = cos 30°]

I got Vb = (27.56,15.91), |Vb| = 31.82 km/h, distance between ship A and B 40 minutes later = 12.14 km

Are these equations correct? Is their solution?
 
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