- #1
anurag07
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Homework Statement
Two ships A and B are 4 km apart. A is due west of B. If A moves with uniform velocity of 8 km/hr due east and B moves with a uniform velocity of 6 km/hr due south. Calculate 1) the magnitudeof the velocity of A in relation to B 2) the closest distance apart of A and B
Homework Equations
The first solution is pretty simple but the second one, I don't get it. I know how to just solve it but without any intuition. Can someone help me get an intuitive idea of the solution of 2) ? The solution will be down there.
The Attempt at a Solution
In the example BD is drawn from B to relative velocity of A with respect to B represented by AE. From the given distance between A and B is 4km i.e. AB = 4km. In ABE sin x = BE/AE = 6/10 ( x= Angle EAB, BE = Velocity of B in opposite direction, AE = relative velocity of A in respect to B )
x=36.67
Closest distance apart from A to B is BD.
Now in BDA,
BD = sin x * AB = sin 36.87 * 4 = 2.4 km
This is the closest distance.