Ship A is located 4.0 km north and 2.6 km east of ship B. Ship A has a velocity

Question

Ship A is located 4.0 km north and 2.6 km east of ship B. Ship A has a velocity of 24 km/h toward the south and ship B has a velocity of 38 km/h in a direction 34° north of east. What are the (a) x-component and (b) y-component of the velocity of A relative to B? (Axis directions are determined by the unit vectors i Overscript EndScripts and j Overscript EndScripts, where i Overscript EndScripts is toward the east.) (c) At what time is the separation between the ships least? (d) What is that least separation?

Explanation / Answer

The position vector of
A = (2.6, 4.0) + t(24cos(-90), 24sin(-90))
B = (0, 0) + t(38cos34, 38sin34)

The distance vector from A to B = vector B - vector A =
d = (2.6, 4.0) + t(0-38cos34, 24sin(-90)-38sin34)
d = (2.6, 4.0) + t(-31.5, -45.24)
or d = (2.6 - 31.5t)i + (4.0 - 45.24t)j
----------
the distance is
d = sqrt(2.6 - 31.5t)^2 + (4.0 - 45.24t)^2) ---> find the minimum: set derivative = 0
t(min) = 0.0932879 h
d(min) = 0.70589 km

Minimum distance = 0.706 km at t = 0.093 h

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