Ship A is located 3.7 km north and 2.3 km east of ship B. Ship A has a velocity

Question

Ship A is located 3.7 km north and 2.3 km east of ship B. Ship A has a velocity of 19 km/h toward the south and ship B has a velocity of 41 km/h in a direction 36° 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 = (2.4, 4.7) + t(22cos(-90), 22sin(-90))
B = (0, 0) + t(40cos37, 40sin37)

The distance vector from A to B = vector B - vector A =
d = (2.4, 4.7) + t(0-40cos37, 22sin(-90)-40sin37)
d = (2.4, 4.7) + t(-31.945, -46.07)
or d = (2.4 - 31.945t)i + (4.7 - 46.07t)j
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the distance is
d = sqrt(2.4 - 31.945t)^2 + (4.7 - 46.07t)^2) ---> find the minimum: set derivative = 0
d' = (-11727880 + 125717117t)/40 000d > set = 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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