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October 26th, 2011, 05:15 PM   #1
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help with 2d kinematic problem

Ship A is located 4.0 km north and 2.1 km east of ship B. Ship A has a velocity of 22 km/h toward the south and ship B has a velocity of 40 km/h in a direction 37░ north of east.

(c) At what time is the separation between the ships least?

(d) What is that least separation?

any help is greatly appreciated i have a test tomorrow and i was able to solve all of the other problems except for this one
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October 26th, 2011, 08:08 PM   #2
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Re: help with 2d kinematic problem

Let ship B's starting point be the origin on the xy-plane. Ship A(x, y): (2.1, 4 - 22t); ship B(x, y): (40tcos(37), 40tsin(37) )

Distance formula:

d = ?((2.1 - 40tcos(37))▓ +(4 - 22t - 40tsin(37))▓)

d▓ = (2.1 - 40tcos(37))▓ +(4 - 22t - 40tsin(37))▓

Shortcut:
Since what you get in the numerator after differentiating the expression for d is the same as the derivative of the expression for d▓, find minimal t by taking the derivative of the expression for d▓, setting it equal to zero, and solving for t:

2(2.1 - 40tcos(37))(-40cos(37)) + 2(4 - 22t - 40tsin(37))(-22 - 40sin(37)) = 0

(2.1 - 40tcos(37))(-40cos(37)) + (4 - 22t - 40tsin(37))(-22 - 40sin(37)) = 0

-84cos(37) + 1600tcos▓(37) - 88 + 484t + 880tsin(37) - 160sin(37) + 880tsin(37) + 1600tsin▓(37) = 0

-21cos(37) + 400tcos▓(37) - 22 + 121t + 220tsin(37) - 40sin(37) + 220tsin(37) + 400tsin▓(37) = 0

t(400 + 121 + 440sin(37)) = 21cos(37) + 40sin(37) + 22

t = (21cos(37) + 40sin(37) + 22)/(521 + 440sin(37)) ? 0.0799 hours, about 4 min 47 sec. when the separation is least; applying formula for distance, minimum d ? 0.5535 km.
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