October 26th, 2011, 06:15 PM  #1 
Newbie Joined: Aug 2011 Posts: 25 Thanks: 0  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 
October 26th, 2011, 09:08 PM  #2 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,912 Thanks: 1110 Math Focus: Elementary mathematics and beyond  Re: help with 2d kinematic problem
Let ship B's starting point be the origin on the xyplane. Ship A(x, y): (2.1, 4  22t); ship B(x, y): (40tcos(37), 40tsin(37) ) Distance formula: d = ?((2.1  40tcos(37))² +(4  22t  40tsin(37))²) d² = (2.1  40tcos(37))² +(4  22t  40tsin(37))² Shortcut: Since what you get in the numerator after differentiating the expression for d is the same as the derivative of the expression for d², find minimal t by taking the derivative of the expression for d², 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) + 1600tcos²(37)  88 + 484t + 880tsin(37)  160sin(37) + 880tsin(37) + 1600tsin²(37) = 0 21cos(37) + 400tcos²(37)  22 + 121t + 220tsin(37)  40sin(37) + 220tsin(37) + 400tsin²(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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