Vector Equations Of LinesDate: 6/28/96 at 1:59:19 From: Anonymous Subject: Vector Equations Of Lines Dr. Math, I can't solve this problem. Can you help? A boat leaves a harbour, O, position vector (0i+0j) at 9 a.m. and travels with velocity (12i+4j) km/h. After 3 hours the boat develops engine problems and, with a storm forecast, the captain decides to head directly for shelter at harbour A, position vector (42i+9j) km. With the engines off the wind and current would cause the boat to drift with ocity (6i+2j) km/h. With the faulty engine working the boat could travel at a speed of 10 km/h in still water. With the engines working in this way at what velocity should the boat be set so that this, together with the wind and the current, causes the boat to arrive at A? At what time will it arrive? Date: 6/28/96 at 9:41:38 From: Doctor Anthony Subject: Re: Vector Equations Of Lines If you plot the position of the boat after 3 hours (12 noon) and the harbour at A, you will see that the vector from the boat to A will be 6i - 3j. Now if the boat is driven with velocity vector ai + bj where the magnitude of this vector sqrt(a^2+b^2) = 10, and if t is the time from present position to A we have the vector equation t{(ai+bj) + (6i+2j)} = 6i - 3j Now equating i and j components t(a+6) = 6 t(b+2) = -3 and dividing these equations (a+6)/(b+2) = -2 a+6 = -2b-4 so a+2b = -10 also a^2 + b^2 = 100 a = -2b-10 and a^2 = 4b^2 + 40b + 100 substitute into a^2+b^2=100 4b^2+40b+100+b^2 = 100 5b^2 + 40b = 0 b(b+8) = 0 b=0 is not a viable result since this would require a = -10, so we choose the answer b = -8, and so a = -2b-10 = 16-10 = 6 And so the velocity vector of the boat in still water would be 6i - 8j The time is given by the equation t(a+6) = 6 t(6+6) = 6 12t = 6 and so t = 1/2 hour The boat therefore arrives at A at 12.30 p.m. -Doctor Anthony, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
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