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October 25, 2014

October 25, 2014

Posted by **Steven** on Monday, October 22, 2012 at 4:37pm.

- Math Calculus 2 -
**Steve**, Monday, October 22, 2012 at 5:59pmat time t=0,

let the plane be at (0,0,.5)

let the car be at (0,0,0)

let the +x axis be due east, so +y is due north

at time t hours,

plane is at (240t,0,.5)

car is at (60/√2 t,-60/√2 t,0)

the distance d between car and plane is

d^2 = (240t - 60/√2 t)^2 + (60/√2 t)^2 + .5^2

36 seconds = 0.01 hours, so

d^2 = (2.4-.6/√2)^2 + (.6/√2)^2 + .5^2

d = 2.0817

2d dd/dt = 7200(17-4√2)t

2(2.0817) dd/dt = 7200(17-4√2)(.01)

dd/dt = 196.16 mph

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