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January 31, 2015

January 31, 2015

Posted by **eve** on Sunday, July 31, 2011 at 7:32am.

- Math -
**Reiny**, Sunday, July 31, 2011 at 10:01amlet's look at t hrs after the above condition

let the distance between them be D miles

then

D^2 = (40t+8)^2 + (15+60t)^2

2D dD/dt = 2(40t+8)(40) + 2(15+60t)(60)

dD/dt = [ 40(40t+8) + 60(15+60t) ]/D

when t = 0 , the above condition,

D = √8^2+15^2) = 17

and

dD/dt = [ 40(8) + 60(15) ]/17 = 71.76

At that moment they are separating at 71.8 mph

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