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- anonymous

Two drivers begin at the same location. The first drives north at 30mph, the second at 50mph to the west. How fast is the distance between them changing 4 hours into the drive?

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- anonymous

- jamiebookeater

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- anonymous

distance = speed x time

- anonymous

therefore the distance the first driver goes is 30t , similarly the second driver goes 50t

- anonymous

now, the distance between them is a hypotenuse of right angle triangle, which can be found by pythagorus

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- anonymous

Their distance at any time is \[\sqrt{x^{2}+y^2}\], where x is the first divers distance relative to the start, and y, the second divers...

- anonymous

d= sqrt ( 900t^2 + 2500t^2 ) = sqrt ( 3400t^2 ) = sqrt(3400) t

- anonymous

so would it be d(distance)/dt= Sqrt[(dN/dt)^2 + (dW/dt)^2]?

- anonymous

let D equal the distance , which is D= sqrt(3400) t from above
dD / dt = sqrt(3400) // doesnt depend on the time

- anonymous

which is what you would expect, none of the speed of the drivers are changing

- anonymous

okay... that makes sense... I thought there was more to it.

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