anonymous
  • 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?
Mathematics
jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
distance = speed x time
anonymous
  • anonymous
therefore the distance the first driver goes is 30t , similarly the second driver goes 50t
anonymous
  • anonymous
now, the distance between them is a hypotenuse of right angle triangle, which can be found by pythagorus

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anonymous
  • 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
  • anonymous
d= sqrt ( 900t^2 + 2500t^2 ) = sqrt ( 3400t^2 ) = sqrt(3400) t
anonymous
  • anonymous
so would it be d(distance)/dt= Sqrt[(dN/dt)^2 + (dW/dt)^2]?
anonymous
  • anonymous
let D equal the distance , which is D= sqrt(3400) t from above dD / dt = sqrt(3400) // doesnt depend on the time
anonymous
  • anonymous
which is what you would expect, none of the speed of the drivers are changing
anonymous
  • anonymous
okay... that makes sense... I thought there was more to it.

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