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anonymous

  • 5 years ago

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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  1. anonymous
    • 5 years ago
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    distance = speed x time

  2. anonymous
    • 5 years ago
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    therefore the distance the first driver goes is 30t , similarly the second driver goes 50t

  3. anonymous
    • 5 years ago
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    now, the distance between them is a hypotenuse of right angle triangle, which can be found by pythagorus

  4. anonymous
    • 5 years ago
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    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...

  5. anonymous
    • 5 years ago
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    d= sqrt ( 900t^2 + 2500t^2 ) = sqrt ( 3400t^2 ) = sqrt(3400) t

  6. anonymous
    • 5 years ago
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    so would it be d(distance)/dt= Sqrt[(dN/dt)^2 + (dW/dt)^2]?

  7. anonymous
    • 5 years ago
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    let D equal the distance , which is D= sqrt(3400) t from above dD / dt = sqrt(3400) // doesnt depend on the time

  8. anonymous
    • 5 years ago
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    which is what you would expect, none of the speed of the drivers are changing

  9. anonymous
    • 5 years ago
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    okay... that makes sense... I thought there was more to it.

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