You say goodbye to your friend at the intersection of two perpendicular roads. At time t=0 you drive off North at a (constant) speed v and y

Question

You say goodbye to your friend at the intersection of two perpendicular roads. At time t=0 you drive off North at a (constant) speed v and your friend drives West at a (constant) speed ????. You badly want to know: how fast is the distance between you and your friend increasing at time t?

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Nick 5 years 2021-08-11T22:35:35+00:00 1 Answers 233 views 0

Answers ( )

  1. quangkhai
    0
    2021-08-11T22:37:31+00:00
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    Answer:

    d'=\sqrt{v^2+w^2}

    Explanation:

    Rate of Change

    When an object moves at constant speed v, the distance traveled at time t is

    x=v.t

    We know at time t=0 two friends are at the intersection of two perpendicular roads. One of them goes north at speed v and the other goes west at constant speed w (assumed). Since both directions are perpendicular, the distances make a right triangle. The vertical distance is

    y=v.t

    and the horizontal distance is

    x=w.t

    The distance between both friends is computed as the hypotenuse of the triangle

    d^2=x^2+y^2

    We need to find d’, the rate of change of the distance between both friends.

    Plugging in the above relations

    d^2=(v.t)^2+(w.t)^2

    d^2=v^2.t^2+w^2.t^2=(v^2+w^2)t^2

    Solving for d

    d=\sqrt{(v^2+w^2)t^2}

    d=\sqrt{(v^2+w^2)}.t

    Differentiating with respect to t

    \boxed{d'=\sqrt{v^2+w^2}}

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