An automobile travels past the farmhouse at a speed of v = 45 km/h. How fast is the distance between the automobile and the farmhouse increa

Question

An automobile travels past the farmhouse at a speed of v = 45 km/h. How fast is the distance between the automobile and the farmhouse increasing when the automobile is 3.7 km past the intersection of the highway and the road?

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Adeline 1 week 2021-10-10T19:37:57+00:00 1 Answer 0

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    2021-10-10T19:39:56+00:00

    Answer:

    \frac{ds}{dt} = 39.586 km/h

    Step-by-step explanation:

    let distance between farmhouse and road is 2 km

    From diagram given

    p is the distance between road and past the intersection of highway

    By using Pythagoras theorem

    s^2 = 2^2 +p^2

    differentiate wrt t

    we get

    \frac{d}{dt} s^2 = \frac{d}{dt} (4 + p^2)

    2s \frac{ds}{dt}  =2p \frac{dp}{dt}

    \frac{ds}{dt} = \frac{p}{s}\frac{dp}{dt}

    \frac{ds}{dt} = \frac{p}{\sqrt{p^2 +4}} \frac{dp}{dt}

    putting p = 3.7 km

    \frac{ds}{dt} = \frac{3.7}{\sqrt{3.7^2 +4}} 45

    \frac{ds}{dt} = 39.586 km/h

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