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Math Help - Am I Reading Too Much Into This?

  1. #1
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    Am I Reading Too Much Into This?

    [Problem] [Show Your Work]

    The average distance between the earth and the sun is 93 million miles. Using the velocity of light, find an approximate value for the amount of time it takes a light ray to travel from the sun to the earth.

    (Velocity of Light = 186,000 miles per second)
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  2. #2
    Super Member angel.white's Avatar
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    Quote Originally Posted by ITTTechstudent View Post
    [Problem] [Show Your Work]

    The average distance between the earth and the sun is 93 million miles. Using the velocity of light, find an approximate value for the amount of time it takes a light ray to travel from the sun to the earth.

    (Velocity of Light = 186,000 miles per second)
    Um, I could be reading too little into it, but I'd say:

    average~time = \frac{93,000,000 ~miles}{186,000 ~\frac {miles}{seconds}}
    miles cancel out, seconds go to numerator, and 93000/186=500 so the average time is 500 seconds.
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  3. #3
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    hmm... the answer u ended up with seems more rational than mine...i had no where near 500 secs lol...so that could be right..

    thanks!
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  4. #4
    MHF Contributor Reckoner's Avatar
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    Quote Originally Posted by ITTTechstudent View Post
    hmm... the answer u ended up with seems more rational than mine...i had no where near 500 secs lol...so that could be right..

    thanks!
    Some calculus might make it a bit clearer, if you know that the average value of a function f on an interval [a,\,b] is \overline f = \frac1{b - a}\int_a^bf(x)\,dx.

    Suppose at a given time t the Sun is a distance s(t) miles from the Earth, and further suppose that the Earth revolves around the Sun exactly once between times t = a and t = b.

    Then any light ray hitting the Earth at a time t will have traveled for \frac{s(t)}{186\cdot10^3} seconds. As s should be a continuous function of t, we may determine the average time to be

    \overline t = \frac1{b - a}\int_a^b\frac{s(t)}{186\cdot10^3}\,dt

    =\frac1{\left(186\cdot10^3\right)(b - a)}\int_a^b s(t)\,dt

    =\frac1{186\cdot10^3}\left(\frac{\int_a^b s(t)\,dt}{b - a}\right)

    =\frac1{186\cdot10^3}\cdot\overline s

    =\frac{\overline s}{186\cdot10^3}

    =\frac{93\cdot10^6}{186\cdot10^3} = \frac{10^3}2 = 500

    Carrying the units through, the answer is in seconds, and this agrees with angel.white's solution.
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  5. #5
    Super Member angel.white's Avatar
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    Quote Originally Posted by Reckoner View Post
    Some calculus might make it a bit clearer, if you know that the average value of a function f on an interval [a,\,b] is \overline f = \frac1{b - a}\int_a^bf(x)\,dx.

    Suppose at a given time t the Sun is a distance s(t) miles from the Earth, and further suppose that the Earth revolves around the Sun exactly once between times t = a and t = b.

    Then any light ray hitting the Earth at a time t will have traveled for \frac{s(t)}{186\cdot10^3} seconds. As s should be a continuous function of t, we may determine the average time to be

    \overline t = \frac1{b - a}\int_a^b\frac{s(t)}{186\cdot10^3}\,dt

    =\frac1{\left(186\cdot10^3\right)(b - a)}\int_a^b s(t)\,dt

    =\frac1{186\cdot10^3}\left(\frac{\int_a^b s(t)\,dt}{b - a}\right)

    =\frac1{186\cdot10^3}\cdot\overline s

    =\frac{\overline s}{186\cdot10^3}

    =\frac{93\cdot10^6}{186\cdot10^3} = \frac{10^3}2 = 500

    Carrying the units through, the answer is in seconds, and this agrees with angel.white's solution.
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