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Time problem

  1. Aug 5, 2005 #1
    Hello. Im stuck on this physics problem that i have for homework. Its the wording of the question thats odd to me. Im not sure exactly where to begin.

    So after one century, the length of each day will be 24 hours and .001 s? And after the second century the length of each day will be 24 hours and .002 s? Is there a quicker way to do this than doing a 20-step summation!!?!? I started off by multiplying .001 x 365, because that will give you the total amount of added time after a year ... however after that it would be long summation, and Im sure that there must be a quicker way.

    (FYI, the answer is 2 hours if you want to check your work :P)

    thanks in advance for any help/leads.
    Last edited: Aug 5, 2005
  2. jcsd
  3. Aug 5, 2005 #2

    One thing that worries me is the word continously, but disregarding that you are on the right track. But you will be off by a factor of 100.

    So you have 100 years in every century with 365 days in every year (we are disregarding the fractional day here).

    [tex] 36500 \times .001 + 36500 \times .002 + 36500 \times .003 \ldots 36500 \times .019 [/tex]

    notice it only goes up to .019 since the first year has no additional time.

    This is then equal to:

    [tex] 36500 \times (.001 + .002 + .003 \ldots .019) [/tex]

    Then divide this number by 3600 (the number of seconds in an hour) and you should get the correct answer.
  4. Aug 6, 2005 #3

    Andrew Mason

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    The proper way to do this is by integration since the change is continuous.
    Let the length of a day in seconds D be a continuous function of time t. D(t) is linear. The number of days in a century is about 365.25*100 = dc. So if the length of the day at t=0 is 24*3600, after dc days D = 24*3600+.001. The slope of the graph of D(t) vs. t is:

    [tex]\frac{D(dc)- D(0)}{dc}[/tex] so:

    [tex]D(t) = D(0) + \frac{D(dc)- D(0)}{dc}t [/tex]

    Integrating D(t) from 0 to 20 centuries gives us the length (L(t)) in days:

    [tex]L(t) = \int_0^t (D(0) + \frac{D(dc)- D(0)}{dc}t) dt [/tex]

    [tex]L(t) = D(0)t + \frac{D(dc)- D(0)}{2dc}t^2 [/tex]

    [tex]L(20dc) = D(0)(20dc) + \frac{D(dc)- D(0)}{2dc}(20dc)^2[/tex]

    If you work that out and subtract D(0)*20dc (the measured time), you will get the difference between actual accumulated time and measured time.


    [tex]L(20dc) = D(0)(20dc) + \frac{.001}{2dc}(20dc)^2[/tex]

    [tex]L(20dc) = D(0)(20dc) + .001(200dc)[/tex]

    [tex]L(20dc) - D(0)(20dc) = 7305[/tex]

    So the difference is 7305 seconds or just over 2 hours.


    Edit: My earlier post used seconds rather than days.
    Last edited: Aug 6, 2005
  5. Aug 6, 2005 #4
    Yeah, I realized that this is quite easy once you look at it this way:

    [tex] 36500 \times (.001 + .002 + .003 \ldots .020) = 36500 \times 10^{-3} \times \sum_{x=0}^{20} x \rightarrow 36500 \times 10^{-3} \int_0^{20} x dx [/tex]

    all we are doing here is going from the discrete sum to the continuous integration (I know, I know this is not mathematically rigorous, I left out the limiting procedure, but hey, I am a physicist)

    You get the same answer as Andrew.
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