Average Rate of Change (word problem)

  1. I'm having trouble understanding this word problem.

    A construction worker drops a bolt while woring on a high-rise building 320m above the ground. After t seconds, the bolt has fallen a distance of s metres, where
    s(t) = 320-5t^2, o < t < 8

    a. Find the average velocity during the first, third and eighth seconds.

    I have to use f(x + h) - f(x)/h

    I got 5(t +h)

    The answer is 5m/s for the first one.

    How do I get the rest of them (third and eighth second)?
  2. jcsd
  3. HallsofIvy

    HallsofIvy 41,256
    Staff Emeritus
    Science Advisor

    the "first second" is from t= 0 to t=1, of course: average speed during the first second is s(1)- s(0) since here h= 1. The "third second" is between t= 2 and t= 3 so average speed during the third second is s(3)- s(2) and the "eighth" second is between t= 7 and t= 8 so average speed for the eighth second is s(9)- s(8).
  4. Thank you :)

    so I plug in h into 5(t+h)^2????
  5. [​IMG]

    this is what I have so far
  6. You are making the problem too difficult. Velocity is the change of position over time. The formula you list is the formula for position.
  7. Saladsamurai

    Saladsamurai 3,016
    Gold Member

    [tex]\frac{s(t+h)-s(t)}{h}[/tex] is the formula for average velocity.

    It is a very wordy way of saying [tex]\frac{\Delta x}{\Delta y}[/tex],

    which is the change in y over the change in x. (<--slope)

    Now in order to see this, you know that [tex]\Delta y =y_{final}-y_{initial}[/tex] right? Can you see how that is the same as [tex]s(t+h)-s(t)[/tex] ?

    It is saying: take the position of some initial time (t) PLUS some amount h and subtract the original initial time from it.

    So in your second problem after 3 seconds, since your initial time is 0 and your final time is 3 then t+h means 0+3 which is of course 3.

    You see h is the change in time from 0 to 3.

    So now back to the problem; you should evaluate it as
    Last edited: Sep 12, 2007
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