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Average Rate of Change (word problem)

  1. Sep 11, 2007 #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. Sep 11, 2007 #2


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    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. Sep 11, 2007 #3
    Thank you :)

    so I plug in h into 5(t+h)^2????
  5. Sep 11, 2007 #4

    this is what I have so far
  6. Sep 12, 2007 #5
    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. Sep 12, 2007 #6
    [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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