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Rate of change ball thrown

  1. Feb 16, 2015 #1
    1. The problem statement, all variables and given/known data
    10pdx1D.png


    3. The attempt at a solution
    As far as I know those are correct. Could I have some assistance on how to solve for the rate of change? I know the formula is f(b)-f(a)/b-a but not sure how to go about it
     
  2. jcsd
  3. Feb 16, 2015 #2

    SteamKing

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    What if f(t) = h(t)? Could you pick two times, t1 = a and t2 = b, and calculate the rate of change, knowing h(t) as a function of t?
     
  4. Feb 17, 2015 #3
    That formula is saying take the change in height f(b) - f(a) and divide by the change in time b - a. Essentially, avg rate = total displavement over time for that displacement
     
  5. Feb 17, 2015 #4

    HallsofIvy

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    Unfortunately most of the answers you give here are wrong. You are given that [itex]h(t)= 56t- 16t^2= -16(t^2- (7/2)t)[/itex]. Completing the square gives [itex]h(t)= -16(t- 7/2)^2+ 196[/itex].
     
  6. Feb 17, 2015 #5

    mfb

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    @HallsofIvy: You got a factor of 2 wrong for the last equation.
     
  7. Feb 21, 2015 #6

    Mark44

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    Use parentheses!
    What you wrote is the same as ##f(b) - \frac{f(a)}{b} - a##.

    If you write a fraction such as the one above in single-line text, it has to be written as (f(b) - f(a))/(b - a).
     
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