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Math Help - Velocity problem

  1. #1
    Junior Member
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    Velocity problem

    Hello everyone,

    I am looking to find out where I am going wrong on the following problem.

    If a ball is thrown up into the air with an initial velocity of 40 ft/sec, its height in feet t seconds later is given by y=40t-16t^2.

    a. Find the average velocity for the time period beginning at t=1 and lasting for .
    1.) 0.5 sec
    2.) 0.1 sec
    3.) 0.05 sec
    4.) 0.01 sec
    I know that average velocity is equal to change in height/change in time so this how I set it up for the first case.

    {((40(1.5)-16(1.5^2)-40(1)-16(1^2))}/(1.5-1)

    This gives me an answer of 0 ft/sec which doesn't seem to make sense. Can somebody point out where I am going wrong. BTW I did the other calculations the same way and they come out to "better" numbers but I still wonder if I am not making a fundamental error.

    Thanks,
    juventinoalex
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  2. #2
    MHF Contributor

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    Why does it not make sense? y= 40t- 16t^2= 24 is the same as 16t^2- 40t+ 24= 16(t- 1)(t- 1.5)= 0. The ball goes up to a maximum height at t= 1.25, then back downward. From t= 1 to t= 1.25, the velocity is positive. From t= 1.25 to t= 1.5, the velocity is negative. The average velocity is 0.
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  3. #3
    Junior Member
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    OK, thanks for clearing this up for me. I guess I ditn't really think it out all the way.
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