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    sim0nz12345's Avatar
    sim0nz12345 Posts: 77, Reputation: 2
    Junior Member
    		  
    #1

    Aug 22, 2007, 02:41 AM
    Rates of Change Part 2
    Hi there
    I could really use some help on this particular questions

    Show that the particle moving with position equation x(t)=24t^2-3t^4 turns around after 2 seconds ( find values for position at t=0, t=1, t=2 and t=3 seconds and use a 0.01 second interval to show the velocity is 0 at t=2 seconds.)

    I know how to find the values of the positions but not the velocity section

    Thanks
    Capuchin's Avatar
    Capuchin Posts: 5,255, Reputation: 656
    Uber Member
    		  
    #2

    Aug 22, 2007, 04:02 AM
    The velocity is the rate of change of position. You know exactly how to do it, it's the same as the other one I showed you.

    It wants you to show that at t=2 the sign of velocity changes from positive to negative.

    Again, this is a very unintuitive way to lookat it, I think, but you have to live with what you're given :)
    sim0nz12345's Avatar
    sim0nz12345 Posts: 77, Reputation: 2
    Junior Member
    		  
    #3

    Aug 22, 2007, 04:21 AM
    Hmmmm I see that I got lazy with the questions
    Thanks again
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