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Math Help - Motion Problem

  1. #1
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    Motion Problem

    Starting at t = 0 a particle moves along the x axis so that it's position at time t is given by x(t) =  t^{4} - 5t^{2} + 2t. For what values of t does that particle take while moving to the left?

    Answer: 0.203 < t < 1.470

    Any starting hints on how this came to be? Note: This is the complete problem given.

    Perhaps the answer is to simply graph the function Then you can see what values the function takes as it goes left from 0.

    But the problem is that a graph goes on forever to the left. So how can there be a left interval?

    But perhaps this is some kind of flipped graph, where time is on the y axis rather than the x one. Yeah, I think that's the problem.
    Last edited by Jason76; August 21st 2013 at 10:44 PM.
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  2. #2
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    Re: Motion Problem

    hey Jason76.

    Hint: Do you know how to use calculus to find the minimum and maximum of a function in a given region?
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  3. #3
    MHF Contributor ebaines's Avatar
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    Re: Motion Problem

    If the particle is moving to the left that means its velocity is negative. The function you were given is the position of the particle; hence the derivative of the function with respect to time gives its velocity. So take the derivative, plot that and and see where it is negative. The derivative will be a cubic function, which you should know takes a general shape of a "sideways S." It appears from the answer that t is limited to positive values only.
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