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Thread: Motion Problem

  1. #1
    MHF Contributor Jason76's Avatar
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    Motion Problem

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

    Answer: $\displaystyle 0.203 < t < 1.470$

    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 $\displaystyle 0$.

    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 $\displaystyle y$ axis rather than the $\displaystyle x$ one. Yeah, I think that's the problem.
    Last edited by Jason76; Aug 21st 2013 at 09:44 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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