
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? http://www.freemathhelp.com/forum/im...s/confused.png Note: This is the complete problem given.
Perhaps the answer is to simply graph the function http://www.freemathhelp.com/forum/im...n_confused.gif 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? http://www.freemathhelp.com/forum/im...s/confused.png
But perhaps this is some kind of flipped graph, where time is on the $\displaystyle y$ axis rather than the $\displaystyle x$ one. (Happy) Yeah, I think that's the problem.

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?

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.