# Motion Problem

• Aug 21st 2013, 09:28 PM
Jason76
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.
• Aug 22nd 2013, 02:14 AM
chiro
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?
• Aug 22nd 2013, 03:36 AM
ebaines
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.