Question

If a particle is located at x=2 when t=2.5, can
it reach the location x=1 at any later time?
Hint: dx/dt = (t^3 -x^3) =(t-x)(t^2 + xt + x^2)
Attempt:
I know that since the particle was at x=2 when t = 2.5 and that t will always be greater than 2.5 (and thus positive) since it is at a "later time".
I think that the answer has something to do with the two factors: (t-x) and (t^2 + xt + x^2). One of these two factors has to be negative some time after t = 2.5. Setting these equal to 0, I believe the 2nd equation has imaginary roots so that leaves (x-t) = 0 -> x = t. Not sure where to go now

Answer 1

I assume the particle is traveling on the x-axis? For the particle to move to the left (from a larger to smaller value of \(x\)), the velocity would have to be negative, as this would mean it's moving in the negative direction.

Since \(x(t)\) could describe the particle's position on the axis at time \(t\), \(\dfrac{dx}{dt}\) must describe its velocity as it moves along the axis. Negative velocity means \(\dfrac{dx}{dt}<0\) as well.

You're right about having \(t>0\). Since
\[\frac{dx}{dt}=(t-x)(t^2+xt+x^2)\]
should be negative, you're basically solving the inequality
\[(t-x)(t^2+xt+x^2)<0\]
The factor \(t-x\) will be negative whenever \(x>t\), which means \(x\) is also positive. If this is the case, then \(t^2+xt+x^2\) is necessarily positive.

To tell the truth, I'm not entirely sure what you're supposed to do :/