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

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

anonymous · Answer

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 :/