Correct me if I'm wrong, but isn't this just
How did you get the square root expression?
No, I get it now. Yeah, if F is a force function (which can be used to define change in potential energy--i.e., work, no?), then F = ma. We can solve for acceleration as a function of x, which is the derivative of velocity. So, write acceleration as a function of x with what you're given, take its integral. Velocity is the change in x for a change in time. Therefore, we should be able to answer the question. What did you get for v(x)?
Question 2: I never said that. Go back and read my reply with more care.
I have told you exactly how to answer the question you posted. But it would appear that you need to go back an review calculus as it applies to kinematics eg. a = dv/dt, v = dx/dt, a = v dv/dx = d/dx(v^2/2) etc.
Mr F, one of the first q's was this. The particle is released from rest at x = 1. Show that the position of P in its subsequent motion always lies between x = 1 and x = 9.
I integrated a(x) to get v(x) and since this is the derivative of P(x), I solved it to find the max and min. I got x=1 and x=9.
What Mr. F is trying to tell you is how what you want and what you have are related.the time it takes for P to travel from x=1 to x=9 is with limits 9 and 1.
By the chain rule in post #8 Mr. F gives
So this gives you the relationship you need. Try to sort this out.
I have said that you need to thoroughly review calculus as it relates to kinematics and I am even more convinced now that this advice is warranted.
Read what has been posted! Ask specific questions related to the help you have been given. People are not going to keep hitting their heads against brick walls saying the same things over and over.
ok done that. Thanks for your patience.
Next part is show that there is a unique equilibrium position for P. Find the energy the particle P has in this state. Show that the equilibrium is stable and find the frequency of small oscillations near this point.