Calculate displacement when acceleration is dependent on displacement?

Jasoni22

I really have no idea where to start with this.

I have [itex]a=\frac{2.071\cdot10^{11}A}{mx^{2}}[/itex] m/s², where [itex]A[/itex] and [itex]m[/itex] are constants. Since the acceleration is dependent on position, and the position is obviously dependent on acceleration, I don't know how to separate the two in order to do any calculations.

Basically, what I'm trying to do is for any initial position [itex]x_{0}[/itex], calculate the position at time [itex]t[/itex] of an object being accelerated from rest by a force that is decreasing as the square of the distance from the object.

sweet springs

Hi.

You should solve the differential equaiton


d^2 x/dt^2 = C x^-2

Regards.

Ken G

An alternative approach that is often useful when you have a(x) is to note that a=v*dv/dx = 1/2*dv²/dx. Equating that to a(x) and integrating over dx is the source of the concept of kinetic energy and its connection to potential energy, for example. Solving an equation like 1/2*dv²/dx = a(x) is then a first-order differential equation in x, rather than second order in t. You can often get v(x), which is equivalent to calculating the change in potential energy and attributing the change in kinetic energy to it. Once you have v(x), you write v(x) = dx/dt and solve the first-order differential equation in t to get x(t), if all the functions are integrable and invertiable as needed. This way you only have to deal with first-order differential equations that can be solved by integration (if you are lucky). Or, you can just do it sweet_springs' way, if you can see by inspection the function you need. I mention the other way because of its important connection to the concept of conservation of energy, which is merely a shortcut for the math I described.

sweet springs

Hi. We are saying the same thing.

2v dv/dt = 2vC x^-2

d (v^2) = 2C x^-2 dx

v^2 = -2C /x + c

v^2 + 2C /x = v0^2 + 2C /x0

Regards.