Is This a Differential Equations Problem?

[NJJ289]

Given the acceleration of a particle as a function of its position (x) along a straight line, how can one create a function to represent its acceleration as a function of time (t)?

I haven't covered differential equations yet and was wondering if this is an example of a kind of problem that subject deals with.
For example,

Let's say the particle moves with A(x)=M/X^2

taking d(A)/d(x) can give the change in Acceleration at any distance x, but what about time?

In calc 1 we did some 'related rates' problems that were similar to this, but they always had a special relationship that let you solve it (right triangles etc).

So: how do you solve this case in particular, what subject matter deals with this, and where can I get more info on other questions like it?

Thanks for the help! :)

 

[chiro]

Typically what happens is that you parametrize the system.

I'll give an example of a something going in a circle.

Say we have a system x^2 + y^2 = 1.

We can use the substitution

x = cos(theta)
y = sin(theta)

Now using trigonometric identities we can verify this works since

x^2 + y^2 = cos^2(theta) + sin^2(theta) = 1 as required.

You basically do the same sort of thing and put things in terms of a parameter (namely time), but its not always easy to do so (as the circle example)

 

[HallsofIvy]

Given the acceleration of a particle as a function of its position (x) along a straight line, how can one create a function to represent its acceleration as a function of time (t)?

I haven't covered differential equations yet and was wondering if this is an example of a kind of problem that subject deals with.
For example,

Let's say the particle moves with A(x)=M/X^2

taking d(A)/d(x) can give the change in Acceleration at any distance x, but what about time?

A is the velocity at position x? Then the "acceleration" is the derivative with respect to t, not x, whether at a given x or t. Using the chain rule, the acceleration would be given by
$$\frac{dA}{dt}= \frac{dA}{dx}\frac{dx}{dt}$$
Now, since A is velocity, you are saying that $dx/dt= A$
so that equation would be the same as
$$\frac{dA}{dt}= A\frac{dA}{dx}$$

In particular, if A(x)= M/x^2, as you give, then
$$\frac{dA}{dt}= \left(\frac{M}{x^2}\right\left(\frac{-3M}{x^3}\)= -\frac{3M}{x^5}$$

In calc 1 we did some 'related rates' problems that were similar to this, but they always had a special relationship that let you solve it (right triangles etc).

So: how do you solve this case in particular, what subject matter deals with this, and where can I get more info on other questions like it?

Thanks for the help! :)

That's just Calculus- in particular the application of the chain rule.