Need help converting a function of displacement to a function of time.

[lylos]

Homework Statement

Assume we have some mass, m, that varies it's distance as v(x)=1/x^2. Assume v(x=0)=0 at t=0.

What is the force F(x) that causes this motion?
Find x(t).
What is F(t)?

Homework Equations

$$F=m\ddot{x}$$

The Attempt at a Solution

I think I need to take the derivative of my velocity with respect to x to find my acceleration. With that I can find the force as a function of x? I think this may be wrong, but I'm really not sure. Could someone help me out as to where to begin?

 

[Hootenanny]

HINT: By the chain rule

$$a = \frac{dv}{dt} = \frac{dv}{dx}\overbrace{\frac{dx}{dt}}^{v}$$

$$\Rightarrow a = v\frac{dv}{dx}$$

 

[lylos]

HINT: By the chain rule

$$a = \frac{dv}{dt} = \frac{dv}{dx}\overbrace{\frac{dx}{dt}}^{v}$$

$$\Rightarrow a = v\frac{dv}{dx}$$

By this, I find the following, by the way... in the problem I'm solving v(x)=ax^-3, where a is a constant.

$$v(x)=ax^{-3}$$
$$F(x)=m\ddot{x}$$
$$F(x)=m\frac{dv}{dt}$$
$$F(x)=m\frac{dv}{dx}\frac{dx}{dt}$$
$$\frac{dv}{dx}=-3ax^{-4}$$
$$F(x)=m*-3ax^{-4}*ax^{-3}=-3ma^2x^{-7}$$
$$F(t)=m\frac{dv}{dt}=-3ma^2x^{-7}$$
$$\frac{dv}{dt}=-3a^2x^{-7}$$
$$dv=-3a^2x^{-7}dt$$
$$\int dv=\int -3a^2x^{-7}dt$$
$$v(t)=\frac{dx}{dt}=-3a^2x^{-7}t$$
$$x^7 dx=-3a^2t dt$$
$$\int x^7 dx=\int -3a^2t dt$$
$$\frac{x^8}{8}=\frac{-3}{2}a^2t^t$$
$$x^8=-12a^2t^2$$
$$x(t)=(12a^2t^2)^{1/8}$$

 

[Hootenanny]

By this, I find the following, by the way... in the problem I'm solving v(x)=ax^-3, where a is a constant.

$$v(x)=ax^{-3}$$
$$F(x)=m\ddot{x}$$
$$F(x)=m\frac{dv}{dt}$$
$$F(x)=m\frac{dv}{dx}\frac{dx}{dt}$$
$$\frac{dv}{dx}=-3ax^{-4}$$
$$F(x)=m*-3ax^{-4}*ax^{-3}=-3ma^2x^{-7}$$
$$F(x)=m\frac{dv}{dt}=-3ma^2x^{-7}$$

You're good upto here, after this things get a bit hairy. When integrating don't forget that x is a function of time, therefore you should note that

$$\int x(t) dt \neq xt+ const.$$

Notice that the question asks you to determine x(t) before finding F(t), therefore it might be easier to find x(t) next.

 

[lylos]

You're good upto here, after this things get a bit hairy. When integrating don't forget that x is a function of time, therefore you should note that

$$\int x(t) dt \neq xt+ const.$$

Notice that the question asks you to determine x(t) before finding F(t), therefore it might be easier to find x(t) next.

So to find x(t)?:

$$v(x)=ax^{-3}=\frac{dx}{dt}$$
$$x^3 dx = a dt$$
$$\int x^3 dx = \int a dt$$
$$\frac{x^4}{4}=at$$
$$x(t)=(4at)^{1/4}$$

 

[Hootenanny]

So to find x(t)?:

$$v(x)=ax^{-3}=\frac{dx}{dt}$$
$$x^3 dx = a dt$$
$$\int x^3 dx = \int a dt$$
$$\frac{x^4}{4}=at$$
$$x(t)=(4at)^{1/4}$$

Looks good to me