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