Advanced mechanics - x(t) from v(x)

[tourjete]

Homework Statement

A particle of mass m's velocity varies according to bx^-n

Find the position as a function of time, setting x = x₀ at t=0

Homework Equations

v(x) = bx^-n

possibly relevant: f(x) = -b²mnx^-2n-1

The Attempt at a Solution

The first part of the question asked me to find the force acting on the particle as a function of x, which I did using the chain rule. I'm a little unclear as to whether I need f(x) to get x(t).

Anyway, here's my attempt at a solution:
dv/dt = (dv/dx)*(dx/dt)
dv/dt = (dv/dx) * v(x)

Both of these quantities are known so I plugged them in and got an expression for dv/dt. I then tried to integrate that expression twice, once to get v(t) and another time to get x(t). However, when I do that I just get the expression times t²/2, which would make x(0) = 0, not x₀ as the problem statement gives.

Am I doing this the complete wrong way or am I on the right track and just not understanding calculus?

 

[Dickfore]

<br /> v = \frac{d x}{d t} = b x^{-n}<br />
This is a 1st order ODE with separable variables. The variables separate as:
<br /> x^{n} \, dx = b \, dt<br />
which can be integrated by elementary tabular integrals. The one arbitrary constant is found from the initial condition.

 

[tourjete]

Thanks!

I solved the ODE by integrating both sides and got xⁿ⁺¹/(n+1) = bt. I don't see where the constant comes into play here.

 

[rude man]

Finish the expression for the integration of your ode. What must one add to every indefinite integral of df/dx?

 

[tourjete]

so I have:

xⁿ⁺¹/(n+1) = bt + C

then I plug in the initial condition x=x₀ at t = 0

C = x₀ⁿ⁺¹/(n+1)

I just need to solve this for x now, correct?

 

[Dickfore]

yes.

 

[rude man]

Molto bene, molto bene!