Motion in 1D under a resistive force

[pleasemindthe]

Homework Statement

A particle moving on a straight line is subject to a resistive force of the magnitude kv^n, where v is the velocity at time t and k is a positive real constant.

Find the times and distances at which the particle comes to rest i.e. v=0, for the following cases (this assumes at t=0, x=0, v=v0.)

n<1 ?
1<n<2 ?
n=2 ?

Homework Equations

F=ma

The Attempt at a Solution

I've tried by getting two equations, one for time and one for distance.

So I set F=ma= m(dv/dt) = -kv^n for time
and F = ma = mv(dv/dx) = -kv^n for distance

I then plodded along, and took v=0 to get two similar equations for time and distance...

(m*v0^(1-n)) / (k (1-n)) = t and
(m*v0^(2-n)) / (k (2-n)) = x

but I'm not sure if that's at all right! They'd be fine for n<1, but the time one screws up for 1<n<2 and the distance is special for n=2.

Basically I'm not sure I've gone the right way at all... I think I need an e^f(t) so I can take limits but I've just got my head in a muddle.

 

[anathema]

Your approach is on the right track, but there are a few things that need clarification.

Firstly, your equations for time and distance are not quite correct. For time, you should have:

t = (1/k)∫(v0 - v)^-n dt

And for distance:

x = (1/k)∫(v0 - v)^-n v dt

You can use integration by parts to solve these integrals, which will give you a general solution for t and x in terms of v0 and n.

Secondly, your equations are actually fine for all values of n. If you take the limit as n approaches 1 or 2, you will find that the equations still hold.

Finally, you are correct in thinking that you need an exponential function in order to take limits as n approaches 1 or 2. To do this, you can use the fact that:

lim x→1 (1 - x)^-1 = e

lim x→2 (1 - x)^-2 = e^2

By using these limits in your equations for time and distance, you can get solutions that are valid for all values of n, including 1 and 2.

I hope this helps. Good luck with your problem!