# Newton's Law

1. Jun 12, 2013

### An1MuS

1. The problem statement, all variables and given/known data

A ball is being pulled by a force given by $F=kv^2$. Where "k" already includes the mass of the ball. After 10m it will run into a wall. The initial speed is 1m/s and there are no other forces involved. How much time does it take for the ball to hit the wall?

2. Relevant equations

$kv^2=\frac{dv}{dt}$

3. The attempt at a solution

So i know that this is a differential equation of separable variables.
$\frac{dv}{v^2}=kdt$
I probably should integrate both sides, but i don't know what limits i should use if any, and if i'm on the right track

$\int \frac{dv}{v^2}=\int kdt$

Last edited: Jun 12, 2013
2. Jun 12, 2013

### Staff: Mentor

If the velocity at time t = 0 is equal to zero, then the ball never moves, and an infinite time is required to reach the wall. If the initial velocity is v0, then you can use that as the initial condition.

3. Jun 12, 2013

### Andrew Mason

??? I thought F = kv2. dv/dt = a = F/m

You want to express position as a function of time. You have to know its initial speed. Is the initial speed (at 10 m from the wall) = 0? Is there any other force on the ball (eg. friction)? Can you state the whole problem as given?

AM

4. Jun 12, 2013

### Staff: Mentor

Don't forget to include the mass when you replace F with ma. Yes, k contains the mass, but it's a separate equation from F = ma. Thus $m a = k v^2$., and so $a = (k/m)v^2$.

5. Jun 12, 2013

### An1MuS

The whole problem as given is actually a fluid mechanics problem where a ball is dropped and the drag force increases with speed squared. I just stated this one, because if i know how to solve this simpler case i should know how to solve the fluid mechanics one.

You guys are right, I updated the OP with the info you asked.

Last edited: Jun 12, 2013
6. Jun 12, 2013

### Andrew Mason

If it is a drag force, the acceleration is negative. Are you sure that F = kv2 and not F = -kv2?

AM

7. Jun 12, 2013

### An1MuS

In the fluid problem it's negative, but for the sake of learning, i put it positive to be simpler. Then it's just a matter of me putting a minus sign in the fluid exercise where it's needed. I made up this exercise avoiding fluid mechanics jargon so that more people could understand (and reply to) the exercise, and also because i just need to understand the underlying principle of solving this (simpler) exercise for me to solve the fluid mechanics one. Hope this makes sense...

8. Jun 12, 2013

### Arkavo

absolutely correct if initial velocity is 0 the force never gets a chance to act

9. Jun 12, 2013

### Andrew Mason

You have to find the solution to this differential equation:

dv/dt = kv2/m

dv/v2 = kdt/m

That should be easily solved using integration, using the value for the initial speed at t=0.

AM

10. Jun 12, 2013

### An1MuS

"m" should already be inside k.

$kt=\frac {-1}{v_f}- (-\frac {1}{v_i})$ where $v_i=1$ But then where do the 10meters come in? And what about $v_f$?

11. Jun 12, 2013

### Staff: Mentor

You should really be integrating with respect to x, rather than with respect to t, by substituting v(dv/dx) for dv/dt. That way you can find the velocity directly at x = 10.

12. Jun 13, 2013

### Andrew Mason

What does that mean? You have written F = kv2. F = ma so a = kv2/m unless you want to dispute Newton's second law.

Take Chestermiller's suggestion and integrate with respect to distance.

AM

13. Jun 13, 2013

### An1MuS

Y, you're right. So it should be

$F=mkv^2$ so that it gives $kv^2 = \frac {dv}{dt}$

Integrating in respect to x, is this it?

$\int kv^2 dx = \int v\frac{dv}{dx}dx$

And if it's this i don't know how to continue

Also i don't understand this substitution $\frac{dv}{dt}=v\frac{dv}{dx}$

14. Jun 13, 2013

### Staff: Mentor

No. That's not it. If

$kv^2 = v\frac{dv}{dx}$

then you can cancel a v out from both sides of the equation, and obtain:

$kv = \frac{dv}{dx}$

The solution to this differential equation is

$v=v_0e^{kx}$

The substitution $\frac{dv}{dt}=v\frac{dv}{dx}$ comes from $v=\frac{dx}{dt}$ so that $dt=\frac{dx}{v}$

$\frac{dx}{dt}=v_0e^{kx}$

So, $$dt=\frac{e^{-kx}}{v_0}dx$$
So, $$t=\frac{1-e^{-kx}}{kv_0}$$

Last edited: Jun 13, 2013
15. Jun 13, 2013

### Andrew Mason

You have:

F = md2x/dt2 = mdv/dt = mkv2

That can be rearranged to:

dv/v2 = kdt

Since v = dx/dt, dt = dx/v, so this becomes:

dv/v = kdx

Find the solution by integration and solve for xf-xi = 10m.

AM