# Dynamics Problem

1. Jan 23, 2008

### ncr7

Hey I have been looking at the problem for a long time and for some reason I can't think of how to solve for it... I know I have to do some Calc to figure it out.

Ok so here it is

The acceleration of a particle is defined by the relation a = -0.05$$[v]^{2}$$, where
a is expressed in m/$$^{2}$$ and v in m/s. The particle starts at s=0m with a velocity of 5 m/s. Determine (a) the velocity v of the particle after it travels 10 m, (b) the distance s the particle will travel before its velocity drops to 2 m/s, (c) the distance s the particle will
travel before it comes to rest.

so far I think what I have to do is take the derivative of a which ends up being -.1v but after that I am stuck because now I have velocity as a function of velocity... any ideas? I realize I am most likely doing this totally wrong.

2. Jan 23, 2008

### Staff: Mentor

Since $a = dv/dt$, you'll need to integrate to find v. Start there.

3. Jan 23, 2008

### ncr7

so I'll get
$$\int$$dv=$$\int$$a dt
(v2-v1) = -0.05$$v^{2}$$*t - 0 (since $$t_{0}$$=0)

but I now have 2 unknowns though.
v and t

since I know the change in position shouldn't I do an integration over that? I'm just unsure how exactly it should be done

4. Jan 23, 2008

### Staff: Mentor

:yuck:

Before you can integrate properly, you must separate the variables:
$$\frac{dv}{dt} = -0.05v^2$$

$$\frac{dv}{v^2} = -0.05 dt$$

Now try integrating. (And don't forget the integration constants.)

5. Jan 23, 2008

### ncr7

ok, I'm starting to have some wierd mistakes because I have been on this problem to long lol... but I don't have two integration constants for t or v. I have the change in s from 0m to 10m. This would only really be able to give me where it's time when it comes to a complete stop which is part of the problem. But right now I need to find velocity at 10m.

6. Jan 23, 2008

### Staff: Mentor

One step at a time. First complete the integration to find v as a function of t. Then, using v = dx/dt, integrate once more.

7. Jan 23, 2008

### Shooting Star

In this case, it would save a lot of trouble if he writes a = v(dv/dx). All the answers want some x at some v.

8. Jan 23, 2008

### Staff: Mentor

Even quicker! (I was trying to keep it as straightforward and simple as possible.)

9. Jan 23, 2008

### ncr7

ok I think I got it.

so taking

$$\int_0 ^{10m} -0.05v^2 dx = \int_0^v v dv$$

I get

$$-0.05x|_0 ^{10} = ln|v||_{5m/s} ^v$$

$$e^{1.11} = v$$

getting 3.03 m/s

I think that's right...

Last edited: Jan 23, 2008
10. Jan 24, 2008

### Staff: Mentor

$$v\frac{dv}{dx} = -0.05v^2$$
$$\frac{dv}{v} = -0.05dx$$
$$ln(v) = -0.05x + C$$