# One-dimensional motion with w/ aceleration dependant on speed.

1. Jun 11, 2009

### Brasi333

1. The problem statement, all variables and given/known data
The accelration of a marble in a certain fluid is proportional to the speed of the marble squared and given (in SI units) by a=-3.00v2 for v>0. If the marble enters this fluid with a speed of 1.50 m/s, how long will it take before the marble's speed is reduced to half of its initial value?

2. Relevant equations

vf=vi+at

Average of f(x)=integral of f(x) with respect to x from a to b, divided by b-a.

3. The attempt at a solution

*Everything is in meters and seconds*. As this problem is very straightforward except for the acceleration, I began with using the kinematic equation above that involves vf, vi, a, and t. Since the acceleration varies with respect to speed, I figured that I would need the average acceleration. Integrating -3.00v2 from 1.5 (vi) to 0.75 (vf; half of vi) I got 2.953125. Then, dividing this by -0.75 (0.75-1.5) I got -3.9375, which should be the average acceleration. Going back to the kinematic equation, solving for t yields t=(vf-vi)/a, where I have the following values:
vf= 0.75
vi= 1.5
a=-3.9375 (average acceleration, calculated above).

Using these values I get a time of t=0.190 seconds. The answer in the book is 0.222 seconds. At this point I've gone over the problem multiple times and I can't see what I'm doing wrong. However, this is my first time doing physics in a long while, and my method is of my own creation, not from the book or anything, so I could be off the ball on that. By working backwards I can see that the average acceleration should be -3.375. Any help would be greatly appreciated, thank you.
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Jun 11, 2009

### Cyosis

It goes wrong, because you're using equations that are valid for constant acceleration, which is not the case here.

You have been given a relation between a and v^2. Remember that a=dv/dt, can you solve the differential equation $dv/dt=-3.00v^2$?

3. Jun 11, 2009

### Brasi333

Thanks for the help Cyosis.

You got me reviewing differential equations sooner that I figured :P. I got the right answer, but Im showing how I solved it in case others want to look at this example. (As an aside, could you explain to me how you know when to resort to differential equations? Is it just because all they gave essentially was a non-constant acceleration, and we had to work backwards from that?)

Ok, dv/dt=a=-3.00v2

dv/(-3v2)=dt (separation of variables)

1/(3v)=t+C (integrate with respect to t)

solving for v gives v=1/[3(t+C)]

v(0)=1.5=1/(3C) (initial conditions)

solving for C gives C=1/4.5 or 0.222

now we have v(t)=1/[3(t+0.222)]

solving for t gives t=1/(3v)-0.222

The question asks for the time when v=0.5vi, or 0.5*1.5. So, using v=0.75 gives t=0.222s.

Thanks Cyosis!

4. Jun 11, 2009

### Cyosis

You're welcome.

It's not so much of when to resort to using differential equations, but more of a lot of equations in physics actually being differential equations. For example F=ma, I am sure you know that equation but a=dv/dt and v=dx/dt so we can write F=ma as a first order differential equation with respect to v $F=m \frac{dv}{dt}$ or a second order differential equation in x, $F=m \frac{d^2x}{dt^2}$. In fact all equations of motion are differential equations.

Lets do an example quickly, calculating the equation of motion for a falling particle. The force is mg so we have $mg=m \frac{dv}{dt}$, integrating on both sides yields v=gt+c, c being the velocity at t=0.