# Velocity and Acceleration Question

1. Jul 10, 2004

### faust9

OK, got a test Monday, and my professor e-mailed us our review for siad test this past Friday. His tests are very similar to his reviews so did I do this right?

Question:
The motion of a particle is described as $v=2x-\frac{9}{x} \ {\rm for}\ (0 \leq t \leq 10)s\ {\rm where}\ t_0=0,x_0=0$

Ok, this is what I did:

a) Determine the distance traveled in t=4 seconds

$$v=2x-\frac{9}{x}=\frac{dx}{dt}$$

$$\frac{2x^2-9}{x}dx=dt$$

$$\int \frac{2x^2-9}{x}dx=\int dt$$

$$\int \frac{1}{4U}dU=t+c$$

$$\frac{1}{4}\ln (2x^2-9)=t+c$$

$${\rm let }\ t=0,\ {\rm and }\ x=3$$

$$c=\frac{ \ln (9)}{16}$$

thus:

$$x^2=\frac{9}{2}e^{4t}+\frac{9}{2}$$

$$x(4)\approx 6323.567 m$$

b)How long will it take the particle to reach 0.02c?

$$0.02c \approx 1.8 \times 10^8$$

so I plugged that into the original equation
$$1.8 \times 10^8 =\frac{2x^2-9}{x}$$

$$2x^2-1.8 \times 10^8x-9=0$$

$$x=\{ -5 \times 10^{-8},9.0 \times 10^7\}$$

the first was a trivial solution so:

$$t=\frac{1}{4}\ln (2(9.0\times 10^7)^2-9)-\frac{\ln 9}{16}$$

$$t=8.7816s$$

c) Express acceleration as a function of x

$$a=v\frac{dv}{dx}$$

$$a=(2x-\frac{9}{x})(2x-\frac{9}{x})\frac{d}{dx}$$

$$a=(2x-\frac{9}{x})(2+\frac{9}{x^2})=4x-\frac{81}{x^3}$$

d) Determine the acceleration at 0.02c

$$a=4(9.0\times 10^7)-\frac{81}{(9.0\times 10^7)^3}=3.6\times 10^8$$

Is the above correct? I was talking to another student who did it differently and got a different answer. I want to make sure the above is correct so I don't miss this question on the test. Any insight would be greatly appreciated. Thanks.

Last edited: Jul 10, 2004
2. Jul 10, 2004

### AKG

You did it wrong. It seems like a weird question though. Notice that at t=0, x=0, and velocity is undefined (it approaches $-\infty$).

$$v = 2x - \frac{9}{x}$$

$$\frac{dx}{dt} = 2x - \frac{9}{x}$$

$$\frac{dx}{2x - \frac{9}{x}} = dt$$

$$\frac{x}{2x^2 - 9}dx = dt$$

$$\int _{x(0)} ^{x(4)} \frac{x}{2x^2 - 9}dx = \int _0 ^4 dt$$

Now look at the function on the left that we'll be integrating. It is not defined at $x = \sqrt{4.5}$, however, we don't even know if this thing shows up. Assuming it doesn't, we can proceed as follows:

$$u = 2x^2 - 9$$

$$du = 4xdx$$

$$dx = \frac{du}{4x}$$

$$\frac{1}{4} \int _{-9} ^{2[x(4)]^2 - 9} \frac{du}{u} = 4$$

$$\ln \frac{2[x(4)]^2 - 9}{-9} = 16$$

$$x(4) = \sqrt{\frac{-9e^{16} + 9}{2}}$$

The above is undefined. This question is very strange. Normally, you have velocity given as a function of time, and you can integrate that to find x(t), but given v(x), it seems you have to solve some sort of differential equation. I tried this integration, doubtful it would work, just to see what it would give. I might have done something wrong, but in general this question seem weird. Are you sure the question you've given us is the actual question?

3. Jul 10, 2004

### faust9

Oh I'm sorry, $x_0=3$ not $x_0=0$.

Other than that, the question was posed as I described. Velocity was initially given as a function of x.

4. Jul 10, 2004

### AKG

I think I know what's going on here

I think you meant $x_0 = 3$, not $x_0 = 0$. That would explain one of your lines of work, and the problem seems to make a little more sense.

Also, you mean $c = \frac{\ln (9)}{4}$. Try to take a little more care, no need to rush, because it's too confusing otherwise, and makes it difficult to help you. Now, although you got the answer that you did for (a), and it seems right, you've used indefinite integration, and I don't know if that is "allowed." I suppose you have to assume that it is, however, if $x = \sqrt{4.5}$ on the interval $0 \leq t \leq 4$, you wouldn't be allowed to integrate like that, I believe.

(c) seems okay, as long as you did the final simplification right (I didn't check).

As long as you plugged stuff in right (for b and d), and are allowed to do that integration for (a), everything seems fine. Well, your answers seem fine, you made some strange mistakes along the way though:

* I think you meant that x(0) = 3
* The integration you started to do was upside down, however, midway through your work you flipped it around again. In this case, two wrongs made a right, but I don't know if your markers will like that
* c = ln(9)/4

5. Jul 10, 2004

### AKG

Looking back, with x(0) = 3, my approach gives the same answer as yours. I would suggest that in situations like these, it is better to use definite integration, not indefinite. For one, it eliminates the need to solve explicitly for "c." Even if you just want a generally formula for x(t), and not specifically x(4), for example, all you need to do is replace the limits of integration x(4) with x(t) and 4 with t (on the right side, where you're integrating 1dt.

Also, I don't know if it is necessary, but the equation you arrived at for x never gets $\sqrt{4.5}$ as it's value, and given that x, dx/dt is indeed the given v, so there is no inconsistency in using that x, so it's safe to use it.

6. Jul 10, 2004

### faust9

Thank you. I'll go over it once again and see if there's another way.

The flipped integral--I did it correclty on paper; however, I inadvertently flipped it when I was typing it in. I'm not a profecient keyboard artist.

 Cool beans (in regards to your post immediatly above). Again thanks.

Last edited: Jul 10, 2004
7. Jul 10, 2004

### AKG

You've confused me again. You know that

$$c = \frac{\ln 3}{2}$$

... is wrong also, right? And beans??? You're welcome.

8. Jul 10, 2004

### faust9

I know I fixed it. I redid it using definate integration and saw my faux pas.

9. Jul 10, 2004

### AKG

Beans? I don't get it.

10. Jul 10, 2004

### faust9

Its a colloquialism. It's kind of like aloha in that it has many meanings depending on the context used. In this case it ment I got it and thanks.