# Homework Help: Velocity as a function of position problem (simple i think)

1. Apr 6, 2013

### joshmccraney

1. The problem statement, all variables and given/known data
If we know that velocity $v$ of a particle as a funciton of position $x$ is: $$v(x)=20-2/3x$$
then i am asked to determine the acceleration when $x=15$. I am then asked to show that the particle never actually reaches $x=30$

2. Relevant equations
$$vdv=adx$$
$$vdt=dx$$
$$adt=dv$$

and $a$ is the acceleration (not necessarily constant) and $t$ is time.

3. The attempt at a solution
I have already reduced the problem to what we see in $1.$ I have tried to work with the relevant equations but i cant make sense of the second two since they are time-dependent, which is a variable i have no information on. However, intuitively, i feel to answer the second part i would take a limit as t approaches infinity, so I am conflicted.

Thanks everyone for the help! you people are awesome!!

2. Apr 6, 2013

### SammyS

Staff Emeritus
v(x)=20-2/3x\$
By the way, is that
$\displaystyle \ v(x)=20-\frac{2}{3x}\ ?$​
... or is that
$\displaystyle \ v(x)=20-\frac{2}{3}x\ ?$​

You need to find the velocity, v, when x = 15 ?

Then, take the derivative of v with respect to t . (Both sides of the equation.)

3. Apr 6, 2013

### joshmccraney

i thought order of operations made it linear, that is,
$\displaystyle \ v(x)=20-\frac{2}{3}x\$​
but i guess youre probably used to posts getting sloppy?

wait, no, i need to find acceleration when x = 15. i think that's what i posted

yes, this i know, but there are no t's. i assume position is a function of time, and thus i have to reckon with the chain rule. can you elaborate on what you mean?

4. Apr 6, 2013

### SammyS

Staff Emeritus
Yes. That's it exactly.

If you work this out first;
$\displaystyle \frac{d}{dt}v=\frac{d}{dt}\left(20-\frac{2}{3}\,x\right)\,,$

then you will see that you also need: the velocity, v, when x = 15 ?

5. Apr 6, 2013

### joshmccraney

ahhh yes that makes perfect sense (although i would have never thought of that). the answer is -20/3---thanks! but do you have any recommendations for showing that x = 30 is impossible? i was thinking of:

$\displaystyle \ 30=\int_0^\infty (20-\frac{2}{3}x)dt\$​
it seems if this is true we would have shown that the only time we will reach x = 30 is in the limit (i can finesse this at the end). any suggestions as to how i proceed?

thanks so much for the advice!

6. Apr 6, 2013

### Curious3141

$\displaystyle \ 30=\int_0^\infty (20-\frac{2}{3}x)dt\$​

How would you go about showing that?

I would just solve the differential equation $\displaystyle \frac{dx}{dt} = 20 - \frac{2}{3}x$ to get $x$ in terms of $t$ then prove that $x<30$ for all $t$.

7. Apr 6, 2013

### joshmccraney

you guys are too smart. thanks!