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!