Velocity as a function of position problem (simple i think)

[member 428835]

Homework Statement

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

Homework Equations

$$vdv=adx$$
$$vdt=dx$$
$$adt=dv$$

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

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 can't 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!

 

[SammyS]

Homework Statement

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

Homework Equations

$$vdv=adx$$
$$vdt=dx$$
$$adt=dv$$

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

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 can't 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!

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.)

 

[member 428835]

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?

v(x)=20-2/3x$

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

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

v(x)=20-2/3x$
Then, take the derivative of v with respect to t . (Both sides of the equation.)

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?

 

[SammyS]

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?

Yes. That's it exactly.

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?

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 ?

 

[member 428835]

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!

 

[Curious3141]

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!

\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 $t$.

 

[member 428835]

you guys are too smart. thanks!