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How the heck do you find the position as function of time??

It seems so simple but really has me stumped. It must be possible.

I have played around with vdv=ads and integrated trying to unlock time..

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- #1

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How the heck do you find the position as function of time??

It seems so simple but really has me stumped. It must be possible.

I have played around with vdv=ads and integrated trying to unlock time..

- #2

SammyS

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Write [itex]\ a=2x\ [/itex] as [itex]\displaystyle\ \frac{dv}{dt}=2x\ .[/itex]say you are given the acceleration as a=2x and you know it has an initial velocity of 5m/s at x=0,t=0

How the heck do you find the position as function of time??

It seems so simple but really has me stumped. It must be possible.

I have played around with vdv=ads and integrated trying to unlock time..

Then use the chain rule to write [itex]\displaystyle\ \frac{dv}{dt}\ [/itex] as [itex]\displaystyle\ \frac{dv}{dx}\,\frac{dx}{dt}\,,\ [/itex] but realize that [itex]\displaystyle\ \frac{dx}{dt}=v\ .[/itex]

That leaves you with a separable differential equation with variables, x and v .

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Ya I agree, and once you separate variables and integrate you can obtain v=sqrt(2*x^2+25)Write [itex]\ a=2x\ [/itex] as [itex]\displaystyle\ \frac{dv}{dt}=2x\ .[/itex]

Then use the chain rule to write [itex]\displaystyle\ \frac{dv}{dt}\ [/itex] as [itex]\displaystyle\ \frac{dv}{dx}\,\frac{dx}{dt}\,,\ [/itex] but realize that [itex]\displaystyle\ \frac{dx}{dt}=v\ .[/itex]

That leaves you with a separable differential equation with variables, x and v .

That leaves me no closer to finding the position as function of time...

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SammyS

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That's not quite right.Ya I agree, and once you separate variables and integrate you can obtain v=sqrt(2*x^2+10)

That leaves me no closer to finding the position as function of time...

It gives v = √(10) when x = 0 .

Once you fix that, remember, [itex]\displaystyle\ v=\frac{dx}{dt}\ .[/itex]

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My bad I meant to write 25. My initial question was asking for the position as a function of time as the particle moves down the positive x-axis given the specific initial conditions stated above.That's not quite right.

It gives v = √(10) when x = 0 .

Once you fix that, remember, [itex]\displaystyle\ v=\frac{dx}{dt}\ .[/itex]

I don't care about the velocity as a function of x nor the velocity at x=0. Those are all easily obtainable..

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- #7

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say your are giving the acceleration as a=2x and you know it has an initial velocity of 5m/s at x=0,t=0

How the heck do you find the position as function of time??

It seems so simple but really has me stumped. It must be possible.

Moderator's Note: Multiple threads merged

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TSny

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The trick is to use the chain rule to write a = dv/dt = dv/dx ##\cdot## dx/dt = v dv/dx.

See if you can use this to find v as a function of x. Then see if you can figure out a way to determine x as a function of t.

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Pengwuino

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Are you familiar with differential equations? Are you sure it's not [itex]a(t) = 2t[/itex]? If you really mean [itex]a(t)= 2x(t)[/itex], then you have a second order differential equation.

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oh my god I am such an idiot for not recognizing the diffeq... Thank you

I will put my head in a hole now

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SammyS

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As I said, [itex]\displaystyle\ \frac{dx}{dt}=v\ .[/itex]My bad I meant to write 25. My initial question was asking for the position as a function of time as the particle moves down the positive x-axis given the specific initial conditions stated above.

I don't care about the velocity as a function of x nor the velocity at x=0. Those are all easily obtainable..

That gives you the differential equation:

[itex]\displaystyle\ \frac{dx}{dt}=\sqrt{2\,x^2+25\,\ .}[/itex]

... and you should care about the velocity at x=0 . That's one of your boundary conditions.

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TSny

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Yes, if you are familiar with 2nd order differential equations, Pengwuino's suggestion is the way to go. My suggestion involves solving a couple of first-order differential equations using separation of variables and takes more steps. Not as good.

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