Easiest problem that I can't figure out

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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.

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

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  • #2
SammyS
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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..
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 .
 
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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 .
Ya I agree, and once you separate variables and integrate you can obtain v=sqrt(2*x^2+25)
That leaves me no closer to finding the position as function of time...
 
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  • #4
SammyS
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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...
That's not quite right.

It gives v = √(10) when x = 0 .

Once you fix that, remember, [itex]\displaystyle\ v=\frac{dx}{dt}\ .[/itex]
 
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That's not quite right.

It gives v = √(10) when x = 0 .

Once you fix that, remember, [itex]\displaystyle\ v=\frac{dx}{dt}\ .[/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..
 
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I am going to move this post to advanced physics. Please don't get mad at me moderators as I think there is a simple solution to this.
 
  • #7
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Hardest problem that I can't figure out...

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​
 
Last edited by a moderator:
  • #8
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.
 
  • #9
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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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.
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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This thread can be deleted
 
  • #12
SammyS
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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.
I don't care about the velocity as a function of x nor the velocity at x=0. Those are all easily obtainable..
As I said, [itex]\displaystyle\ \frac{dx}{dt}=v\ .[/itex]

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.
 
  • #13
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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