# Easiest problem that I can't figure out

• MotoPayton

#### MotoPayton

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..

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 $\ a=2x\$ as $\displaystyle\ \frac{dv}{dt}=2x\ .$

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

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

Write $\ a=2x\$ as $\displaystyle\ \frac{dv}{dt}=2x\ .$

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

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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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, $\displaystyle\ v=\frac{dx}{dt}\ .$

That's not quite right.

It gives v = √(10) when x = 0 .

Once you fix that, remember, $\displaystyle\ v=\frac{dx}{dt}\ .$

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.

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.

Last edited by a moderator:

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.

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

Are you familiar with differential equations? Are you sure it's not $a(t) = 2t$? If you really mean $a(t)= 2x(t)$, 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

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, $\displaystyle\ \frac{dx}{dt}=v\ .$

That gives you the differential equation:
$\displaystyle\ \frac{dx}{dt}=\sqrt{2\,x^2+25\,\ .}$​

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

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.