Non-constant acceleration, solving for velocity

[turtles123]

Hello!

I am trying to solve for the velocity in terms of position of a particle moving with non-constant acceleration.
a=c*t (where c is a constant)

I can easily solve for velocity in terms of t.
dv/dt=a
dv/dt=c*t
I differentiate and get v=1/2*c*t^2+v0 (where v(0)=v0)

However I am not sure how to solve for velocity in terms of only position. I would know how to do this if acceleration was proportional to velocity, but since it is proportional to time, I am not sure what to do to get rid of the variable t.

Let me know if anyone has any suggestions.

 

[PeroK]

What's position as a function of time (for this motion)?

 

[.Scott]

In general, knowing the acceleration function will only give you changes in velocity.
To know the velocity function, you will also need to know the initial velocity - or the velocity at some point in time.

 

[turtles123]

We are not given the position as a function of time. We are only given acceleration and are to assume that v(0)=v0 (a constant) and x(0)=x0 (a different constant).

 

[PeroK]

We are not given the position as a function of time.

If you are not given position as a function of time, you could always calculate it!

 

[turtles123]

Yes, I did solve for position as a function of time, but then to make velocity of a function of position is very messy. When I plug in time= from the velocity equation into the position equation I get a very funky and long result, that I can't put in terms of x=

 

[turtles123]

Alternatively if I try to solve for t from the x(t) equation it is very hard to do so because I have #t^3+#t.

 

[PeroK]

Alternatively if I try to solve for t from the x(t) equation it is very hard to do so because I have #t^3+#t.

The general problem does look a bit gnarly. Why are you doing this?

 

[.Scott]

Alternatively if I try to solve for t from the x(t) equation it is very hard to do so because I have #t^3+#t.

I thought you had a=ct and v0 = v0 ... c and V0 are constants.

If this is the case, you do not need to compute position as a function of time. If you did, you would need to include a p0.

dv(t)/dt = a(t)
So integrate.

 

[turtles123]

But then I get velocity in terms of time and I need velocity in terms of position, x.

 

[.Scott]

But then I get velocity in terms of time and I need velocity in terms of position, x.

So you will need a p0. So you solve p=f(t) - which will be a quadratic. The you solve for the quadratic.

 

[kuruman]

1. You know that $v=v_0+\frac{1}{2}ct^2.$
2. Solve this equation for $t$ in terms of $v$ and the constants.
3. Observe that $$a=\frac{dv}{dt}=\frac{dv}{dx}\frac{dx}{dt}=v\frac{dv}{dx}~\Rightarrow~\frac{dv}{dx}=\frac{a}{v}=\frac{ct}{v}.$$4. 4. Replace $t$ with what you got in step 2.
5. Separate variables and integrate.