# Homework Help: Non-uniform acceleration

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1. Jan 18, 2016

### Magnets Yo

• post moved into HH forum, so required template is absent
I thought i would do some digging into the wonderful world of magnets, and I found some things that... well are real head scratchers...
I was wondering if anyone here could help me wrap my brain around non uniform acceleration. I'm trying to calculate velocity using distance and varying acceleration (force, technically, but i have a mass value selected soooo...) and I'm getting results that don't make sense using the simple √2ad.
I expected to hit a roadblock using it, but I wasn't expecting my results to be backwards.
the set of variables with lower net force is resulting in higher final velocity, which most certainly is not correct.
yes, I tried google, but no combination of words seemed to bring me to the equations i needed.

2. Jan 18, 2016

### Staff: Mentor

At any time t, the area under the acc-time graph gives the change of velocity up to that time.

3. Jan 18, 2016

### Magnets Yo

right, but my variables are acc and distance, not acc and time, so would the answer be; √(2 times area under graph)?

Last edited: Jan 18, 2016
4. Jan 18, 2016

### SteamKing

Staff Emeritus
Your simple formula for velocity, $v = \sqrt{2 ad}$, works only for constant acceleration.

If acceleration is varying with time or distance, you'll have to use calculus to find velocity and whatever else you're interested in.

5. Jan 18, 2016

### Staff: Mentor

Oops, I read into what you wrote something different.

You could divide your graph into say, ten regions, approximating the acceleration to a constant value within each region and applying your formulae to each to find how velocity changes over each region.

6. Jan 18, 2016

### Staff: Mentor

If you are looking for velocity when acceleration is not constant, and the object in question is moving in response to a conservative field (like gravity, or electric, or magnetic fields), then often the most expedient approach is to consider the change in potential energy. The change in potential energy is tied to the change in kinetic energy (work-energy theorem).

7. Jan 18, 2016

### Magnets Yo

thats what i thought to do, but it seems i'm using the wrong formulae for the problem, on account of the backwards reults

8. Jan 18, 2016

### Magnets Yo

care to go into a little more depth? i can do math, i just don't know what math i have to do, ya know?

9. Jan 18, 2016

### Magnets Yo

hmm... that might end up being useful... got some equations for me?

10. Jan 18, 2016

### SteamKing

Staff Emeritus
If the acceleration is not constant, then your formula is no good. You'll have to use calculus to integrate acceleration with respect to time or position to obtain velocity.

This link shows how to analyze acceleration as a function of position to obtain velocity:

http://physics.stackexchange.com/qu...-of-position-to-acceleration-as-a-function-of

11. Jan 18, 2016

### Staff: Mentor

The equations will depend upon the "shape" of the magnetic field and any materials that might be involved in conducting the field. You haven't mentioned what your setup looks like, or any other details.

12. Jan 18, 2016

### Magnets Yo

listen, gneill, i already calculated the force/acceleration. i just need the equations to get velocity from distance and varying acceleration. dont worry about my magnets, i already have that sorted. i just dont know what equations i need for varying acceleration. and the equation i've been using is apparently not applicable because its giving me incoherent results

13. Jan 18, 2016

### Ray Vickson

You need to be more specific. Do you have acceleration $a$ given as a function of position $x$, that is, $a(x)$? If so, the link in post #10 has the complete answer. If that is not what you need (perhaps because your inputs are not of the form $a(x)$) then you need to spell out what form they actually take.

14. Jan 18, 2016

### Magnets Yo

.

15. Jan 18, 2016

### Magnets Yo

i have a set of different acceleration values that stop and start at set distances, and i need the final velocity. i made a graph that might be helpful... MIGHT BE...

16. Jan 18, 2016

### Staff: Mentor

Is the stair-step effect an artifact of a measurement process, or a feature of the actual system?

Assuming that the curves of F versus d are the actual force curves and that the force always acts along the line of motion of the object, then the change in potential energy between two locations is equal to the area under the curve between those two distances.

By the work-energy theorem the change in potential energy will be the negative of the change in kinetic energy. That is, $ΔKE = -ΔPE$.

17. Jan 18, 2016

### Ray Vickson

On an interval $(t_0,t_1)$ where the force $F$ is a constant, the differential equation for position $x = x(t)$ is
$$m \frac{d^2 x}{dt^2} = F \; \Longrightarrow \: x = a + b t + \frac{1}{2} \frac{F}{m} t^2,$$
so if you know the initial position $x_0 = x(t_0)$ and initial velocity $v_0 = v(t_0)$, you can solve for the "constants of integration" $a,b$ in terms of $x_0, v_0, F$. Then you have a quadratic equation that can be solved to get $t$ in terms of $x$. If you know that $F$ = constant on the interval $x_0 \leq x \leq x_1$, you can find the time $t_1$ that gives you $x(t_1) = x_1$. If you are given both $x_0$ and $x_1$, your expression for $t_1$ will depend on the value of the initial velocity $v_0$, so you need to know that before you can continue. You can then also calculate the final velocity $v_1 = v(t_1)$, which you will need in the next part of the solution.

Now, if in the interval $x_1 \leq x \leq x_2$ the force has the form $F = \alpha - \beta x$, (with constant $\alpha>0$ and $\beta > 0$, as in your graph), the position $x(t)$ obeys the DE
$$m \frac{d^2 x}{dt^2} = \alpha - \beta x \: \Longrightarrow \: x = \frac{\alpha}{\beta} + A \cos(\omega t + \phi),$$
where $\omega = \sqrt{\beta/m}$. Here, $A, \phi$ are "constants of integration". If you know $x(t_1) = x_1$ and $v(t_1) = v_1$ from the previous interval, you can fit the values of $A$ and $\phi$ to that data. Then you can express $t$ in terms of $x$ as
$$t = \frac{\pi - \phi}{\omega} - \frac{1}{\omega} \arccos\left( \frac{\alpha - \beta x}{A \beta} \right)$$
and so solve for $t = t_2$ that gives you $x(t_2) = x_2$. You can also figure out the value of velocity $v(t_2) = v_2$, which you will need for the next part of the trajectory.

For the next part of the trajectory you have constant force again, so repeat the first step, starting at $t = t_2$ with initial conditions $x(t_2) = x_2$ and $v(t_2) = v_2$, etc., etc. Of course, the reason you want $x$ and $v$ to be continuous across the boundaries is that you cannot have a jump discontinuity in $x(t)$ and you do not have any infinite forces, so $v(t)$ cannot jump either.

You can keep going like that, switching between x-intervals of constant force to intervals of linear (in x ) force, fitting initial data in the next interval to final data in the previous interval. It is going to be messy and time-consuming, but essentially straightforward.

18. Jan 18, 2016

### Magnets Yo

It is a feature of the actual system, yes.
The object in question receives "bursts" of acceleration across a given distance that weaken from one terminal to the next

19. Jan 18, 2016

### Staff: Mentor

This thread is closed due to a heated exchange