# If acceleration is stated is terms of velocity...

1. Dec 8, 2016

### k_squared

1. The problem statement, all variables and given/known data
This is not a particular problem but a generic one. If one has acceleration stated in terms of velocity (or position) such as a=v, how do we convert these values into time?

2. Relevant equations

And of course, the usual time-based derivative relations for acceleration, velocity and positions.

3. The attempt at a solution
Sorry, guys, this one is killing me right now.

2. Dec 8, 2016

### Orodruin

Staff Emeritus
First of all, this cannot be the relationship as it is dimensionally inconsistent.

Second, what you have is just a differential equation that you need to solve.

3. Dec 8, 2016

### cnh1995

You can use the integral-differential relationship between the two quantities. Also, a=v is dimensionally incorrect. The equation could be a=kv, where k has the unit of time-1.

4. Dec 11, 2016

### k_squared

Allright, clarified question... let's saw we have $$\alpha=50\sqrt{\omega}$$. This is, I believe, equivalent to the diffyQ $$y''=50\sqrt{y'}$$. The book solves the equation by taking from $$dw/dt=a$$ $$dt=dw/a$$, and integrating with bounds that are suggested by the problem. Wolframalpha solves the problem by giving $$y=\frac{625t^3}{3}+c_1^2+25c_1t^2+c_2$$

So my real question is.. how do I reconcile these two, or am I wrong about something? (I can't find appropriate values of the constants of integration in the case of the last equation!)

5. Dec 11, 2016

### fpsulli3

In that case, $y$ is $\theta$ and the rest is the same, although I think Wolfram Alpha must have had a factor of $t$ on the $c_1^2$, did it not?

Since $\alpha = d\omega / dt$, you can rewrite the first equation as:

$$\frac{d\omega}{ dt} = 50\sqrt{\omega}$$

And then through separation of variables:

$$\frac{d\omega}{50\sqrt{\omega}} = dt$$

Then integrate both sides:

$$\frac{1}{50}\int\frac{d\omega}{\sqrt{\omega}} = \int dt$$

$$\frac{1}{25}\sqrt{\omega} = t + c_1$$

$$\omega= 625t^2 + 1250c_1t + 625c_1^2$$

Note that if we roll a factor of 25 into $c_1$, this becomes:

$$\omega= 625t^2 + 50c_1t + c_1^2$$

This is ok to do because the constants are arbitrary. And since $\omega = d\theta / dt$, you can repeat the process:

$$\frac{d\theta}{dt} = 625t^2 + 50c_1t + c_1^2$$

$$d\theta = (625t^2 + 50c_1t + c_1^2) dt$$

$$\int d\theta = \int (625t^2 + 50c_1t + c_1^2) dt$$

$$\theta = \frac{625}{3}t^3 + 25c_1t^2 + c_1^2 t + c_2$$

As for the boundary conditions, you can redo the above using the boundary conditions as the limits of integration. Or, you can use the above equation to plug the values for $y$ and $t$ that you are given, and solve for the arbitrary constants. You'll need two conditions to solve for both constants.