# Non uniform acceleration

1. Jul 30, 2015

### SammyD97

1. The problem statement, all variables and given/known data
A particle is moving along a straight line such that when it is at the origin it has a velocity of 5m/s. If it begins to decelerate at a rate of a=(-2.5v^(1/2)) m/s^2, where v is in m/s, determine the distance it travels before it stops

2. Relevant equations

3. The attempt at a solution
I integrated the above expression twice with respect to time, to get an expression of s=(2(v^(5/3) t^2))/3+C1+C2. Where C1 ns C2 are constants. I've ended up with more variables than I have the know how to handle. I don't know what to do with the initial velocity. I thought of integrating with respect to v but it didn't make sense. Please help.

2. Jul 30, 2015

### SteamKing

Staff Emeritus

How did you wind up with two constants of integration without one being multiplied by something else?

You know that when x = 0, v = 5 m/s. Did you apply these initial conditions to determine what the constants of integration must be?

3. Jul 30, 2015

### RUber

It stops when v = 0, right?
Graphically, I get an answer of less than 3m.

I would attack it like this:
since acceleration is the derivative of velocity and $a = -2.5 v^{1/2}$
assume that velocity is of the form $v(t) = c(t+a)^2$
Then $v' = 2c(t+a) = -2.5 \sqrt{ c (t+a)^2 }$.
Using initial information, you should be able to find a suitable solution for c and a, then solve your velocity equation for the time when v = 0.
Last, integrate velocity from 0 to t_final to get total change in position.

4. Jul 30, 2015

### SammyD97

Thanks guys. will try it and see what I end up with.

5. Jul 30, 2015

### Staff: Mentor

What's wrong with solving the differential equation

$$\frac{dv}{dt}=-2.5v^{\frac{1}{2}}$$

for v as a function of t, subject to the initial condition?

Chet

6. Jul 30, 2015

### rcgldr

Why twice, you only need to integrate dv once to end up with v, and you're solving for v = 0.

I missed that the question was about distance traveled, not time, so you will need to integrate twice.

So that would be

$$\frac{dv}{v^{\frac{1}{2}}} = -2.5 \ dt$$

Last edited: Jul 30, 2015
7. Jul 30, 2015

### Staff: Mentor

I would have had the -2.5 associated with the dt.

8. Jul 30, 2015

### SammyD97

I thought I needed to s end up with an expression for position / distance

9. Jul 30, 2015

### rcgldr

You're correct. I lost a update to my prior post; and it's fixed now. So yes, you can start with dv/(v^(1/2) = -2.5 dt, and noting that v = ds/dt, this results in some function f(ds/dt) + 5 = t. (5 being the constant of integration, in this case the initial velocity). Again you can separate the equation so ds and the s terms are on one side and dt and the t terms are on the other and integrate again.

Last edited: Jul 31, 2015
10. Jul 31, 2015

### ehild

You can use the distance travelled as independent variable instead of t. dv/dt=(dv/ds)(dv/dt)(ds/dt)=0.5 d(v2)/ds.You need to integrate with respect to s.

Last edited: Jul 31, 2015
11. Jul 31, 2015

### rcgldr

If using chain rule, shouldn't that be dv/dt = (dv/ds)(ds/dt) = v dv/ds ?

12. Jul 31, 2015

### ehild

Yes, it was a typo, but the last formula is true. a=dv/dt= 0.5d(v2)/ds.

13. Jul 31, 2015

### rcgldr

OK, but is it any simpler than just continuing with the chain rule result:

$$v \ \frac{dv}{ds} = -2.5 v^{\frac{1}{2}}$$

$$v^{\frac{1}{2}} \ dv = -2.5 \ ds$$

Which only requires a single integration, while the previously mentioned equation

$$\frac{dv}{v^{\frac{1}{2}}} = -2.5 \ dt$$

requires two integrations.

As for the constant(s) of integration, the problem states that v = 5 m/s at t = 0 and s = 0, and that the initial position is s = 0, which allows you to solve for the constant(s) of integration with either approach. I got the same result for s using both methods.

Last edited: Jul 31, 2015
14. Aug 25, 2015

### SammyD97

Ok it took long for me to get back to this thread but I did get it eventually.

15. Aug 25, 2015

### rcgldr

If anyone is curious the distance is

$$\frac{4}{3} \ \sqrt{5} \approx 2.981424$$

Last edited: Aug 25, 2015
16. Aug 25, 2015

Yep.