# Function of force with respect to v & t; find final v and distance

1. Sep 13, 2013

### oddjobmj

1. The problem statement, all variables and given/known data
A particle ( m = 1 kg ) moves on the x axis. The initial position is x = 0 and the initial velocity is v(0) = 5 m/s. The particle experiences a force,

F(v,t) = − βv^2 e^(−αt)

(a) Determine the final velocity of the particle.
(b) Determine the distance it has moved at t = 10 seconds.
[DATA: α = 0.27 s−1 ; β = 1.19 kg/m .]

2. Relevant equations
a=F/m

3. The attempt at a solution

This is part of the calculus review of my physics homework. What is confusing me is the function which is dependent on two variables.

For part (a) I divided the F function by the mass 1 kg which should then mean that a(v,t)=F(v,t)/m, correct? Beyond that I'm not sure how to differentiate the acceleration function to get velocity. If I differentiate with respect to t or with respect to v I still have a v in the equation which means I cannot find the limit of the resulting function as t approaches infinity to find the final velocity.

Once I figure out how to find the acceleration function I can probably differentiate that to find the position function for part (b).

Any suggestions? Thank you!

Last edited: Sep 13, 2013
2. Sep 13, 2013

### SteamKing

Staff Emeritus
Velocity is not equal to the derivative of acceleration. Check your definitions of position, velocity, and acceleration again.

3. Sep 13, 2013

### TSny

What is the interpretation of v2? Is it v*2 or v^2 or something else?

Yes

Note that the acceleration can be written as a = dv/dt. Use this for the left hand side of a(v,t)=F(v,t)/m. See if you can somehow integrate this to find v(t). Hint: separation of variables.

4. Sep 13, 2013

### oddjobmj

*facepalm* Thanks for pointing that out. That would explain the confusion I suppose.

Yes, TSny, I missed the carrot and have adjusted the original. It is supposed to be v^2.

I'll get back after going in the -correct- direction.

Last edited: Sep 13, 2013
5. Sep 13, 2013

### oddjobmj

So, here it goes:

dv/dt=-1.19(v^2)e^(-.27t) =>

v^(-2)dv=-1.19e^(-.27t)dt (integrate both sides) =>

-v^(-1)=4.40741e^(-.27t)+C =>

How do I do this without ending up with a limit depending on C? I could integrate the right half from 0 to infinity but what about the left half?

EDIT: Going to try to solve for C by leaving it on the left and plugging in the initial conditions.

EDIT 2: Plugging in 5 for v and 0 for t I get a C of ~4.6074. If I then solve the equation for v and take the limit as t approaches infinity with that C value the result is -.217 m/s.

Taking the integral of the above equation solved for v to find x(t) and plugging in t=10 seconds I get x(10)=-2.22

I'm not sure which, if not both, are incorrect because the solution is dependent on both being correct and it is not accepting my answers. Suggestions?

Last edited: Sep 13, 2013
6. Sep 13, 2013

### yands

At Infiniti the exponential goes to zero therefore, 1/v = + c

7. Sep 13, 2013

### yands

To find c apply the given initial condition.
After applying
1/5 =-β/αm + c

8. Sep 13, 2013

### oddjobmj

Thank you, yands, yes. That should be a positive 0.217 m/s.

9. Sep 13, 2013

### yands

Just keep calm while solving problems dear:p

10. Sep 13, 2013

### TSny

Note the negative sign on the left side of the equation. Should C be a positive or negative number?

[EDIT: Sorry, I see you found the mistake.]

11. Sep 13, 2013

### oddjobmj

It should be negative. So the resulting velocity would be 0.217 m/s. That is still not correct, though.

EDIT: I mean to say, that the combination of the two results is not correct even if 0.217 m/s is. I have to try the integral again with the negative.

12. Sep 13, 2013

### yands

You are making signs mistakes dude
V^-2 dv = -β/me^-αt dt
-V^-2 dv = β/me^-αt dt
V^-1 = -β/αm e^-αt +c

Last edited: Sep 13, 2013
13. Sep 13, 2013

### yands

Try to substitute the numerical values after you done manipulations

14. Sep 13, 2013

### yands

Now again apply nicely the initial condition

15. Sep 13, 2013

### yands

C is a constant to be determined so negative or positive it does not matter

16. Sep 13, 2013

### yands

Just be sure to take the reciprocal of the answer

17. Sep 13, 2013

### oddjobmj

Yeah, that's what I did. C=4.6074. Are you interpreting the ~ as a negative?

18. Sep 13, 2013

### yands

I got 0.217 too . Why are you so sure it is not correct

19. Sep 13, 2013

### oddjobmj

The answer is in two parts. Both are required for it to be correct. I need to find the distance traveled at t=10.

20. Sep 13, 2013

### yands

Why are you so sure the answer is incorrect