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

  • Thread starter oddjobmj
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Homework Statement


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 .]


Homework Equations


a=F/m


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!
 
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Answers and Replies

  • #2
SteamKing
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Velocity is not equal to the derivative of acceleration. Check your definitions of position, velocity, and acceleration again.
 
  • #3
TSny
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Homework Statement


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) = − βv2 e^(−αt)
What is the interpretation of v2? Is it v*2 or v^2 or something else?

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?
Yes

Beyond that I'm not sure how to differentiate the acceleration function to get velocity.
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
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*facepalm* Thanks for pointing that out. That would explain the confusion I suppose.

What is the interpretation of v2? Is it v*2 or v^2 or something else?
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.
 
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  • #5
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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?
 
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  • #6
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At Infiniti the exponential goes to zero therefore, 1/v = + c
 
  • #7
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To find c apply the given initial condition.
After applying
1/5 =-β/αm + c
 
  • #8
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Thank you, yands, yes. That should be a positive 0.217 m/s.
 
  • #9
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Just keep calm while solving problems dear:p
 
  • #10
TSny
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-v^(-1)=4.40741e^(-.27t)+C =>

Plugging in 5 for v and 0 for t I get a C of ~4.6074.
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.]
 
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  • #11
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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
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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
 
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  • #13
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Try to substitute the numerical values after you done manipulations
 
  • #14
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Now again apply nicely the initial condition
 
  • #15
35
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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.]
C is a constant to be determined so negative or positive it does not matter
 
  • #16
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Just be sure to take the reciprocal of the answer
 
  • #17
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Yeah, that's what I did. C=4.6074. Are you interpreting the ~ as a negative?
 
  • #18
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I got 0.217 too . Why are you so sure it is not correct
 
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  • #19
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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
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Why are you so sure the answer is incorrect
 
  • #21
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Can you show us the second part solution
 
  • #22
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When I submit the answers it says incorrect, haha.

At this point I'm not even sure what to try. We just solved the equation for v so I should be able to integrate that, solve for the new C by plugging in x=0 and t=0 and then solving for t=10 but there is a log of a negative number in the answer so I'm getting imaginary numbers.
 
  • #23
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I doubt that the first answer is incorrect. And I do think you have done some mistake in the second part
 
  • #24
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I've solved the second part and the log seems fine
 
  • #25
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Well, v=-1/(4.40741(0.763379)^t+4.60741).

If I integrate that to find x(t) I get x(t)=-.80386*ln(1.30996^(-.27t)+.956594)+C.

Plugging in t=0 and t=0 to find C I find that C=.539555

Plugging in t=10 and C=.539555 to find x(t) I find that x(t)=-1.68.

One of the answers is incorrect.
 

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