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

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  • #1
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!
 
Last edited:

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
oddjobmj
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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
oddjobmj
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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?
 
Last edited:
  • #6
yands
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At Infiniti the exponential goes to zero therefore, 1/v = + c
 
  • #7
yands
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To find c apply the given initial condition.
After applying
1/5 =-β/αm + c
 
  • #8
oddjobmj
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Thank you, yands, yes. That should be a positive 0.217 m/s.
 
  • #9
yands
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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.]
 
  • #11
oddjobmj
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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
yands
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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
 
Last edited:
  • #13
yands
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Try to substitute the numerical values after you done manipulations
 
  • #14
yands
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Now again apply nicely the initial condition
 
  • #15
yands
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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
yands
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Just be sure to take the reciprocal of the answer
 
  • #17
oddjobmj
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Yeah, that's what I did. C=4.6074. Are you interpreting the ~ as a negative?
 
  • #18
yands
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I got 0.217 too . Why are you so sure it is not correct
 
  • #19
oddjobmj
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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
yands
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Why are you so sure the answer is incorrect
 
  • #21
yands
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Can you show us the second part solution
 
  • #22
oddjobmj
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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
yands
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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
yands
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I've solved the second part and the log seems fine
 
  • #25
oddjobmj
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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.
 
  • #26
yands
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Im new on this forum I don't know how to use math symbols I will try to attach onenote file
 
  • #27
oddjobmj
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Welcome to the forum, Yands. Thanks for your help so far!

If you want to insert symbols you will have to go to advanced mode instead of quick reply. There are 'quick symbols' to the right and there are more in the latex reference menu if you click the Ʃ button in the ribbon all the way to the right of the bar with the text formatting options e.g bold and italic.
 
  • #28
yands
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hopefully you will enjoy
 

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  • #29
TSny
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C is a constant to be determined so negative or positive it does not matter

I was referring to the numerical value of C in the equation -v^(-1)=4.40741e^(-.27t)+C. C must be negative in this equation in order to give the correct numerical value of v at t = 0.
 
  • #30
oddjobmj
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Thank you yands but after plugging in the known values to find c I find that c is an imaginary number. I am assuming c in your equation is -4.60741 and m=1.

Wolfram alpha result (shortened the URL):
http://tinyurl.com/mjq2xmt

I just arbitrarily plugged in X for c to solve for it.
 
  • #31
oddjobmj
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Nevermind! Ah, I forgot C is actually positive. That worked.

Thank you both!
 
  • #32
yands
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Your welcome
 

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