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Resistance proportional with velocity problem

  1. Feb 8, 2017 #1
    1. The problem statement, all variables and given/known data
    If we have object with mass 10 kg traveling at starting velocity of 50 km/s and one force of resistance that is equal to v^2 of object's velocity how can we calculate distance and time in which that object travels until it gets to velocity of 10 km/s.


    2. Relevant equations


    3. The attempt at a solution

    Kinetic energy at start - energy lost due to work of resistance = final kinetic energy
    1/2 m v0^2 - ??? = 1/2 m vf^2

    As you can see I have trouble calculating work done by resistance. At start it will be starting velocity squared times distance but it changes every moment as velocity changes...
     
  2. jcsd
  3. Feb 8, 2017 #2

    Ray Vickson

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    Force = mass × acceleration, so if the force of resistance is ##k v^2## (for some constant ##k##) then we have ##m\, dv/dt = -k v^2##, a simple differential equation to determine ##v = v(t).## When you know ##v(t)## you can integrate to get ##x(t)##.
     
  4. Feb 10, 2017 #3
    I solved it, first I wrote function v=f(t) then found ▲t by simply putting in final velocity. Then I integrated that function with domain ▲t to get distance.
     
  5. Feb 10, 2017 #4

    Ray Vickson

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    For the sake of interest, what did you get?
     
  6. Feb 10, 2017 #5
    I got that time it took object to slow from 50 to10 km/s is 4 s and distance it passed during that time is 62.76 m.
     
  7. Feb 10, 2017 #6

    Ray Vickson

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    I got ##t = 4/5## sec and distance = ##10 \ln(5) \doteq 16.09## m.
     
  8. Feb 10, 2017 #7
    I think you got something wrong.
     
  9. Feb 10, 2017 #8

    Ray Vickson

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    Show your work, so we can decide.
     
  10. Feb 10, 2017 #9
    Can you try calculate time it would take it to come to 0 velocity. If your technique is valid you should get error or infty.
    EDIT: ok I did something bad, it never comes to infinity it goes to limes.
     
  11. Feb 10, 2017 #10
    Ok with my calculations it comes again that time is infinite until velocity reach 0, what did you get?

    I solved it like this: v=f(t), v₁=50, v₂=10, m=10, Fr=f(t)²

    f(t)=v₁-a·∆t => f(t)=v₁- (f(t)²/m)·∆t => (∆t/m)⋅f(t)²+f(t)-v₁ = 0

    So when velocity is f(t)=v₂ then: ∆t=(-10⋅f(t)+500)/f(t)² = 4s

    And for distance: s=∫{0 to 4}(-1+√(1+4⋅(∆t/m)⋅v₁)/(2⋅∆t/m) d∆t ≈ 62.76
     
  12. Feb 11, 2017 #11

    Ray Vickson

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    Your solution makes no sense and is incorrect. Do you know what is meant by a differential equation? Do you know how to solve the differential equation for v(t) in this problem? That was given in post #2.

    If you do not know how to solve a differential equation, you can develop a finite-difference scheme to solve the problem approximately. For small ##\Delta t > 0## we have ##v(t+\Delta t) \doteq v(t) - \Delta t \cdot v(t)^2## (which becomes more and more nearly exact as ##\Delta t## gets smaller and smaller), so
    $$ \begin{array}{rcl}
    v(\Delta t) & \doteq &50 - \Delta t \cdot 50^2 \\
    v(2 \Delta t)&\doteq& v(\Delta t) - \Delta t \cdot v(\Delta t)^2 \\
    v(3 \Delta t) &\doteq& v(2 \Delta t) - \Delta t \cdot v(2 \Delta t)^2 \\
    \vdots &\vdots & \vdots \\
    v(N \Delta t) &\doteq & v((N-1) \Delta t) - \Delta t \cdot v((N-1) \Delta t)^2
    \end{array}
    $$
    Be warned, however, that we need to take ##\Delta t ## very small to get any kind of decent accuracy. For example, we can take ##\Delta t = 1/1000 = 0.001## and ##N = 1000##, to get a good numerical solution over the interval ##0 \leq t \leq 1##, that is, at values ##t = 0, 0.001, 0.002, \ldots, 0.999, 1.000.## You could do all this in a spreadsheet.

    There is also a simple exact formula for the solution ##v(t)##, but I think PF rules forbid me from telling you what it is.

    Below is a plot of the 1000-point curve of ##v(t)##, and the constant function v=10; you can see that ##v(t)## reaches 10 at about ##t = 0.8## sec. That is exact when we use the true formula for the exact solution.

    BTW: the solution never reaches 0 exactly, just as you claimed.
     

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    Last edited: Feb 11, 2017
  13. Feb 11, 2017 #12


    Ok I see, but I don't see mass in your equation.
    I quote from your post:
    "v(Δt)≐50−Δt⋅50^2
    v(2Δt)≐v(Δt)−Δt⋅v(Δt)^2
    ..."
    I believe it should be modified like this, after all acceleration is force/mass.
    v(Δt)≐50−Δt⋅(50^2)/m
    v(2Δt)≐v(Δt)−Δt⋅(v(Δt)^2)/m
    ...
    Tell me if I am mistaken.
     
  14. Feb 11, 2017 #13

    Ray Vickson

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    You are correct.

    The graph I attached is correct because it used the correct finite-difference method.
     
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