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Solving real life problem with differential equation

  1. Oct 30, 2013 #1
    1. The problem statement, all variables and given/known data
    A car starts from rest.When it is at a distance s from its starting point,its speed is v and it acceleration is a = (25v + v^3).

    Show that dv = (25 + v^2)ds and find its speed when s = 0.01

    2. The attempt at a solution

    a = v(dv/ds) = (25v + v^3) divide both sides by v and cross multiply s

    dv = (25 + v^2) ds

    1/(25+v^2) dv = 1 ds

    integrating both sides

    (1/5)tan-inverse(v/5) = s

    using limits s = 0 when v = 0 and s = 0.01 when v=v

    1/5(tan-inverse(v/5) = .01

    tan-inverse(v/5) = .05

    v/5 = tan.05

    v = 5tan.05 = .0044

    My book says the answer is 1.28m/s. I think I might have gone wrong with the limits?
    Any help would be appreciated.

    If I solve the last part using radians the answer is still only 0.25
  2. jcsd
  3. Oct 30, 2013 #2


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    You should certainly use radians. Standard formulae such as the one you used to integrate to get arctan assume radians. I also get 0.25. The book answer matches 5 tan(.25). Maybe it was supposed to be s =.05.
  4. Oct 30, 2013 #3
    Thanks it might be wrong so I think I'll move on to the next question then and when I first started differentiation I remember my book making a very big deal about only using radians in calculus but it just never seems to stick, thanks anyway.
  5. Oct 30, 2013 #4

    Ray Vickson

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    There is something seriously wrong with the original question: if the acceleration is a = v^3 + 25 v, then v(t) obeys the differential equation
    [tex] \frac{dv}{dt} = v^3 + 25 v, [/tex]
    whose possible solutions are ##v(t) = v_1(t), v_2(t) \text{ or } v_3(t)##, where
    [tex] v_1(t) = 0\:\: \forall t \\
    v_2(t) = \frac{ 5e^{25(t+c)} }{ \sqrt{1 - e^{50(t+c)} } }\\
    v_3(t) = -\frac{ 5e^{25(t+c)} }{ \sqrt{1 - e^{50(t+c)} } }
    [/tex] and where c is a constant. For solutions v_2 and v_3 there are no values of c that make v(0) = 0; in fact, there is no t at all that makes v(t) = 0, so the car could never, ever be at rest! It can, of course, be at rest for solution v_1(t), but in that case it remains at rest forever.
  6. Oct 30, 2013 #5


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    Good point. If a = v*f(v) and v(t0) = 0 then a and all higher derivatives are zero at t = t0. No movement can occur.
    To make the question work, the initial condition needs to be lim s→0 v = 0, or somesuch. A 'real world' example is an object nudged away from unstable equilibrium, like a pencil balanced on its point.
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