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Solving real life problem with differential equation 
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#1
Oct3013, 12:08 AM

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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)taninverse(v/5) = s using limits s = 0 when v = 0 and s = 0.01 when v=v 1/5(taninverse(v/5) = .01 taninverse(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
Oct3013, 06:00 AM

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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.



#3
Oct3013, 07:18 AM

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#4
Oct3013, 11:17 AM

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Solving real life problem with 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. 


#5
Oct3013, 03:48 PM

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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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