Solving real life problem with differential equation

1. Woolyabyss

102
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

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3. haruspex

14,069
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. Woolyabyss

102
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. Ray Vickson

6,281
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
$$\frac{dv}{dt} = v^3 + 25 v,$$
whose possible solutions are ##v(t) = v_1(t), v_2(t) \text{ or } v_3(t)##, where
$$v_1(t) = 0\:\: \forall t \\ v_2(t) = \frac{ 5e^{25(t+c)} }{ \sqrt{1 - e^{50(t+c)} } }\\ v_3(t) = -\frac{ 5e^{25(t+c)} }{ \sqrt{1 - e^{50(t+c)} } }$$ 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. haruspex

14,069
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.