(adsbygoogle = window.adsbygoogle || []).push({}); 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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# Solving real life problem with differential equation

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