- #1
Woolyabyss
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Homework Statement
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) ds1/(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 = .0044My 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