F/m = a
The Attempt at a Solution
Fc = constant force
(Fc-cv'^2)/m = a
(Fc-cv'^2)/m = dv' / dt' * using the primes to differentiate between v and v' during integration
dt '(Fc-cv'^2) dv'*(m)
dt' = (m/ (Fc-cv'^2)) dv'
dt' = m / [ c ((Fc/c) - v'^2) ] dv'
∫c dt' = ∫ m / ((Fc/c) - v'^2) dv'
I need a method to integrate the right hand side that will allow me to get v by itself.
*u-substitution won't work
*I can't integrate it straight through
* I can use ln
* I can't use partial fractions
* I don't think I can use any trig identities here either
*I'm not sure where to take this.
***More work added on,
right hand side
(m*c)/Fc = ∫1/ (1- ((c*v^2)/Fc)) dv'
so I found that
∫ 1/ (A^2 - x^2) dx = 1/ (2A) * ln | (x+A)/(x-a)| + C
I can rewrite my right hand side as the below to fit the equation but what do I do with the constant (c/Fc)? Once I figure that out I can use e to get rid of ln and solve for v
∫1/ (1^2- (c/Fc)*(v^2) dv'
6.4 KB Views: 241
6.3 KB Views: 442