- #1

Ashley1nOnly

- 132

- 3

## Homework Statement

## Homework Equations

F=ma

F/m = a

## The Attempt at a Solution

Fc = constant force

fquad= cv'^2

[/B]

(Fc-cv'^2)/m = a

dt '(Fc-cv'^2) dv'*(m)

dt' = (m/

(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' = ∫

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'

∫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'

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