How Do You Solve a Particle's Velocity Equation with Variable Acceleration?

[Il-J]

Homework Statement

A particle starts from rest and travels along a straight line with an acceleration a = (30-0.2v)ft\s^(2), where v is in ft/s. Determine the time when the velocity of the particle is v = 90 ft/s.

Homework Equations

1) ads=vdv
2) if a=g-kv
3) then v(t)=g/k[1-e^(-kt)]

The Attempt at a Solution

i solved it, the answer is 4.58 s but the question expected me to be able to manipulate the first two equations to acquire the third, or maybe introduce another equation, regardless could someone please explain the steps leading up to the third equation?
my first instinct was to divide everything by a to acquire 1=g/a-kt and integrate implicidly with respect to t, but realized i didnt know what i was doing when it came down to the a.

 

[Mark44]

a = dv/dt, so you could rewrite your given equation as dv/dt = 30 - .2v, v(0) = 0, and solve this differential equation for v.

 

[Il-J]

thanks, but I am not quite there, i do that and v=g/k(1-(dv/dt)/g) so that implies that dv/dt=ge^(-kt), but now where does that come from? sorry, but I am really having trouble wrapping my head around the concepts here

 

[Mark44]

Where did you get v=g/k(1-(dv/dt)/g)? Are you just solving for v in the equation dv/dt = 30 - .2v?

I solved the differential equation and got v as a function of t.