differential a(v) problem

Il-J

1. The problem statement, all variables and given/known data
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.


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


3. 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 aquire 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 aquire 1=g/a-kt and integrate implicidly with respect to t, but realised 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 im 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 im 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.