00:01
So in this question, we're told that the rate of change of the mass of a forest fighting airplane is m dot equals sigma, and the initial mass is m -0 is the initial mass.
00:16
So this tells us that m of t equals m -0 plus sigma t.
00:24
And now we're told that there's a friction force, f equals minus bv.
00:29
And the force is the rate of change of the momentum, d by dt of p, but the momentum is m of t times v.
00:42
So what we have here is that minus bv equals d by dt of m of t times v, so m0v plus sigma t v.
00:58
So this is going to be m0 dt d t, dv by d t, plus sigma v plus sigma t v v v v v by d t is m n b b b b by d t is v plus sigma v plus sigma v.
01:20
So we have that dv by d t times m nought plus sigma t is equal to minus b plus sigma times v.
01:32
So now the integral dv 1 over v, so i divide by v, is equal to the integral d t of minus b plus sigma over m0 plus sigma t.
01:54
And when t equals zero, v is v0.
02:00
So now, and then when t is t, we've got v of t.
02:03
So now integrating this between v of t and v of zero, we get log v of t.
02:08
Over v of 0...