Solving the Drag Differential Equation

Start from m dv/dt = -kv. This says the boat's acceleration depends on its own current speed: the faster it moves, the harder drag pulls it back, so slowing down itself slows down. You cannot integrate this directly because v and t are tangled together on the same side. Separating variables fixes that: divide both sides by v and multiply by dt, giving dv/v = -(k/m) dt. Now each side has only one variable, so you integrate both sides freely.

Integrating dv/v from v0 to v, and dt from 0 to t, gives ln(v/v0) = -(k/m)t. Exponentiate both sides and you get v(t) = v0·exp(-(k/m)t). This is exponential decay: v drops fast at first, then levels off, approaching zero but never touching it, because the rate of decrease shrinks right along with v. That is why a boat coasting under pure drag seems to stop but mathematically coasts forever at ever-tinier speeds. The quantity m/k is the time constant: it has units of time and tells you how quickly the decay happens, with v dropping to about 37 percent of v0 every time t advances by one time constant.

To find total distance, integrate v(t) over all time from 0 to infinity, since the boat technically never fully stops. That integral converges to a finite number, v0·m/k, because the exponential decay shrinks fast enough. This links directly to the earlier kinematics of Unit 1 and to the force analysis of Unit 2: once you know the net force law, calculus turns it into motion, exactly like constant-acceleration kinematics did for constant forces.