A motorboat going downstream overcame a raft at a point A (Kinematics question)

Question

A motorboat going downstream overcame a raft at a point A. $T$ = $60$ min later it turned back and after some time passed the raft at a distance $l$ = $6$ km from the point $A$. Find the flow velocity. Assuming duty of motorboat is constant.

This is a question from IE irodov (mechanics).

Now i tried doing this problem without using relative motion/ velocity:

Let $$\begin{align}v_r &= \text{ velocity of raft}\\ v_f &= \text{ velocity of flow}\\ v_b &= \text{ velocity of boat}\end{align}$$

Let raft reach $C$ when boat reaches $B$.
$$\therefore AB=(v_b+v_f)60\\ AC = (v_r+v_f)60\\ \implies CB = 60(v_b - v_r)$$

Now boat turns back. Let boat and raft meet at $D$ after time $t$ after the bot reaches B. Therefore:

$$(v_r + v_f)t + (v_b - v_f)t = 60(v_b - v_r)\\ \implies t = \dfrac{60(v_b - v_r)}{v_r + v_b}$$

Now, $AC + CD = 6$ $$(v_r + v_f)60 + (v_r + v_f)60\left(\dfrac{v_b - v_r}{v_b + v_r}\right) = 6$$

Simplifying for $v_f$, we get:

$$v_f = \dfrac{v_b +v_r - 10 v_b^2}{10v_b} {km\over min}$$

I cant simplify further. The answer given is 3km/hr

EDIT

I just realised that $v_r = 0$. Therefore last equation gives:

$$v_f = \frac{6(1-10v_b) km}{hr}$$

Answer 1

All that is quite unnecessary, really, just change the frame of reference to the river, and visualize it as a moving conveyor belt on which the raft is merely a marked point A.

The motorboat moves away from A for for $1$ hr on the "conveyor belt" , so it will take exactly the same time to get back to it.

In the meantime, A has moved downstream $6$ km w.r.t. the bank,
thus flow velocity $= \dfrac62 = 3\;$km/hr.

Answer 2

I have already given a creative solution, but if your faculty is intolerant of unorthodoxy, here is a simplified algebraic one.

I take the velocity of flow as $v$, and that of the boat as $kv$, where $k$ is a multiplier (often a very useful ploy in simplifying), and count time in hours to further simplify.

Following your diagram,

Going with the flow and using relative velocities,

$AB = (k+1)v, \quad AC = v, \quad CB = kv$

For return meeting at $D$ against the flow,

$[v+ v(k-1)]t = kv \Rightarrow t = 1\quad$ (in fact we already knew this !)

Finally, as $(1+t)= 2$ hrs have elapsed from start,

$v + v = 6, \Rightarrow v = 3$