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