The speed of a boat in still water is 15 km/hr. It can go 30 km upstream and return downstream to the original point in 4 hours 30 minutes. Find the speed of the stream.

Explanation

Case I : If `v_R lt v_B`, the boat can cross river along a path perpendicular to flow.
Case II : If `v_B lt v_R`, drift cannot be zero, apply calculus in this case
Case I : If `v_R lt v_B`
Shortest path : `(D)/(v_B sin theta) = 3 T_0` ….(i)
Quickest path : `(d)/(v_B) = T_0` ...(ii)
Also `v_R - v_B cos theta = 0` for shortest path. ....(iii)
Thus, `sin theta = (1)/(3)` from (i) and (ii) or `v_R = v_B cos theta`
=`(d)/(T_0) sqrt(1 - (1)/(9)) =(2 sqrt(2 d))/(3 T_0)`
Case II : If `v_B lt v_R`
`x = ((d)/(v_B sin theta))(v_R - v_B cos theta)`
=`(d)/(v_B)(v_R cosec theta - v_B cot theta)`
For minimum `x, (d x)/(d theta) = -0`
`v_B(- cosec^2 theta) + v_R cosec theta cot theta = 0`
`cos theta = (v_B)/(v_R)`
Time taken in this case is given by `3 T_0 = (v_R d)/(v_B sqrt(v_R^2 - v_B^2))`
Also `v_B = (d)/(T_0)`
On solving, we get `v_R = sqrt((3)/(2)) (d)/(T_0)`.
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