QUESTION:
A motorboat going downstream overcame a raft at a point A; \tau = 60\rm \ min later it turned back and after some time passed the raft at a distance l = 6.0\rm \ km from the point A. Find the flow velocity assuming the duty of the engine to be constant.
To solve this problem, you must have an idea of Kinematics.
SOLUTION:
Let’s choose a coordinate system such that the x-axis is along the flow of the river and the x-y plane is the horizontal plane as shown in the video. Let B is a point at which the motorboat turned back towards the raft and C is a point which is at a distance l = 6.0\rm \ km from point A, where it passes the raft.
Given:
AC = l = 6\rm \ km and
the time taken by the motor boat to travel from point A to point C is \tau = 60\rm \ min = 1 \ hr
Let the flow velocity of the river with respect to the ground is u and the speed of the motorboat with respect to the river is v.
Since the raft will with the flow of the river so, the velocity of the raft with respect to the ground will be u.
The velocity of the motorboat with respect to the ground, while it is going downstream, will be v + u in a direction along the x-axis, and the velocity of the motorboat with respect to the ground while it going upstream will be v – u in a direction along the -x-axis.
Let the time taken by the motorboat to travel from point B to point C is \tau_{1}and the time taken by the raft to travel from point A to point C is \tau_{2}.
From the diagram, we can write
\begin{aligned}AC &= AB – BC \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (1)\\\end{aligned}
Since the time taken by the raft to travel from point A to point C is equal to the time taken by the motorboat to travel from point A to point B plus the time taken by the motorboat to travel from point B to point C.
So, we can also write
\begin{aligned}\tau_{2}&= \tau + \tau_{1} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (2)\\\end{aligned}
Since AC = l = u\tau_{2}, AB = (v+u) \tau and BC = (v-u)\tau_{1} so, from equation 2, we have
\begin{aligned}\dfrac{l}{u} &= \tau + \dfrac{BC}{v-u}\\\end{aligned}
From equation 1, we have BC = AB-AC, so the above equation becomes
\begin{aligned}\dfrac{l}{u} &= \tau + \dfrac{AB-AC}{v-u}\\\Rightarrow \dfrac{l}{u} &= \tau + \dfrac{(v+u) \tau – l}{v-u}\\\Rightarrow \dfrac{l}{u} &= \dfrac{2v\tau – l}{v-u}\\\Rightarrow vl-ul &= 2vu\tau – ul\\\Rightarrow u &= \dfrac{l}{2\tau} \\&= \dfrac{6\rm \ km}{2\times 1\rm \ hr}\\&= 3\rm \ km/h\\\end{aligned}
Therefore, the flow velocity of the river is;
u = 3\rm \ km/h
