91Ó°ÊÓ

In case a body\(A\)is moving relative to any other body\(B\) and the body\(B\)be moving relative to some other body\(C\), then the relative velocity of body\(A\)with respect to body\(C\)is given by:

\({\overrightarrow v _{{A \mathord{\left/

{\vphantom {A C}} \right.

\kern-\nulldelimiterspace} C}}} = {\overrightarrow v _{{A \mathord{\left/

{\vphantom {A B}} \right.

\kern-\nulldelimiterspace} B}}} + {\overrightarrow v _{{B \mathord{\left/

{\vphantom {B C}} \right.

\kern-\nulldelimiterspace} C}}}\)

It is given that two students are canoeing on a river and, they accidentally drop an empty bottle while heading upstream. Yet they continued paddling for \(60\) minutes.

Let the whole scenario be illustrated as:

Assuming the velocity of the stream is cat is \({v_x}\) and that of students be \({v_y}\).

Then, relative velocity of student with respect to upstream will be:

\({v_x} - {v_y} = \frac{{2{\rm{ km}}}}{{1{\rm{ hr}}}} = 2{\rm{ km/hr}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,......\left( 1 \right)\)

Now, time took by bottle to cover distance \(d = 5 - 2 = 3{\rm{ km}}\) will be:

\({t_a} = \frac{3}{{{v_y}}}{\rm{ hr}}\)

This will be equal to the time taken by students to travel the distance \(2{\rm{ km}}\) upstream plus \(3{\rm{ km}}\) downstream, that is:

\({t_b} = 1 + \frac{5}{{{v_x} + {v_y}}}{\rm{ hr}}\)

Now,

\(\begin{aligned}{t_a} = {t_b}\\\frac{3}{{{v_y}}} = 1 + \frac{5}{{{v_x} + {v_y}}}\\3\left( {{v_x} + {v_y}} \right) = \left( {{v_x} + {v_y}} \right){v_y} + 5{v_y}\\{v_y}^2 + {v_x}{v_y} + 2{v_y} - 3{v_x} = 0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,......\left( 2 \right)\end{aligned}\)

From equation (1) and (2):

\(\begin{aligned}{v_y}^2 + {v_x}{v_y} + 2{v_y} - 3{v_x} = 0\\{v_y}^2 + \left( {{v_y} + 2} \right){v_y} + 2{v_y} - 3\left( {{v_y} + 2} \right) = 0\\{v_y}^2 + {v_y} - 6 = 0\\{v_y} = 1.5{\rm{ km/hr}}\end{aligned}\)

Hence, the speed of the river is \(1.5{\rm{ km/hr}}\).