question_answer

                A boat which has a speed of 5 km/h in still water crosses a river of width 1 km along the shortest possible path in 15 min. The velocity of the river water in km/h is:                                                             

A)                 1                             

B)                 3             

C)                 4             

D)                 \[\sqrt{41}\]     

Correct Answer: B

Solution :

                                The shortest possible path is straight line AC. Let u be the velocity of river water and v that of boat making an angle \[\theta \] with line AC.                 Magnitude of velocity of boat along AC                 \[{{v}_{y}}=v\cos \theta \]                 Time \[=15\min =\frac{15}{60}h=0.25\,h\]                 \[{{v}_{y}}=\] Velocity along \[AC\]                 \[=\frac{Displacement}{Time}\]                 \[=\frac{1}{0.25}=4\,km/h\]                 but  \[{{v}_{y}}=v\cos \theta \]                 \[\Rightarrow \cos \theta =\frac{{{v}_{y}}}{v}=\frac{4}{5}\]                 Velocity of river water                 \[u=v\sin \theta \]                 \[=v\sqrt{1-{{\cos }^{2}}\theta }\]                 \[=5\times \sqrt{1-{{\left( \frac{4}{5} \right)}^{2}}}\]                 \[=5\times \frac{3}{5}=3\,km/h\]                 Alternative: From above figure                 \[{{v}^{2}}={{u}^{2}}+v_{y}^{2}\] \[\Rightarrow \]               \[u=\sqrt{{{v}^{2}}-v_{y}^{2}}=\sqrt{{{5}^{2}}-{{4}^{2}}}\]                 = 3 km/h