Question

A man wants to cross a river by a boat. If he crosses the river in minimum time, then he takes 10 minutes with a drift distance of 120m. If he crosses the river taking the shortest path, he takes 12.5 minutes. The velocity of the boat with respect to water is:

A) 10 m/min

B) 15 m/min

C) 20 m/min

D) 22.5 m/min

Answer 1

Hint: When he crosses the river with minimum time then he moves in a direction perpendicular to the boundaries of the river. And when he crosses a river taking the shortest path, he moves such that some component of velocity of boat is opposite to the velocity of water in river and other component is perpendicular to boundaries of river and the path is perpendicular to river flow.  

Complete step by step answer:

Let $x$ be the distance between two ends of the river, $V_R$ be the velocity of water of the river and $V_B$ be the velocity of the boat.

When he crosses the river in minimum time, drift distance is equal to time taken multiplied by velocity of water of river,

$drift = T \times {v_R}$ or ${v_R} = \dfrac{{\text{Drift}}}{{\text{Time}}}$, here drift=120m and time = 10 min=600 s.

${v_R} = \dfrac{{120}}{{600}} = 0.2m/s$  and

$x = {v_B} \times 600$   ...................(1)

When he crosses a river with shortest time one component of boat velocity is opposite to velocity of water and other component is perpendicular to velocity of water.

Let angle between water flow and velocity of water is $\theta $, then 

${v_B}\cos \theta  = {v_R}$ and \[{v_B}\sin \theta  = \dfrac{{\text{shortest path}}}{{\text{Time}}}\],

 here shortest path=$x$ and time=$12.5$ min =$ 750$ sec.

\[\sqrt {v_B^2 - v_R^2}  = \dfrac{x}{{750}}\] or $x = 750.\sqrt {v_B^2 - v_R^2} $   -(2)

Divide equation (2) by (1), we get

\[600{v_B} = 750\sqrt {{v_B}^2 - {v_R}^2} \] 

\[\dfrac{4}{5} = \dfrac{{\sqrt {v_B^2 - {{\left( {\dfrac{1}{5}} \right)}^2}} }}{{{v_B}}}\]   or $\dfrac{4}{5} = \sqrt {1 - \left( {\dfrac{{{v_B}^2}}{{25}}} \right)} $

${v_B} = \dfrac{1}{3}m/s = 20m/\min $.

Hence, the correct answer is option (C).

Note: When he crosses river with minimum time the distance travelled is $\sqrt {drif{t^2} + {x^2}} $ and the velocity of boat is $\sqrt {{v_B}^2 + {v_R}^2} $ with respect to ground and when he crosses river taking shortest distance, the distance travels is $x$ and the velocity of boat is $\sqrt {{v_B}^2 - {v_R}^2} $ with respect to ground.