Question

Two swimmers start a race. One who reaches the point $C$ first on the other bank wins the race. A makes his strokes in a direction of $37 °$ to the river flow with velocity $5 {kmhr}^{- 1}$ relative to water. $B$ makes his strokes in a direction $127 °$ to the river flow with same relative velocity. River is flowing with speed of $2 {kmhr}^{- 1}$ and is $100 m$ wide. Who will win the race? Compute the time taken by $A$ and $B$ to reach the point $C$ if the speed of $A$ and $B$ on the ground are $8 {kmhr}^{- 1}$ and $6 {kmhr}^{- 1}$ respectively.

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Accepted Answer

Given,

$37 °$

Velocity of river flow $2 {kmhr}^{- 1}$

Angle made by $A$ $= 37 °$

Angle made by $B$ $8 {kmhr}^{- 1}$

From figure, $∠ P B X = 127^{\circ}$

$∠ P B O = 180^{\circ} - 127^{\circ} = 53^{\circ}$

Speeds of $A$ and $B$ on the ground are

$v_{A} = 8 {kmhr}^{- 1}, v_{B} = 6 {kmhr}^{- 1}$

Speed of $A$ and $B$ w.r.t river flow $v_{A} = v_{B} = 5 {kmhr}^{- 1}$

Velocity of $A$ in the direction perpendicular to river flow $= v_{A} \sin 37^{\circ}$

Velocity of $B$ in the direction perpendicular to river flow $= v_{B} \sin 53^{\circ}$

Time taken by $A$ to cross the river is

$t_{1 A} = \frac{O C}{v_{A} \sin 37^{\circ}} = \frac{0.1 k m}{5 {kmhr}^{- 1} \times \sin 37^{\circ}} = \frac{0.1}{3} h r = \frac{0.1}{3} \times 3600 \sec, t_{1 A} = 120 \sec$

Time taken by $B$ to cross the river is

$t_{1 B} = \frac{O C}{v_{A} \sin 53^{\circ}} = \frac{0.1 km}{5 {kmhr}^{- 1} \times \sin 53^{\circ}} = \frac{0.1}{4} h r = \frac{0.1}{4} \times 3600 \sec, t_{1 B} = 90 \sec$

Distance travelled by $A$ in the direction of river flow,

$d_{A} = (v_{A} \cos 37^{\circ} + 2) \times \frac{0.1}{v_{A} \sin 37^{\circ}} km$

$d_{A} = (5 \cos 37^{\circ} + 2) \times \frac{0.1}{5 \sin 37^{\circ}} km$

$d_{A} = 6 \times \frac{0.1}{3} = 0.2 km$

Distance travelled by $B$ in the direction of river flow,

$d_{B} = (v_{B} \cos 53^{\circ} - 2) \times \frac{0.1}{v_{A} \sin 53^{\circ}} km$

$d_{B} = (5 \cos 53^{\circ} - 2) \times \frac{0.1}{5 \sin 53^{\circ}} km \Rightarrow d_{B} = 0.025 km$

Time taken by $A$ to reach point $C$ $= \frac{0.2 km}{8 {kmhr}^{- 1}} = 90 \sec$

Time taken by $B$ to reach point C $C = \frac{0.025 km}{6 {kmhr}^{- 1}} = 15 \sec$

$∴$ Total time taken by $A = t_{1 A} + 90 \sec = 120 + 90 = 210 \sec$

Total time taken by $B = t_{1 B} + 90 \sec = 90 + 15 = 105 \sec$

So, $B$ will win the race.

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