Speed of a man in water is 5 km ther, mantories to cross a vive filowi

1.	Speed of a man in water is 5 km ther, mantories to cross a vive filowing towardscast at 3km/hr of width Toom.find the Shortest time in which man crossesriver.Aut what angle from flow velocity manShould swim to reach directly opposite PointMan starts swimming forum Point A makingS3 angle with flow velocity. Find thePosition where he will reach also findthe time.
Answer» $\huge \underline \green{\sf{ Answer :- }}$ a. Velocity of man with respect to river water, v=5kmh−1. This is greater than the river flow velocity. Therefore, he can cross the river directly (along the shortest PATH or no drift CONDITION from flow velocity). The ANGLE of swim, θ=2π+sin−1(vu) = 90o+sin−1(vu) = 90o+sin−1(53) =90o+37o = 127o w.r.t. the river flow or 37o w.r.t. the perpendicular in upstream direction b. RESULTANT velocity or velocity of mass will be vm = v2 −u2 = 52−32 =4kmh−1 $\small \underline {\sf{In \ the \ direction \ perpendicular \ to \ the \ river \ flow. }}$ c. time taken to cross the river t = d/ v2−u2d = 1km / 4kmh ^ −1 = 1 / 41h $\small \underline {\sf{ q \ = \ 15min}}$

Speed of a man in water is 5 km ther, mantories to cross a vive filowing towardscast at 3km/hr of width Toom.find the Shortest time in which man crossesriver.Aut what angle from flow velocity manShould swim to reach directly opposite PointMan starts swimming forum Point A makingS3 angle with flow velocity. Find thePosition where he will reach also findthe time.

Answer»

$\huge \underline \green{\sf{ Answer :- }}$

a. Velocity of man with respect to river water, v=5kmh−1. This is greater than the river flow velocity. Therefore, he can cross the river directly (along the shortest PATH or no drift CONDITION from flow velocity). The ANGLE of swim,

θ=2π+sin−1(vu)

= 90o+sin−1(vu)

= 90o+sin−1(53)

=90o+37o

= 127o w.r.t. the river flow or 37o w.r.t. the perpendicular in upstream direction

b. RESULTANT velocity or velocity of mass will be vm = v2 −u2

= 52−32

=4kmh−1

$\small \underline {\sf{In \ the \ direction \ perpendicular \ to \ the \ river \ flow. }}$

c. time taken to cross the river

t = d/ v2−u2d

= 1km / 4kmh ^ −1

= 1 / 41h

$\small \underline {\sf{ q \ = \ 15min}}$

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