A river 1 km wide is flowing at 3 km/h . A swimmer whose velocity in still water is 4 km/h can swim only for 15 minutes.In what direction should he strike out in order reach the other bank ? What is the total distance covered ?
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Answered by
76
tan theta = 4/3
theta = tan⁻¹4/3= 53.1⁰
magnitude of resultant velocity
OB'= √ OA² + OB²
= √3² + 4²
=5 km/h
Distance of swimmer= OB' x time
= 5 x 15/60
=1.25 km
theta = tan⁻¹4/3= 53.1⁰
magnitude of resultant velocity
OB'= √ OA² + OB²
= √3² + 4²
=5 km/h
Distance of swimmer= OB' x time
= 5 x 15/60
=1.25 km
Answered by
47
The swimmer should always head towards the point just across the river. That is his/her velocity wrt river is always perpendicular to the river flow. This is the way for the swimming across a river in minimum time duration.
Then the time taken to cross = 1 km / 4 kmph = 1/4 hour = 15 min.
velocity of swimmer wrt stationary observer = √(3²+4²) = 5 kmph
Actual distance covered in 15 min = 5/4 = 1.25 km
Then the time taken to cross = 1 km / 4 kmph = 1/4 hour = 15 min.
velocity of swimmer wrt stationary observer = √(3²+4²) = 5 kmph
Actual distance covered in 15 min = 5/4 = 1.25 km
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