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A lifeguard who can swim at 1.2 m/s in still water wants to reach a dock positioned perpendicularly directly across a 550 m wid...Asked by grey
A lifeguard who can swim at 1.2 m/s in still water wants to reach a dock positioned perpendicularly directly across a 550 m wide river.
If the current in the river is 0.80 m/s, how long will it take the lifeguard to reach the dock?
If instead she had decided to swim in such a way that will allow her to cross the river in a minimum amount of time, where would she land relative to the dock?
If the current in the river is 0.80 m/s, how long will it take the lifeguard to reach the dock?
If instead she had decided to swim in such a way that will allow her to cross the river in a minimum amount of time, where would she land relative to the dock?
Answers
Answered by
MathMate
Her resultant speed is perpendicular to the flow of the river.
Thus resultant velocity, using Pythagoras theorem,
V=√(1.2²-0.8²)
=√(0.8)
=0.8944 m/s
Time to cross river
=550m/0.8944 m/s
=614.9 s
=10 minutes 15 seconds, approx.
to cross as rapidly a possible, the crossing speed is exactly 1.2 m/s,
time = 550/1.2
= 458.3 s
= 7min. 38.3 sec.
Thus resultant velocity, using Pythagoras theorem,
V=√(1.2²-0.8²)
=√(0.8)
=0.8944 m/s
Time to cross river
=550m/0.8944 m/s
=614.9 s
=10 minutes 15 seconds, approx.
to cross as rapidly a possible, the crossing speed is exactly 1.2 m/s,
time = 550/1.2
= 458.3 s
= 7min. 38.3 sec.
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