SophiaRivera007 Posted January 16, 2017 Share Posted January 16, 2017 Can anyone help me to solve this? 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.a) If the current in the river is 0.80 m/s, how long will it take the lifeguard to reach the dock?b) 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? Link to comment Share on other sites More sharing options...
Klaynos Posted January 16, 2017 Share Posted January 16, 2017 What have you done to try and answer this? Drawn a diagram? Link to comment Share on other sites More sharing options...
AshBox Posted January 17, 2017 Share Posted January 17, 2017 My attempt at solving it: a)1.2 sin(α)= 0.8sin(α)= 0.8 / 1.2sin(α)= 2/3α= sin^-1(2/3)α= 41.8 degreesSo his crossing velocity is:1.2 cos41.80= 0.894 m/sAnd the time taken is:distance = speed * time550= 0.894 * tt= 550 / 0.894t=615.21 secondsb)550 / 1.2 = 458.3 secondsdistance = speed x timedistance = 0.80 m/s x 458.3s = 366.6 m Link to comment Share on other sites More sharing options...
SophiaRivera007 Posted January 17, 2017 Author Share Posted January 17, 2017 Thank You AshBox, This will help me. Link to comment Share on other sites More sharing options...
Klaynos Posted January 17, 2017 Share Posted January 17, 2017 My attempt at solving it: a) 1.2 sin(α)= 0.8 sin(α)= 0.8 / 1.2 sin(α)= 2/3 α= sin^-1(2/3) α= 41.8 degrees So his crossing velocity is: 1.2 cos41.80= 0.894 m/s And the time taken is: distance = speed * time 550= 0.894 * t t= 550 / 0.894 t=615.21 seconds b) 550 / 1.2 = 458.3 seconds distance = speed x time distance = 0.80 m/s x 458.3s = 366.6 m We try not to give direct worked answers here but to lead people through doing it themselves correcting and pushing where necessary. Link to comment Share on other sites More sharing options...
AshBox Posted January 19, 2017 Share Posted January 19, 2017 We try not to give direct worked answers here but to lead people through doing it themselves correcting and pushing where necessary. I agree, but in this case we just have to put values in related equations which I have guided here. Link to comment Share on other sites More sharing options...
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