The discussion centers around a problem involving the time it takes for a swimmer to cross a river while accounting for the current. The subject area includes concepts from kinematics and vector resolution.
There is an ongoing exploration of different interpretations of the problem, particularly regarding the swimmer's speed and the effect of the current. Some participants have provided guidance on how to approach the problem, while others express uncertainty about the calculations and assumptions being made.
Participants note discrepancies between their calculations and the textbook answer, suggesting potential errors in the assumptions or methods used. The problem's requirement for the swimmer to arrive directly across from the starting point is highlighted as a critical factor in determining the correct approach.
Well at first I imagined a man swimming across a river, then I imagined the current pulling him downwards, and so I made a triangle and tried to solve for the hypotenuse, proceeding by trying to use t=d/v, but then I realized that it shouldn't be a triangle at all because the problem states that he should arrive at the point EXACTLY opposite from the starting point. (Which I didn't write in the problem, my mistake) And now I'm not sure how to go about itDoc Al said:Please show how you calculated the time and got a different answer.
Drawing a triangle is the right thing to do. But realize that the swimmer's speed is the hypotenuse.alexsphysics said:so I made a triangle and tried to solve for the hypotenuse
Exactly. He needs to swim at an angle (with respect to the moving water) so that the component of his velocity parallel to the river cancels the current.alexsphysics said:He should be swimming upwards to counter.
yepsojsail said:To clarify: when you say upwards, do you mean what I would call upstream (edit: or upriver)?
alexsphysics said:Relevant Equations: My textbook says the answer is 624.3s, but I don't see how that is.
University Physics screws with me sometimes :l What was your answer?hmmm27 said:Me, either.
The textbook answer is way off if he just points himself directly at the far bank and starts swimming.
It's much closer if - as it seems - he wants to end up on the riverbank, directly across ; however it's off enough to think the textbook is playing pretty fast and loose with significant number of decimals in the calculations.
I tried drawing a triangle and solving for the hypotenuse, proceeding by trying to use t=d/v. Where am I going wrong?sojsail said:Yes, there is an error somewhere. Before we knew that the goal was to go straight across the river, I worked it out assuming all of his speed would be dedicated to getting across the river quickly. That would yield the minimum time considering his speed and width of the river. That time was 1080 seconds. The book's time is about 1/2 of that. For him to set a course to get him straight across from the starting point forces him to spend some of his speed on the task of cancelling the current instead of getting across the river.
But, the purpose of this question was for you to exercise your skills. You should work on finding the correct time to get to the point straight across the river. Keep us up to date on how you are doing and we will make certain that you have it right.
The vertical side was labeled with the speed that the river was carrying the water, and the horizontal was labeled with the speed that the man was swimmingsojsail said:OK, please describe your triangle.
Doc Al's "at an angle" means not perpendicular. Your vertical and horizontal sides would be perpendicular, would they not?Exactly. He needs to swim at an angle (with respect to the moving water) so that the component of his velocity parallel to the river cancels the current.
It would, but (600 m)/(4 km/h) = 540 s. The swimmer’s speed relative to the river is 4 kph, not 2 kph.sojsail said:That would yield the minimum time considering his speed and width of the river. That time was 1080 seconds.
alexsphysics said:He should be swimming upwards to counter.
This is not correct. It describes the situation where the swimmer just wants to get to the other side.alexsphysics said:The vertical side was labeled with the speed that the river was carrying the water, and the horizontal was labeled with the speed that the man was swimming
The swimmer’s speed relative to the river is 4 kph, not 2 kph.
So now, we just got to wait for the OP to reverse-engineer how to get there. He does seem to be stuck on ##a^2+b^2=c^2## being the only way to order that equation.neilparker62 said:Textbook answer indicates rounding of the Pythagoras calculation before doing the t = s/v calculation. If you don't round it should be 623.54 s.