Time it takes to swim across a river

[alexsphysics]

Homework Statement

A man wishes to swim across a river 600m wide. If he can swim at the rate of 4 km/h in still water and the river flows at 2 km/h, when will he reach it?

Relevant Equations

My textbook says the answer is 624.3s, but I don't see how that is.

Thanks

 

[Doc Al]

I assume the swimmer must swim directly across to the other side.

 

[Doc Al]

Please show how you calculated the time and got a different answer.

 

[alexsphysics]

Please show how you calculated the time and got a different answer.

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 it

 

[Doc Al]

What direction must the swimmer aim in order to counter the current taking him downstream?

so I made a triangle and tried to solve for the hypotenuse

Drawing a triangle is the right thing to do. But realize that the swimmer's speed is the hypotenuse.

 

[alexsphysics]

He should be swimming upwards to counter.

 

[sojsail]

To clarify: when you say upwards, do you mean what I would call upstream (edit: or upriver)?

 

[Doc Al]

He should be swimming upwards to counter.

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]

To clarify: when you say upwards, do you mean what I would call upstream (edit: or upriver)?

yep

 

[sojsail]

OK, please let us know if Doc Al's last post has helped you see how to make progress on this.

 

[hmmm27]

Relevant Equations: My textbook says the answer is 624.3s, but I don't see how that is.

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.

 

[alexsphysics]

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.

University Physics screws with me sometimes :l What was your answer?

 

[sojsail]

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 (Edit: actual 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.

 

[sojsail]

Here is a nudge: Review Doc Al's post #8.

 

[alexsphysics]

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.

I tried drawing a triangle and solving for the hypotenuse, proceeding by trying to use t=d/v. Where am I going wrong?

 

[sojsail]

OK, please describe your triangle.

 

[alexsphysics]

OK, please describe your triangle.

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

 

[sojsail]

Doc Al said

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.

Doc Al's "at an angle" means not perpendicular. Your vertical and horizontal sides would be perpendicular, would they not?

 

[sojsail]

I need to sign off for tonight. I hope someone else will take up monitoring and aiding your progress.

 

[Orodruin]

That would yield the minimum time considering his speed and width of the river. That time was 1080 seconds.

It would, but (600 m)/(4 km/h) = 540 s. The swimmer’s speed relative to the river is 4 kph, not 2 kph.

He should be swimming upwards to counter.

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

This is not correct. It describes the situation where the swimmer just wants to get to the other side.

 

[sojsail]

The swimmer’s speed relative to the river is 4 kph, not 2 kph.

I am back. I see that Orodruin caught me misreading the problem. He swims at 4 kph, so the minimum time to the opposite shore would be 540 s. So the book's 624.3 s may be right Edit: since the landing point is to be exactly opposite.

Picture yourself at the bank of the river about to jump in and swim. You see the current and know that it will carry you with it. There are 2 velocity vectors involved. (For discussion, let's assume the river flows south and you need to go across from west to east.) In the reference system of dry land, what direction should you aim your 4 kph in order to arrive at the point straight across? Now, as you think about that, does the way you drew your triangle (with your 4 kph perpendicular to the river's 2 kph) seem right?

 

[neilparker62]

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.

 

[hmmm27]

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.

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.