Swimming Across River Question

In summary, the conversation revolved around finding the average speed of a man swimming in a river with a current, both across the river and back and a certain distance downstream and back. After some discussion and calculations, the correct average speed was determined to be 3.8m/s and the man would end up at his starting point for both parts of the question.
  • #1

Homework Statement


A man can swim through still water 4.2m/s. Suppose this man swims in a river with current 1.3 m/s. Find his average speed over a trip:
a) across the river and back
b) a certain distance downstream and back

The Attempt at a Solution



a)
Here is a diagram I made:
106kmlv.png

Left side of picture is there, right side is back

So should I solve for x, then take x, multiply by two, then divide by two. Thats the avg speed? because the question doesn't give any distance or time, so this is all I could think of.

b) is this a vector addition question? so his avg speed would be 2.9m/s?

Thanks
 
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  • #2
a. Avg speed is x, right.
b. Careful with this one. Remember, avg speed = total distance/total time. Does not equal 2.9.
 
  • #3
rude man said:
a. Avg speed is x, right.
b. Careful with this one. Remember, avg speed = total distance/total time. Does not equal 2.9.

So can I choose any distance? Since I am not given one.
 
  • #4
physicsnobrain said:
So can I choose any distance? Since I am not given one.

Right.
 
  • #5
Ok. I chose 16m as my distance downstream.

After computing it I get 4.1m/s as the avg speed for the trip
 
  • #7
ok first i add the vectors. On the downstream his velocity increases to 7.1m/s while the upstream is 2.9 m/s.

Using these two found values i calculate time for the 16m upstream and 16m downstream. using t = d/v

For upstream I get 5.52s and downstream i get 2.25s.

Now I do avg spd. total distance/total time. I add up the distances i get 32m, i add up the times i get 7.77s. 32/7.77s I get 4.12m/s

is this not correct?
 
  • #8
physicsnobrain said:
ok first i add the vectors. On the downstream his velocity increases to 7.1m/s while the upstream is 2.9 m/s.

?
 
  • #9
rude man said:
physicsnobrain said:
ok first i add the vectors. On the downstream his velocity increases to 7.1m/s while the upstream is 2.9 m/s.

?

He swims 4.2m/s through still water. There is a current of 1.3m/s. When he's downstream he goes faster, you have to add them to get 7.1m/s, when upstream you subtract the current from his speed.

The question asks about him swimming directly into the current. He goes straight, not across.
 
  • #10
physicsnobrain said:
rude man said:
He swims 4.2m/s through still water. There is a current of 1.3m/s. When he's downstream he goes faster, you have to add them to get 7.1m/s, when upstream you subtract the current from his speed.

The question asks about him swimming directly into the current. He goes straight, not across.

OK but where I come from, 4.2 + 1.3 does not equal 7.1.
 
  • #11
physicsnobrain said:

Homework Statement


A man can swim through still water 4.2m/s. Suppose this man swims in a river with current 1.3 m/s. Find his average speed over a trip:
a) across the river and back

Here is a diagram I made:
106kmlv.png

Left side of picture is there, right side is back

So should I solve for x, then take x, multiply by two, then divide by two. Thats the avg speed? because the question doesn't give any distance or time, so this is all I could think of.

As I read the problem, "across the river and back" to me means that he ends up at his starting point. Your diagram indicates he would end up 2.6 downstream. In order for him to arrive back at his starting point you would have to interchange the 4.2 m/s and x on your diagram.
 
  • #12
rude man said:
physicsnobrain said:
OK but where I come from, 4.2 + 1.3 does not equal 7.1.

Oops silly error. Now with the correct numbers I get an avg spd of 3.8m/s
 
  • #13
skeptic2 said:
As I read the problem, "across the river and back" to me means that he ends up at his starting point. Your diagram indicates he would end up 2.6 downstream. In order for him to arrive back at his starting point you would have to interchange the 4.2 m/s and x on your diagram.

yes you are right. I now get 4.2m/s as the penguins average speed for part a
 
  • #14
physicsnobrain said:
yes you are right. I now get 4.2m/s as the penguins average speed for part a

Is this number correct?
 
  • #15
physicsnobrain said:
rude man said:
Oops silly error. Now with the correct numbers I get an avg spd of 3.8m/s

That's what I got.
 
  • #16
How did you get 3.8 m/s? Can you show your work? I get a value slightly higher.

What did you get for part b?
 
  • #17
skeptic2 said:
How did you get 3.8 m/s? Can you show your work? I get a value slightly higher.

What did you get for part b?

I got 3.8 for part b. For part a I got 4.2m/s.
 
  • #18
skeptic2 said:
How did you get 3.8 m/s? Can you show your work? I get a value slightly higher.

What did you get for part b?

That was part b. Same as what you got.

I never answered part a but 4.2 looks wrong. 4.2 is his speed in calm water. I think the answer is the same as what you would have gotten in your original post: x^2 = 4.2^2 + 1.3^2.
 
  • #19
rude man said:
That was part b. Same as what you got.

I never answered part a but 4.2 looks wrong. 4.2 is his speed in calm water. I think the answer is the same as what you would have gotten in your original post: x^2 = 4.2^2 + 1.3^2.

Are you sure? I think skeptic may have a point about having to return to the same starting point.
 
  • #20
Yes, but I think the answer's the same. He would start upstream by a certain distance and wind up at the starting point, and the velocity vectors would be the same as if he wound up downstream by the same distance.
 
  • #21
rude man said:
Yes, but I think the answer's the same. He would start upstream by a certain distance and wind up at the starting point, and the velocity vectors would be the same as if he wound up downstream by the same distance.

Ok, well then in this case the avg speed is 4.4m/s
 
  • #22
physicsnobrain said:
Ok, well then in this case the avg speed is 4.4m/s

Agreed.
 
  • #23
I'm sorry, I think my post #16 was a little mixed up. I get the same answer as you, 3.8 m/s.
 

1. How is "Swimming Across River Question" relevant to science?

"Swimming Across River Question" is relevant to science as it involves understanding the principles of fluid dynamics and movement of objects through a fluid medium, which is essential in fields such as physics, engineering, and biology.

2. What factors affect the success of swimming across a river?

The success of swimming across a river depends on several factors such as the speed and direction of the river's current, the swimmer's strength and swimming technique, and the distance and obstacles in the river.

3. Can swimming across a river be dangerous?

Yes, swimming across a river can be dangerous, especially if the river has a strong current or obstacles such as rocks or debris. It is important to assess the risks and use proper safety precautions before attempting to swim across a river.

4. How can we apply the concept of swimming across a river in real-life situations?

The concept of swimming across a river can be applied in real-life situations such as designing watercrafts that can navigate through strong currents, predicting and managing the effects of river flooding, and understanding the movement of aquatic animals in their natural habitats.

5. What are some strategies for successfully swimming across a river?

Some strategies for successfully swimming across a river include finding the calmest and shortest route, using proper swimming techniques to conserve energy, and wearing appropriate safety gear such as a life jacket. It is also important to have a backup plan and to never swim alone.

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