Time of boat completing round trip

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Homework Help Overview

The problem involves a boat making a round trip on a river, with specific conditions regarding the speed of the river's flow and the time taken for the trip. The subject area includes concepts of relative motion, particularly in the context of fluid dynamics and kinematics.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the speed of the boat in still water and how it is affected by the current. There are attempts to derive the boat's speed based on the time taken for the round trip and the flow speed of the river. Questions arise about the interpretation of the phrase "two times smaller" regarding the flow speed.

Discussion Status

The discussion is ongoing, with participants exploring different interpretations of the problem and attempting to derive the necessary speeds. Some have provided hints and equations, while others express confusion about the calculations and the wording of the problem. There is no explicit consensus on the correct approach or solution yet.

Contextual Notes

Participants note discrepancies in the expected answers from the exercise book and question the validity of the book's solution based on their calculations. There is also mention of the original wording of the problem potentially leading to misunderstandings.

Kamataat
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A boat made a round trip of 3km on a river in 2h. The speed of the river's flow was 1m/s. In what time will the boat complete the same round trip, if the speed of the flow is 2 times smaller?

I suck at these kinds of questions, so would anybody please help me out here?

- Kamataat
 
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Assume that the boat would move at a constant speed [tex]v[/tex] if there were no current.

What is the speed of the boat on each leg of the trip?
 
W/o current, it's 1,5km/h. With current it's [itex]v + v_{water} = 1,5 km/h[/itex]?

- Kamataat
 
Last edited:
You need to be a little bit more careful. I am asking, in the situation where, when there is a current of 1 m/s, it takes the boat 2h to complete the trip, what speed [tex]v[/tex] must the boat be capable of traveling at neglecting current?

Hint: On one leg of the trip, the speed of the boat is [tex]v + 1\frac{\mbox{m}}{\mbox{s}}[/tex], and on the other, [tex]v - 1\frac{\mbox{m}}{\mbox{s}}[/tex].
 
I don't get this. It must be able to travel at 1,5km/h? Then it's 1,5km/h-1m/s on one leg and 1,5km/h+1m/s on the other leg. That's what I said in my last post (sort of).

EDIT: I already got as far as your hint on my own, but replacing 1m/s with 0,5m/s in these formulas still gives me 2h for the round trip.

EDIT2: Aaaargh. The 1,4 and 1,6 km/h figures were wrong. I realize this.

- Kamataat
 
Last edited:
1m/s = 3,6km/h

So I have to figure out [itex]v[/itex] and then replace 3,6km/h with 1,8km/h in your hint forulas? And then for each leg of 1,5km I have to do [itex]1,5/v_{hint}[/itex] to get the time it took to complete that leg?

- Kamataat
 
Last edited:
Anybody? :cry:
 
You keep trying to do each leg separately and you can't: you don't know what time each leg took.

If v is the speed of the boat in still water, then going upstream, its speed is v-1 so it would take 1500/(v-1) s to go 1500 m upstream. Going downstream, its speed is v+1 so it would take 1500/(v+1) s to go 1500 m downstream. The total round trip would take 1500/(v+1)+ 1500/(v-1) seconds. We are told that it actually takes 2 hours= 7200 seconds to do that: solve 1500/(v+1)+ 1500/(v-1)= 7200 to find the speed in still water.

Now I have a question: what, exactly, is meant by "2 times smaller"? That's peculiar wording. Two times anything makes it larger, not smaller! I could see that being interpreted as either "1/2 as fast" or "1/3 as fast".
 
You can probably guess that the original exercise wasn't written in English. "Two times smaller" in my language means 1/2 times the original size, "three times smaller" would be 1/3 times the original size and so on...

I solved for "v" in the equation you gave and then plugged it in and by replacing 1m/s with 0,5m/s got an answer that is not right according to the exercise book. The correct answer should be 1,25h.

Since solving for "v" resulted in a quadratic equation, I got two values for "v". One "v" gave 48min as the final answer and the other 98min.

- Kamataat
 
  • #10
I found out at what speed the river must flow for the book's answer to be correct, and it isn't 0,5m/s. Can we conclude from this that the book has the wrong answer?

I'm just asking this because I want to be sure that I've done everything correctly.

- Kamataat
 

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