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Rate problem

  • Thread starter osker246
  • Start date
  • #1
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Homework Statement


A boat takes 2.0 hr to travel 31 km down a river, then 6.0 hr to return. How fast is the river flowing?


Homework Equations


r=d/t


The Attempt at a Solution



first started out by finding the rate the boat traveled in each trip.

31/2=15.5 km/hr

31/6=5.2 km/hr

Now this is where im not sure if im solving this correctly. I then find the difference between the two rates and the time taken to travel both ways.

10.3 (difference of 15.5-5.2)/4(difference of 6-2)=2.6 km/hr

Is this the correct way of solving this problem? Thanks!
 

Answers and Replies

  • #2
alphysicist
Homework Helper
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Hi osker246,

Homework Statement


A boat takes 2.0 hr to travel 31 km down a river, then 6.0 hr to return. How fast is the river flowing?


Homework Equations


r=d/t


The Attempt at a Solution



first started out by finding the rate the boat traveled in each trip.

31/2=15.5 km/hr

31/6=5.2 km/hr

Now this is where im not sure if im solving this correctly. I then find the difference between the two rates and the time taken to travel both ways.

10.3 (difference of 15.5-5.2)/4(difference of 6-2)=2.6 km/hr

Is this the correct way of solving this problem? Thanks!
No, I don't think that's right. Notice is you keep units on your calculation you'll get km/hr^2.

So work with the total rate on each trip that you found. The boat has a speed, and the river has a speed. How do those combine to give 15.5 km/hr on the one trip, and how do they combine to give 5.2 km/hr going the other way?
 
  • #3
35
0
Hi osker246,

So work with the total rate on each trip that you found. The boat has a speed, and the river has a speed. How do those combine to give 15.5 km/hr on the one trip, and how do they combine to give 5.2 km/hr going the other way?
hey alphycist,

Im not sure if I follow what your trying to say. Obviously when the boat travels 15.5 km/hr it is traveling with the current and when it travels 5.2 hm/hr its traveling against the current. I'm at a loss how you find the rate the current moves though.
 
  • #4
alphysicist
Homework Helper
2,238
1
hey alphycist,

Im not sure if I follow what your trying to say. Obviously when the boat travels 15.5 km/hr it is traveling with the current and when it travels 5.2 hm/hr its traveling against the current. I'm at a loss how you find the rate the current moves though.
The boat has a speed [itex]v_b[/itex] (measured in still water), and the water has a speed [itex]v_w[/itex]. How do you combine those to get 15.5 km/h? How do you combine those to get 5.2 km/h? That will give you two equations and then you can solve for both of the unknown speeds. Does that make sense?
 
  • #5
35
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The boat has a speed [itex]v_b[/itex] (measured in still water), and the water has a speed [itex]v_w[/itex]. How do you combine those to get 15.5 km/h? How do you combine those to get 5.2 km/h? That will give you two equations and then you can solve for both of the unknown speeds. Does that make sense?
Ok I think I do understand. Tell me if this is correct.

Vb + Vw = 15.5

and

Vb-Vw = 5.2

Ok so I solved Vb + Vw = 15.5 for Vb. Giving me Vb=15.5 - Vw.

I then pluged in Vb=15.5 - Vw, into the equation Vb-Vw = 5.2 then solving for Vw.

So...

15.5-Vw-Vw=5.2

-2Vw=-10.3

Vw=5.15

Is this correct?
 
  • #6
alphysicist
Homework Helper
2,238
1
That sounds right to me.
 
  • #7
35
0
Alright! The answer was correct! Thank you very much alphysicist. I appreciate it.
 
  • #8
alphysicist
Homework Helper
2,238
1
Sure, glad to help!
 

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