Calculating River Flow Rate from Boat Trip Data

[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 I am not sure if I am 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!

 

[alphysicist]

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 I am not sure if I am 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?

 

[osker246]

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.

 

[alphysicist]

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 v_b (measured in still water), and the water has a speed v_w. 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?

 

[osker246]

The boat has a speed v_b (measured in still water), and the water has a speed v_w. 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?

 

[alphysicist]

That sounds right to me.

 

[osker246]

Alright! The answer was correct! Thank you very much alphysicist. I appreciate it.

 

[alphysicist]

Sure, glad to help!