# Velocity of a Boat and Float vs. Water

1. Jan 27, 2012

### zaper

A boat is heading upstream when a float is dropped overboard and carried downstream. After 1 hour, when the boat is has traveled 2.0 km farther upstream, it turns around and catches up with the float 5.0 km downstream from the turnaround point. If the boat travels with constant velocity:

How fast does the river flow?
What is the boats speed in still water?

I really have no idea where to start with this really. I'm having trouble understanding how to find the velocities and such without a time variable. Any help that you guys can give will be much appreciated.

2. Jan 27, 2012

### Staff: Mentor

Choose a frame of reference first.

You have three choices:
1) the river bank
2) the boat
3) the float

Imagine yourself in one of these frames and from there try to write the equations of motion.

3. Jan 27, 2012

### zaper

Ok so if I'm the river bank then the boat is moving at 2 km/hr assuming up is positive so then will the float be moving at -5 km/hr? So the equation for the boat is d=2t and the float is d=-5t?

4. Jan 27, 2012

### zaper

Sorry I mean -3km/hr not -5 for the float

5. Jan 27, 2012

### Staff: Mentor

what if you were the float what would you see?

6. Jan 27, 2012

### zaper

The bank is moving at 3km/hr and the boat at 5

7. Jan 27, 2012

### Staff: Mentor

the boat traveled 2km up and 5km down in one hour. its total dist is 7km so what is its speed?

from the floats perspective the boat left and came back having traveled 7km in one hour so it went 3.5km up and 3.5km down. Now switch to an observer on the bank he saw the boat go up 2km and at the same time the float go down what? in the 1st half-hour.

Remember for both observers the distance between boat and float must agree.

8. Jan 27, 2012

### zaper

So the velocity of the boat in the water is 7km/hr. And if the boat is 3.5 km away from the float at the farthest point then the observer would see the float go back 1.5 km in 1/2 hour. So this would mean that the river goes 3 km/hr? If the river goes 3 km/hr and the boat goes 2 km in the first half hour then the boat would have gone 4 km in an hour and the boat goes 4 km/hr faster than the water which in 7 km/hr in still water?.

9. Jan 27, 2012

### Staff: Mentor

I think thats right:

the boat in still water travels at 7km/hr so in a half hour it travels 3.5 km/hr

the river travels at 3km/hr so in a half hour the float has drifted 1.5km downstream.

3.5 -1.5 = 2km for the boat going upstream. At that point the float has been carried 1.5km downstream and the distance between float and boat is 3.5km. The boat turns around and travels 3.5 + 1.5 = 5km downstream to meet up with the float.

This is kinda like the movie: "A River Runs Through It" without the waterfalls ending.

10. Jan 27, 2012

### Staff: Mentor

Perhaps I have a slightly difference in interpretation of the problem, but isn't the distance that the boat travels upstream from where the float is dropped measured along the bank? So the float is dropped at some location w.r.t. the bank, and the boat travels 2km further upstream w.r.t. the bank in one hour.

While the boat is traveling upstream the float is carried downstream with the current. The boat catches the float 5km from the point of turnaround, again with respect to the bank.

#### Attached Files:

• ###### Fig1.gif
File size:
2.7 KB
Views:
251
11. Jan 27, 2012

### zaper

Makes sense to me. Thank you very much

12. Jan 27, 2012

### zaper

How would you go about solving this with your interpretation gnell? I think the question in worded rather poorly

13. Jan 27, 2012

### Staff: Mentor

First assign variables to the unknown quantities:

vr = speed of river w.r.t. the riverbed (bank)
vb = speed of the boat in still water
t = time from the turnaround to catching the float

So the speed upriver is vb - vr; the speed downriver is vb + vr. Now write expressions for the various portions of the journey. You have three unknowns so you should be looking for three equations.

14. Jan 27, 2012

### Staff: Mentor

i was trying to show that you can solve it conceptually by being in the floats frame of reference initially so the boat travels away upstream for 3.5km and then back for 3.5km.

From this we determine that the boat travels at 7km/hour.

To the observer on the bank the boat goes 2km upstream so the float must has gone 1.5km downstream in a half hour. The distances both observers measure must be the same.

From this we determine that the river flows at 3km/hour.

15. Jan 27, 2012

### zaper

Here are my 3 possible equations:

1) As gnell says it takes the boat 1 hr to go 2 km so vb-vr = 2km/hr or vb = (2-vr) km/hr

2) Since t=d/v then t=5/(vb+vr) hr

3) The float moves at vr so its distance is vr*(1+t)km (because t is the time after the turnaround and the float moves 1 hour more than that

Are these what you were thinking?

16. Jan 27, 2012

### Staff: Mentor

Yes, they're the ones

17. Jan 27, 2012

### zaper

So subbing the first equation into the second we get t=5/2. Now since the boat moves at vb+vr and the float moves at vr can we say that the boat closes the distance at simply vb and vb km/hr*5/2hr=5km?

18. Jan 27, 2012

### Staff: Mentor

One thing, I just noticed that in your first equation you didn't change the sign of vr when you moved it to the other side of the equation. You should have: vb = 2km/hr + vr .

Another thing, if we agree from the outset that all distances are in km and all speeds in km/hr, we can write the equations in unitless form:

(vb - vr)*1 = 2
(vb + vr)*t = 5
vr*(t + 1) = 3

I'd suggest eliminating t from the equations first, leaving you with two equations in two unknowns vb and vr, the reason being that t is not one of the quantities requested for an answer.

Last edited: Jan 27, 2012
19. Jan 27, 2012

### zaper

Oh wow. Can't believe I missed that. What would recommend I do next then? I can't really seem to find an easy substitution

20. Jan 27, 2012