# Relative Velocity of a boat Question

1. Sep 15, 2010

### jt121

1. The problem statement, all variables and given/known data

A boat takes 4.0 hr to travel 25 km down a river, then 5.0 hr to return.

A. How fast is the river flowing?
B. What is the velocity of the boat relative to the water flow?

2. Relevant equations

VAC=VAB+VBC

3. The attempt at a solution

For some reason I have trouble thinking in terms of the equation... I keep thinking that I can simply divide the time by distance traveled which yields a velocity of 6.25 km/hr for one direction and 5.0 km/hr for the return. Any help is appreciated.

Thanks.

EDIT: This should probably be in Introductory Physics, sorry. Any help on this section is still appreciated. Thanks =)

2. Sep 15, 2010

### CompuChip

You are right about everything you have posted.
So in the formula you posted, what do the three variables stand for?
For which of the three are the numbers you calculated (6.25 and 5.0) values?

3. Sep 15, 2010

### jt121

Thanks for the response, good to know I'm headed in the right direction.

To go further, I tried to identify the variables in the equation as you said.

So far;

Let:
Vwg = Velocity of water relative to the ground
Vbw = Velocity of the boat relative to water
Vbg = Velocity of the boat relative to the ground = 6.25 km/hr and 5 km/hr ?

So if I set this up right,

Vwg = Vbw + Vbg?

Maybe I'm lost.

Thanks again.

4. Sep 15, 2010

### CompuChip

You got them mixed up in the final line ;)

Vbg = Vbw + Vwg?

Think of you walking on one of those automatic walkways you see in airports, for example.
Suppose it is moving at Vwg 5 m/s and you are walking at Vyw = 1.5 m/s. Clearly, what is your speed Vyg with respect to someone sitting around, waiting for their flight?
Do you see why you get something like (looking at the subscripts only)
ac = ab + bc​
with the b's in the middle "dropping out"?

Anyway, there are two velocities Vbg, one for the upstream and one for the downstream trip. So you should apply the equation twice. What can you say about Vwg and Vbw in the two cases?

5. Sep 15, 2010

### jt121

Solved the problem, thank you so much for the help! I was hung up on the equation for each "leg" of the trip but you helped me simplify it.

6. Sep 16, 2010

### CompuChip

Good, so you didn't fall into the minus sign traps then (e.g. the velocities are actually +6.25 and -5, if positive direction is downstream.

Well done.