Relative Velocity of a boat Question

[jt121]

Homework Statement

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?

Homework Equations

V_AC=V_AB+V_BC

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 =)

 

[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?

 

[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:
V_wg = Velocity of water relative to the ground
V_bw = Velocity of the boat relative to water
V_bg = Velocity of the boat relative to the ground = 6.25 km/hr and 5 km/hr ?

So if I set this up right,

V_wg = V_bw + V_bg?

Maybe I'm lost.

Thanks again.

 

[CompuChip]

You got them mixed up in the final line ;)

So if I set this up right,

V_bg = V_bw + V_wg?

Think of you walking on one of those automatic walkways you see in airports, for example.
Suppose it is moving at V_wg 5 m/s and you are walking at V_yw = 1.5 m/s. Clearly, what is your speed V_yg 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 V_bg, one for the upstream and one for the downstream trip. So you should apply the equation twice. What can you say about V_wg and V_bw in the two cases?

 

[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.

 

[CompuChip]

Solved the problem

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.