Relative Velocity Homework: Find Speed of River

[Garicup]

Homework Statement

A tugboat releases a barge and continues down the river for t = 64 min. Then it returns and finds the barge 3.9 km downstream from its initial position. Assuming that the speed of the tugboat was constant with respect to the water, the speed of the river is (in km/h)

Homework Equations

Relative Velocity ~~~
I think V_tw = V_tb + V_bw

tugboat wrt water (tw)
tugboat wrt barge (vb)
barge wrt water (bw)

The Attempt at a Solution

I thought the speed of the river would be velocity of barge wrt water?
except even that was true...i have no idea what to do with the information given
Looking for anything to point me in a starting direction. Not really sure where to start.

 

[alphysicist]

Hi Garicup,

Homework Statement

A tugboat releases a barge and continues down the river for t = 64 min. Then it returns and finds the barge 3.9 km downstream from its initial position. Assuming that the speed of the tugboat was constant with respect to the water, the speed of the river is (in km/h)

Homework Equations

Relative Velocity ~~~
I think V_tw = V_tb + V_bw

tugboat wrt water (tw)
tugboat wrt barge (vb)
barge wrt water (bw)

The Attempt at a Solution

I thought the speed of the river would be velocity of barge wrt water?

I believe the barge is just floating along in the water, so it's velocity wrt water would be zero.

So a good starting point is, how long does it take for the tugboat to reach the barge on its return journey? How can you determine that time?

 

[baba89]

To find the speed of the river, we can use the concept of relative velocity. This means that we can find the speed of the river by considering the velocities of the tugboat and the barge with respect to the water.

First, let's define our variables:
Vtw - velocity of tugboat with respect to water
Vtb - velocity of tugboat with respect to barge
Vbw - velocity of barge with respect to water
t - time (64 minutes)

We can use the equation Vtw = Vtb + Vbw to solve for the speed of the river.

Since the tugboat travels for 64 minutes, we can convert this to hours by dividing by 60. So t = 64/60 = 1.067 hours.

Next, we know that the barge is 3.9 km downstream from its initial position when the tugboat returns. This means that the barge has traveled 3.9 km downstream while the tugboat was away for 1.067 hours.

Using the equation Vbw = d/t, where d is the distance traveled and t is the time, we can solve for Vbw:
Vbw = 3.9 km/1.067 hours = 3.655 km/h

Now, we can use the equation Vtw = Vtb + Vbw to solve for Vtw. We know that Vtb is the speed of the tugboat with respect to the barge, and we can assume that this is constant. So we can rearrange the equation to solve for Vtw:
Vtw = Vtb + Vbw
Vtw = Vtb + 3.655 km/h

Since the question tells us that the speed of the tugboat was constant with respect to the water, we can assume that Vtw is the same as the speed of the tugboat. So we can rewrite the equation as:
Vtw = Vtb + 3.655 km/h
Vtw = Vtb + 3.655 km/h

Now, we need to find the value of Vtb. We know that the tugboat traveled for 1.067 hours before finding the barge downstream. So we can use the equation Vtb = d/t, where d is the distance traveled and t is the time, to solve for Vtb:
Vtb = 3.9 km/1.067 hours = 3.655 km/h

Finally, we can