How Do You Calculate the Speed of a Boat Relative to Water?

[Elbobo]

Homework Statement

Two boat landings are 5.1 km apart on the
same bank of a streamthat flows at 3.4 km/hr.
A motorboat makes the round trip between
the two landings in 1.5 hr.
What is the speed of the boat relative to
the water? Answer in units of km/hr.

Homework Equations

The Attempt at a Solution

I tried to get the velocity of the water relative to the shore plus the velocity of the boat relative to shore to equal the velocity of the boat relative to the water.

That would give a right triangle with Vws and Vbs as the legs, and Vbw (what I'm trying to find) as the hypotenuse.

So I get sqrt ( (5.1 / 0.75)^2 + 3.4^2), which is 7.60 and wrong.

Please help me, I don't understand what I am doing wrong...

 

[Doc Al]

I tried to get the velocity of the water relative to the shore plus the velocity of the boat relative to shore to equal the velocity of the boat relative to the water.

velocity(boat/shore) = velocity(boat/water) + velocity(water/shore)

There's no need for any right triangles, since all velocities are in the same (or opposite) direction.

Hint: Use distance = speed*time
Set up that equation for when the boat's going upstream and for when its going downstream.

 

[tomcue]

Your approach is on the right track, but there is a small mistake in your calculations. The correct answer should be 7.68 km/hr. Here's how you can arrive at that answer:

Let Vws be the velocity of the water relative to the shore, Vbs be the velocity of the boat relative to the shore, and Vbw be the velocity of the boat relative to the water.

We know that Vbs = Vbw + Vws (since the boat's velocity relative to the shore is equal to its velocity relative to the water plus the velocity of the water relative to the shore).

We also know that the distance between the two landings is 5.1 km and the boat makes the round trip in 1.5 hours. This means that the boat travels a total distance of 10.2 km in 1.5 hours, giving us a speed of 6.8 km/hr.

Now, using the Pythagorean theorem, we can write:

Vbw^2 + Vws^2 = Vbs^2

Substituting in the values we know, we get:

Vbw^2 + (3.4 km/hr)^2 = (6.8 km/hr)^2

Solving for Vbw, we get:

Vbw = sqrt(6.8^2 - 3.4^2) = 7.68 km/hr

So the speed of the boat relative to the water is 7.68 km/hr.