a river flows due east at 1.50 m/s. A boat crosses the river from the south shore to the north shore by maintaining a constant velocity of 10m/s due north relative to the water. (a) what is the velocity of the boat relative to the shore? (b) if the river is 300 m wide, how far downstream has the boat moved by the time it reaches the north shore?

(a) Well, if the river flows due east at 1.50 m/s and the boat is moving 10m/s due north relative to the water, then we can say the velocity of the boat relative to the shore is the combination of these two movements. So, to calculate that, let me use my mathematical superpowers... Just give me a second... Ta-dah! The velocity of the boat relative to the shore is approximately 10.04 m/s at an angle of about 8.54 degrees north of east. Isn't that "shore-ly" impressive?

(b) Now, the boat needs to cross a 300 m wide river. Since the river flows east, it will cause the boat to drift downstream. Using my trusted humoronometer, I can determine the time it takes for the boat to cross the river is approximately 30 seconds. So, if we multiply the time (30 s) by the river's velocity (1.50 m/s), we find that the boat has moved approximately 45 meters downstream by the time it reaches the north shore. So, if you were to tell the boat it's "shore-ly" good at drifting, it might respond with a grateful chuckle.

To find the velocity of the boat relative to the shore, we need to analyze the vector addition of the velocity of the boat, relative to the water, and the velocity of the river.

Let's break down the problem into its components:

Given:
River velocity (v_river) = 1.50 m/s (due east)
Boat velocity relative to the water (v_boat relative to water) = 10 m/s (due north)

(a) To find the velocity of the boat relative to the shore, we need to determine the resultant velocity vector by adding the velocity of the boat relative to the water and the velocity of the river.

We can use vector addition to find the resultant velocity:

Magnitude of the resultant velocity (v_resultant) = √[(v_boat relative to water)^2 + (v_river)^2]

Direction of the resultant velocity = tan^(-1)(v_river / v_boat relative to water)

Substituting the given values:

Magnitude of v_resultant = √[(10 m/s)^2 + (1.50 m/s)^2]
= √[100 m^2/s^2 + 2.25 m^2/s^2]
≈ √102.25 m^2/s^2
≈ 10.12 m/s

Direction of v_resultant = tan^(-1)(1.50 m/s / 10 m/s)
= tan^(-1)(0.15)
≈ 8.53° (north of east)

Therefore, the velocity of the boat relative to the shore is approximately 10.12 m/s at an angle of 8.53° north of east.

(b) To calculate how far downstream the boat has moved, we need to determine the displacement along the east direction.

Since the boat's velocity is perpendicular to the direction of the river's flow, the boat's displacement will be the same as the width of the river.

Therefore, the boat will have moved 300 meters downstream by the time it reaches the north shore.

In summary:
(a) The velocity of the boat relative to the shore is approximately 10.12 m/s at an angle of 8.53° north of east.
(b) The boat has moved 300 meters downstream by the time it reaches the north shore.