A river flows due east at 1.65 m/s. A boat crosses the river from the south shore to the north shore by maintaining a constant velocity of 10.7 m/s due north relative to the water. What is the velocity of the boat relative to shore?
part B: If the river is 308 m wide, how far downstream has the boat moved by the time it reaches the north shore?
To find the velocity of the boat relative to the shore, we can use vector addition.
The velocity of the river is 1.65 m/s due east. The velocity of the boat relative to the water is 10.7 m/s due north.
To add these velocities, we can use the Pythagorean theorem. The magnitude of the resulting velocity is the square root of the sum of the squares of the individual velocities.
Let's call the velocity of the boat relative to the shore V.
Using the Pythagorean theorem:
V^2 = (10.7 m/s)^2 + (1.65 m/s)^2
V^2 = 114.49 m^2/s^2 + 2.7225 m^2/s^2
V^2 = 117.2125 m^2/s^2
Taking the square root of both sides:
V = √117.2125 m/s
V ≈ 10.82 m/s
Therefore, the velocity of the boat relative to the shore is approximately 10.82 m/s.
To find the velocity of the boat relative to the shore, we need to combine the velocities of the boat and the river. Since the boat is moving north while the river flows east, these velocities are perpendicular to each other, forming a right triangle.
To solve this, we can use vector addition. We'll treat the eastward velocity of the river as the x-component and the northward velocity of the boat as the y-component.
The eastward velocity of the river is given as 1.65 m/s, so we'll represent it as Vx = 1.65 m/s. The northward velocity of the boat is given as 10.7 m/s, so we'll represent it as Vy = 10.7 m/s.
Now, we can use the Pythagorean theorem to find the magnitude of the boat's velocity relative to the shore, which is the resultant of Vx and Vy.
Magnitude of the resultant velocity (v) = √(Vx² + Vy²)
= √(1.65² + 10.7²)
= √(2.7225 + 114.49)
= √117.2125
≈ 10.82 m/s
So, the magnitude of the boat's velocity relative to the shore is approximately 10.82 m/s.
To determine the direction, we can use trigonometry. The angle (θ) between the resultant velocity vector and the positive x-axis can be found using the tangent function:
tan(θ) = Vy / Vx
θ = arctan(Vy / Vx)
θ = arctan(10.7 / 1.65)
θ ≈ 80.14°
Therefore, the velocity of the boat relative to the shore is approximately 10.82 m/s in a direction of 80.14° north of east.