A river flows due east at 1.54 m/s. A boat crosses the river from the south shore to the north shore by maintaining a constant velocity of 11.2 m/s due north relative to the water. What is the velocity of the boat relative to shore?
speed=sqrt(1.54^2 + 11.2^2)
direction: arctan(11.2/1.54) relative to shore
To find the velocity of the boat relative to the shore, we need to combine the velocity of the river and the velocity of the boat relative to the water.
Given information:
- Velocity of the river (Vr) = 1.54 m/s to the east
- Velocity of the boat relative to water (Vbw) = 11.2 m/s to the north
To combine these velocities, we can use vector addition. Since the velocities are at right angles to each other, we can use the Pythagorean theorem to find the magnitude of the resultant velocity, and then use trigonometry to find the direction.
Let's break down the velocities into their x and y components:
- Velocity of the river in the x-direction (Vrx) = 1.54 m/s to the east
- Velocity of the boat relative to water in the x-direction (Vbwx) = 0 m/s (since it's only moving north)
- Velocity of the boat relative to water in the y-direction (Vbwy) = 11.2 m/s to the north
Now we can add the x and y components separately:
- Vx = Vrx + Vbwx = 1.54 m/s + 0 m/s = 1.54 m/s (no change in the x-direction)
- Vy = Vbwy = 11.2 m/s (no change in the y-direction)
Next, we can find the magnitude of the resultant velocity using the Pythagorean theorem:
V = sqrt(Vx^2 + Vy^2)
V = sqrt((1.54 m/s)^2 + (11.2 m/s)^2)
V ≈ 11.29 m/s (rounded to two decimal places)
Finally, we can find the angle θ (direction) of the resultant velocity using inverse trigonometry:
tan(θ) = Vy / Vx
θ = arctan(Vy / Vx)
θ = arctan(11.2 m/s / 1.54 m/s)
θ ≈ 81.46 degrees (rounded to two decimal places)
Therefore, the velocity of the boat relative to the shore is approximately 11.29 m/s to the north and 81.46 degrees east of north.