A ferry is crossing a river. The ferry is headed
due north with a speed of 2.8 m/s relative to
the water and the river’s velocity is 4.9 m/s
to the east.
a) What is magnitude of the boat’s velocity
relative to Earth
Add the two perpendicular vector velocities. The resulant magnitude can be computed using the Pythagorean theorem.
sqrt[2.8^2 + 4.9^2] = ?
67
To solve this problem, we need to use vector addition to find the boat's velocity relative to the Earth. The magnitude of the boat's velocity relative to Earth can be determined by finding the resultant vector of the boat's velocity relative to the water and the river's velocity.
Let's break down the given information:
- The boat's velocity relative to the water is 2.8 m/s directed due north.
- The river's velocity is 4.9 m/s directed to the east.
To find the boat's velocity relative to Earth, we need to add the two velocities using vector addition. We can use the Pythagorean theorem to find the magnitude of the resultant vector.
Let's denote the boat's velocity relative to Earth as V_earth and the boat's velocity relative to the water as V_water.
V_earth = √((V_water)^2 + (V_river)^2)
In this case, V_water = 2.8 m/s and V_river = 4.9 m/s.
V_earth = √((2.8 m/s)^2 + (4.9 m/s)^2)
= √(7.84 m^2/s^2 + 24.01 m^2/s^2)
= √(31.85 m^2/s^2)
≈ 5.64 m/s
Therefore, the magnitude of the boat's velocity relative to Earth is approximately 5.64 m/s.