A ferry is crossing a river. The ferry is headed due north with a speed of 2.6 m/s
relative to the water and the river’s velocity
is 4.4 m/s to the east.
a) What is magnitude of the boat’s velocity relative to Earth?
Answer in units of m/s
b) Find the direction in which the ferry is
moving (measured from due east, with counterclockwise positive).
Answer in units of ◦
.

a.  5.1 m/s b. 149.42°

Explanation:

a. Velocity of boat relative to earth, V = velocity of boat relative to river, V'+ velocity of river relative to earth, v.

We add these vectorially.

Since velocity of boat relative to river is 2.6 m/s due north and velocity of river relative to earth is 4.4 m/s due east, we apply Pythagoras' theorem and add them together since both velocities are perpendicular to each other.

So, V² = V'² + v² = (2.6 m/s)²  + (4.4 m/s)² = 6.76 (m/s)² + 19.36 (m/s)² = 26.12 (m/s)²

V = √[26.12 (m/s)²]

V = 5.1 m/s

b. Since the relative velocities of the ferry relative to the water and the velocity of the river relative to the earth are the components of the velocity of the ferry relative to the earth, the y - component being the velocity of the ferry relative to the earth since it is due north, and the x- component being the velocity of the river relative to the earth since it is due east. So, we have that

tanФ = y - component/x - component = V'/v where Ф is the angle between V and the west direction.

tanФ = V'/v

tanФ = 2.6 m/s ÷ 4.4 m/s

tanФ = 0.5909

Ф = tan⁻¹(0.5909)

Ф = 30.58°

So, the angle measured from due east with counterclockwise direction positive is 180° - Ф = 180° - 30.58° = 149.42°

double the distance = divide strength by 4