A boat crosses a 200 m wide river at 3 ms-1, north relative to water. The river flows at 1 ms-1 as shown.
What is the velocity of the boat as observed by a stationary observer on the river back from which the boat departed?

3.16 m·s⁻¹ at an angle of 71.6°  

Explanation:

Assume that the diagram is like Fig. 1 below.

The boat is heading straight across the river and the current is directed straight downstream.

We have two vectors at right angles to each other.

1. Calculate the magnitude of the resultant

We can use the Pythagorean theorem (Fig. 2).

R² = (3 m·s⁻¹)² + (1 m·s⁻¹)² = 9 m²·s⁻² + 1 m²·s⁻² = 10 m²·s⁻²

R = √(10 m²·s⁻²) ≈ 3.16 m·s⁻¹

2. Calculate the direction of the resultant

The direction of the resultant is the counterclockwise angle (θ) that it makes with due East .

tanθ = opposite/adjacent = 3/1 = 3

θ = arctan 3 = 71.6°

To an observer at point O, the velocity of the boat is 3.16 m·s⁻¹ at an angle of 71.6°.

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d) physical means can be used to separate the components of air.