Can someone please help me no one ever responds to my question and I need the correct answer. I believe the answer is C but i'm
2 answers:
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Answer:
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°.
