Question

(b) A remote-controlled toy aircraft is flying horizontally in a wind. Fig. 3.1 shows the velocity vectors, to scale, of the wind and of the aircraft in still air. A north wind velocity 23ms-¹ 54° aircraft velocity in still air 42ms-¹ Fig. 3.1 The velocity of the aircraft in still air is 42 ms¹ to the north. The velocity of the wind is 23ms-1 in a direction of 54° east of south. Determine the magnitude of the resultant velocity of the aircraft.
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Answer 1

Resposta:

Let $\textbf{Va}$ be the velocity vector of the aircraft in still air:

$\textbf{Va} = 0 \,\textbf{i} + 42\, \textbf{j}\,\,\,\,\,\,\,\,\,\,\,\,(SI)$

Let $\textbf{Vw}$ be the velocity vector of the wind:

$\textbf{Vw} = 23 \cdot sin\,54\º \, \textbf{i} - 23 \cdot cos\, 54\º \textbf{j}\\\\= 18.6\, \textbf{i} - 13.5\, \textbf{j}$

Let's calculate the magnitude of the resultant velocity of the aircraft:

$\textbf{Va} + \textbf{Vw} = \left(0 + 18.6\right) \textbf{i} + \left(42 - 13.5 \right) \textbf{j}\\\\\Longleftrightarrow \textbf{Va} + \textbf{Vw} = 18.6 \, \textbf{i} - 28.5 \, \textbf{j}$

$v = \sqrt{18.6^2 + \left(-28.5\right)^2}\\\\\Longleftrightarrow v = \sqrt{1157.4}\\\\\Longleftrightarrow \boxed{v = 34.0\,\,ms^{-1}}$