Question

A boat is traveling due North at 27mph. A current is flowing at a bearing of 60° west of south at 8mph. Find the actual speed and direction of the boat written in magnitude and direction form.

Answer 1

Answer: $24.02\ mph$

Step-by-step explanation:

Given

Boat is traveling due to north at $v_{bf}=27\ mph$

The current is flowing $60^{\circ}$ west of south at $v_f=8\ mph$

Suppose the actual velocity of boat is $\vec{v_b}$

In vector notation we can write

$\Rightarrow \vec{v_{bf}}=\vec{v_b}-\vec{v_f}\\\\\Rightarrow \vec{v_b}=\vec{v_{bf}}+\vec{v_f}$

Flow of current can be written as

$\Rightarrow \vec{v_f}=-8\sin 60^{\circ}\hat{i}-8\cos 60^{\circ}\hat{j}$

Insert the values of vectors

$\Rightarrow \vec{v_b}=27\hat{j}-8\sin 60^{\circ}\hat{i}-8\cos 60^{\circ}\hat{j}\\\\\Rightarrow \vec{v_b}=-4\sqrt{3}\hat{i}+27\hat{j}-4\hat{j}\\\\\Rightarrow \vec{v_b}=-4\sqrt{3}\hat{i}+23\hat{j}$

Magnitude of the velocity is

$=\sqrt{(4\sqrt{3})^2+(23)^2}\\=\sqrt{48+529}\\=24.02\ mph$