Question

An airplane is flying on a bearing of 330 degrees at 450 mph. Find the component form of the velocity of the airplane.

Answer 1

Answer:

The component form of the velocity of the airplane is $\vec v = 389.711\,\hat{i} -225\,\hat{j}\,\left[\frac{m}{s} \right]$.

Step-by-step explanation:

Let suppose that a bearing of 0 degrees corresponds with the $+x$ direction and that angle is measured counterclockwise. Besides, we must know both the magnitude of velocity ($\|\vec v\|$), in miles per hour, and the direction of the airplane ($\theta$), in sexagesimal degrees to construct the respective vector. The component form of the velocity of the airplane is equivalent to a vector in rectangular form with physical units, that is:

$\vec v = \|\vec v\|\cdot (\cos \theta \,\hat{i}+\sin \theta\,\hat{j})$ (1)

If we know that $\|\vec v\| = 450\,\frac{mi}{h}$ and $\theta = 330^{\circ}$, then the component form of the velocity of the airplane is:

$\vec v = 389.711\,\hat{i} -225\,\hat{j}\,\left[\frac{m}{s} \right]$

The component form of the velocity of the airplane is $\vec v = 389.711\,\hat{i} -225\,\hat{j}\,\left[\frac{m}{s} \right]$.