Question

Vector & Relative Motion: A pilot wishes to fly from City A to City B. City B is exactly due east from A. Her plane has an airspeed of 200km/h. If the wind blowing to the North at 50.0km/h:
a)Ehat Direction does she need to point her plane?
b)Whats her groundspeed?
c)how long will her flight be if A and B are 500km apart?

Answer 1

Answer on Question#38419 – Physics – Mechanics, Kinematics, Dynamics

Suppose we have 3 vectors:

$\overrightarrow{AC}$ – wind blowing to the North, $|\overrightarrow{AC}| = 50$;

$\overrightarrow{CB}$ – plane’s velocity, $|\overrightarrow{CB}| = 200$;

$\overrightarrow{AB}$ – resultant velocity vector.

$$\overrightarrow{AB} = \overrightarrow{AC} + \overrightarrow{CB}$$

Using Pythagorean theorem we get:

$$|\overrightarrow{AB}| = \sqrt{200^2 - 50^2} = \sqrt{37500} = 50\sqrt{15}$$

So the **groundspeed** of the plane is $50\sqrt{15} \approx 194 \, \text{km/h}$.

The **direction** is:

$$\alpha = \atan \frac{50}{50\sqrt{15}} \approx 14.5{}^\circ$$

So the pilot should point the plane $14.5{}^\circ$ to the south of east (see the scheme above).

Since the groundspeed is $50\sqrt{15}$ km/h so she needs $t = \frac{500}{50\sqrt{15}} = \frac{10}{\sqrt{15}} \approx 2.6$ hours (or 2 hours 35 minutes) to cover $500$ km.