Question

the compass of an airplane indicates that it is headed due north, and its airspeed indicator shows that it is moving through the air at 240 km/hr. if there is a wind of 100 km/hr from west to east, what is the velocity of the airplane? 155 km/hr with directional bearing n65.8°e 260 km/hr with directional bearing n67.4°e 260 km/hr with directional bearing n22.6°e 155 km/hr with directional bearing n24.6°e

Explanation:

Step1: Analyze the velocity vectors

The airplane's velocity relative to the air \(v_{a}\) is \(240\) km/hr north (\(v_{a}=\langle0, 240
angle\)), and the wind velocity \(v_{w}\) is \(100\) km/hr east (\(v_{w}=\langle100,0
angle\)). The resultant velocity \(v\) of the airplane is \(v = v_{a}+v_{w}\).

Step2: Calculate the magnitude of the resultant velocity

Using the Pythagorean theorem, \(|v|=\sqrt{100^{2}+240^{2}}=\sqrt{10000 + 57600}=\sqrt{67600}=260\) km/hr.

Step3: Calculate the direction of the resultant velocity

Let \(\theta\) be the angle between the north - direction and the resultant velocity. \(\tan\theta=\frac{100}{240}\), so \(\theta=\arctan(\frac{100}{240})\approx22.6^{\circ}\). The directional bearing is \(N22.6^{\circ}E\).

Answer:

260 km/hr with directional bearing N22.6°E