While flying due east at 130 km/h, an airplane is also carried northward at 41 km/h by the wind blowing due north. What is the plane's resultant velocity? (Draw vector diagram to solve problem.)

Magnitude
km/hDirection (north of east)
72°
18°
18°
72°

72

To solve this problem, we can use vector addition to find the resultant velocity of the airplane. The magnitude of the resultant velocity can be calculated using the Pythagorean theorem, while the direction can be determined using trigonometry.

First, let's draw a vector diagram to visualize the situation. Draw two vectors: one representing the velocity of the airplane (130 km/h due east), and the other representing the velocity of the wind (41 km/h due north). The angle between these two vectors is 90 degrees.

Now, let's calculate the magnitude of the resultant velocity. Using the Pythagorean theorem, we have:

Resultant velocity^2 = (Velocity of airplane)^2 + (Velocity of wind)^2
Resultant velocity^2 = 130^2 + 41^2
Resultant velocity^2 = 16900 + 1681
Resultant velocity^2 = 18581
Resultant velocity ≈ sqrt(18581)
Resultant velocity ≈ 136.2 km/h (rounded to one decimal place)

Next, let's determine the direction of the resultant velocity. We have two angles given: 72° and 18°. Since the airplane is being carried northward by the wind, the resultant velocity will be in the north of the east direction. Therefore, the angle we are looking for is 72°.

Finally, we have our answer:

Magnitude: 136.2 km/h
Direction (north of east): 72°