Question

An airplane undergoes the following displacements, all at the same altitude: First, it flies 59.0 km in a direction 30.0° east of north. Next, it flies 58.0 km due south. Finally, it flies 100 km 30.0° north of west. Use components to determine how far the airplane ends up from its starting point.A) 70.1 km B) 71.5 km C) 68.7 km D) 74.4 km E) 73 km

Answer 1

Answer:

B) 71.5 [km]

Explanation:

To solve this problem we will decompose each of the directions in the x & y axes.

To solve this problem we will decompose each of the directions in the x & y axes. also for a greater understanding of the angles, you should look at the attached image, which contains the orientations for each angle (clockwise or counterclockwise).

59.0 km in a direction 30.0° east of north

$d_{1x}= 59*sin(30) = 29.5[km]\\d_{1y}= 59*cos(30) = 51.09[km]$

58.0 km due south

$d_{2y} = - 58 [km]$

It flies 100 km 30.0° north of west

$d_{3x}= - 100*cos(30) = -86.6[km]\\d_{3y} = 100*sin(30)= 50 [km]$

Now we sum algebraically the components

$d_{x}=29.5-86.6 = -57.1[km]\\d_{y}=51.09 -58+50=43.09[km]$

Using the Pythagorean theorem we can find the magnitude of the displacement.

$d = \sqrt{(57.1)^{2} +(43.09)^{2} } \\d= 71.53[km]$