Question

WILL NAME THE BRAINLIEST! An airplane undergoes the following displacements: It first flies 72 km in a direction of 30° East of north. Next it flies 48 km due south. Finally it flies 100 km 30° north of west. Using analytical methods, determine how far the airplane ends up from its starting point. Show all work!

Answer 1

Answer:

82.1 km

Explanation:

We need to resolve each displacement along two perpendicular directions: the east-west direction (let's label it with x) and the north-south direction (y). Resolving each vector:

$A_x = (72) sin 30^{\circ} =36.0 km\\A_y = (72) cos 30^{\circ} = 62.4 km$

Vector B is 48 km south, so:

$B_x = 0\\B_y = -48$

Finally, vector C:

$C_x = -(100) cos 30^{\circ} =-86.6 km\\C_y = (100) sin 30^{\circ} = 50.0 km$

Now we add the components along each direction:

$R_x = A_x + B_x + C_x = 36.0 + 0 +(-86.6)=-50.6 km\\R_y = A_y+B_y+C_y = 62.4+(-48)+50.0=64.6 km$

So, the resultant (which is the distance in a straight line between the starting point and the final point of the motion) is

$R=\sqrt{R_x^2+R_y^2}=\sqrt{(-50.6)^2+(64.6)^2}=82.1 km$