Question

a plane flies from city A to City B. City B is 1540 km west and 1160 km south of city A. What is the total displacement and direction of the plane?

I know the displacement is 1930 Km but I can't match the degrees on my work. I keep coming up with 37 degrees yet the two answers I can choose from are 1930 km, 223 degrees and 1930 km 217 degrees. Can someone please help me? And show me how you are getting the degrees please?

Answer 1

Remember: east direction is +ve x-axis and North direction is +ve y-axis

Given that City A is 1540 km west from plane A

R1 = 1540 km towards west = (-1540 i) km

City B is 1160 km South from plane A

R2 = 1160 km towards South = (-1160 j) km

Now total displacement of plane from A to B will be

R = R1 + R2

R = -1540 i - 1160 j

|R| = magnitude of displacement = sqrt ((-1540)^2 + (-1160)^2)

|R| = 1928 km

In three significant figures

|R| = 1930 km

Part B.

Since in vector R, both Rx < 0 and Ry < 0, So vector R is in 3rd quadrant, So

Direction = pi + arctan (Ry/Rx) = pi + arctan (1160/1540)

Direction = pi + 37.0 deg = 180 + 37 = 217 deg

See that we need to find direction of vector w.r.t. +ve x-axis, So we add 180 deg in 37 deg

Let me know if you've any query.