An airplane flying parallel to the ground undergoes two consecutive displacements. The first is 75km 30° west of north, and the second is 155km 60° east of north. What is the total displacement of the airplane?
2006-11-02 19:25:39 · 4 answers · asked by Some Guy 2 in Science & Mathematics ➔ Physics
Step 1. Draw the vectors moving from the end of the first to the beginning of the second
Step 2. Establish the angle at the joining point (In this case it is 90°)
Step 3. Use the cosine amd sine rules or, in this case, the properties of a right angled triangle to find both the magnitude and the direction of the resultant vector.
d = √(75² + 155²)
≈ 172.2 km
θ is the clockwise angle between the first vector and the resultant vector
tanθ = 155/75
θ ≈ 64° 10' 44"
So to rotate 64° 10' 44" clockwise from 30" west of north the direction of the resultant position is 24° 10' 44" east of north
ie the plane's total displacement was 172.2 km, 24° 10' 44" east of north
2006-11-02 20:22:21 · answer #1 · answered by Wal C 6 · 0⤊ 0⤋
You shoul to painting the vectors in a booknote, and measured directly. It's more easy to explain.
2006-11-02 23:43:57 · answer #2 · answered by Juan D 3 · 0⤊ 0⤋
172.2 km
2006-11-02 20:17:10 · answer #3 · answered by RunmanJr2 2 · 0⤊ 0⤋
142.45 km
2006-11-02 19:53:33 · answer #4 · answered by ag_iitkgp 7 · 0⤊ 0⤋