Question

A plane is flying in the horizontal plane. At time t1= 0 s with the components of its velocity are vx= 100 m/s and vy= 120 m/s. At time t2= 20 s the components are vx= 80 m/s and vy = 142 m/s.
a. What was the change in the direction of the velocity?
b. What is the change in the magnitude of the plane’s velocity during this time?
c. What are the components of the average acceleration for this interval?
d. What are the magnitude and direction of the average acceleration for this interval?

Answer 1

a. tanA1 = Vy/Vx = 120/100 = 1.20.
A1 = 50.2o = Direction.
Magnitude = Vx/cosA1=100/cos50.2 =156.2
m/s.

tanA2 = 142/80 = 1.775.
A2 = 60.6o = Direction.
Magnitude = 80/cos60.6 = 163 m/s.

60.6 - 50.2 = 10.4o = The change in
direction.

b. 163 - 156 = 7 m/s.

c. a(x) = (80-100)/20 = -1.0 m/s^2.
a(y) = (142-120)/20 = 1.1 m/s^2.

d, tanAr = 1.1/-1 = -1.10.
Ar = -47.7o = Reference angle.
A = 180 + (-47.7) = 132.3o=Direction.

Mag, = -1.0/cos132.3 = 1.49 m/s^2.