displacement

Question

1.1 A light aircraft is 60 km north east from OR Tambo International Airport and it flies north for six hours at a velocity of 96km/h. Determine its position (displacement) with reference to OR Tambo International Airport in magnitude and direction.

Answer 1

The aircraft is at 60 km far from OR Tambo International Airport .

Aircraft travels at rate of  96 km/h for 6 hours . 

Distance travel by the air craft is velocity*time = 96 km/h *  6 hours = 576 km .

So the position of aircraft with respect to OR Tambo International Airport can be determined by the below diagram .

Initially the aircraft is at 60 km north east from OR Tambo International Airport  .

From there Aircraft travels to 576 km in north direction .

So the position of aircraft with respect to OR Tambo International Airport can be calculated using cosine rule .

c² = 60² + 576² -2(60)(576) cos(135°) 

c² = 3600 + 331776 + 48875.22

c² = 384251.22

c = 619.88

The position of aircraft with respect to OR Tambo International Airport is 619.88 km .

 The direction of aircraft with respect to OR Tambo International Airport can be calculated using sine rule .

sin A / 60  = 0.7071 / 619.88

sin A =  0.001140 * 60

sin A = 0.06844

A = arcsin ( 0.06844)

A = 3.924°

 The direction of aircraft with respect to OR Tambo International Airport is 3.924° .