An aircraft travelled from Calabar to Kano as follows. It first flew to illorin with a distance 300km due North west of 30degrees and then moved to kano with a distance of 400km due North East of 60degrees. Calculate the resultant displacement of the aircraft.
what the heck is "due northwest of 30°" ?
due northwest is northwest -- no 30° involved
same for the other.
Did you mean a heading of N 30° W and N 60° E ?
In any case, plot the vectors, get the x- and y-components (using sine and cosine)
then add the components and use the distance formula as usual.
To calculate the resultant displacement of the aircraft, we can use the concept of vector addition.
First, let's analyze the given information:
1. From Calabar to Ilorin: Distance = 300 km, Direction = North West of 30 degrees.
2. From Ilorin to Kano: Distance = 400 km, Direction = North East of 60 degrees.
Step 1: Convert the given directions into Cartesian coordinates.
For the first leg of the trip (Calabar to Ilorin):
North West of 30 degrees = (cos(30°) * 300 km, sin(30°) * 300 km)
= (0.866 * 300, 0.5 * 300)
≈ (259.8 km, 150 km)
For the second leg of the trip (Ilorin to Kano):
North East of 60 degrees = (cos(60°) * 400 km, sin(60°) * 400 km)
= (0.5 * 400, 0.866 * 400)
≈ (200 km, 346.4 km)
Step 2: Add the Cartesian coordinates together.
Resultant displacement = (259.8 km + 200 km, 150 km + 346.4 km)
≈ (459.8 km, 496.4 km)
Step 3: Calculate the magnitude and direction of the resultant displacement.
Magnitude of the resultant displacement = sqrt((459.8 km)^2 + (496.4 km)^2)
≈ 660.4 km
Direction of the resultant displacement = arctan(496.4 km / 459.8 km)
≈ 48.28°
Therefore, the resultant displacement of the aircraft is approximately 660.4 km in a direction of 48.28°.