An aeroplane x whose average speed is 500km/hr leaves kano airport at 7:00am and travels for 2hours on a bearing 050, it then changes it's course and flies on a bearing 120 to an airstrip A. Another aeroplane Y leaves kano airport at 10:00am and flies on a straight course to the airstrip A. Both planes arrive at the airstrip A at 11:30am calculate:
a) the average speed of Y correct to three significant figures
b) the direction of flight Y to the nearest degree
To solve this problem, we need to determine the distance traveled by the airplanes.
For airplane X:
Since its speed is 500 km/hr and it travels for 2 hours, the distance it covers is:
Distance_X = Speed_X * Time_X = 500 km/hr * 2 hr = 1000 km
For airplane Y:
We can calculate the time it takes for airplane Y to travel from 10:00 am to 11:30 am:
Time_Y = 11:30 am - 10:00 am = 1.5 hours
Since both planes arrive at airstrip A at the same time, the distance traveled by airplane Y is the same as the distance traveled by airplane X (1000 km).
a) Average speed of airplane Y:
Speed_Y = Distance_Y / Time_Y = 1000 km / 1.5 hr = 666.67 km/hr (rounded to three significant figures)
Therefore, the average speed of airplane Y is 666.67 km/hr.
b) Direction of flight Y:
We know that airplane X traveled on a bearing of 050 for 2 hours. This means it covered the first leg of a triangle with the angle 050 at the Kano airport. To find the bearing of airplane Y, we need to determine the angle between airplane X's bearing and airplane Y's bearing.
The angle between the bearing of airplane X and airplane Y can be found using the Law of Cosines, as follows:
c^2 = a^2 + b^2 - 2ab*cos(C)
Where:
- c is the distance traveled by airplane Y (1000 km),
- a is the distance traveled by airplane X (1000 km),
- b is the distance from the end point of airplane X to airstrip A (which is unknown), and
- C is the angle between the distance a and distance c.
Since a = c = 1000 km, the equation can be simplified as follows:
1000^2 = 1000^2 + b^2 - 2 * 1000 * 1000 * cos(C)
0 = - 2 * 1000 * 1000 * cos(C)
Using the properties of cosine, we know that cos(C) = 0 when C = 90 degrees. Therefore, the angle between airplane X's bearing and airplane Y's bearing is 90 degrees.
Since airplane X's bearing is 050, and the angle between the bearings is 90 degrees, the bearing of airplane Y would be 050 + 90 = 140 degrees.
Therefore, the direction of flight Y is 140 degrees (nearest degree).