An Aeroplane X Whose Average Speed Is 500km/h Leaves Kano Airport At 7:00am And Travels For 2 Hours On A Bearing 050 Degree. It Then Changes Course And Flies On A Bearing 120 Degree 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 Arrives At The Airstrip A At 11:30am Calculate
(a) The Average Speed Of Y To Three Significant Figure
(b) The Direction Of Flight Y To The Nearest Degree.
Cant make a drawing to show.... So i hope u get it with this
X: 7:00am
Speed of X = 500km/hr for 2hrs
Diatance of X = 1000km i.e. (500×2)km
X change course after 2hrs so time will be 9:00am
X and Y arrive at A at 11:30am
For X and Y to reach A it will take
X to reach for 2.5hrs then distance will be (1250km)
X to A 500km/hr × 2.5hrs = 1250km
While y will be 1.5hrs since Y left by 10:00am
Using cosine rule
|KA|^2 = |KC|^2 + |CA|^2 - 2(|KC||CA|)cos110°
|KA|= 1848.6km
Average speed of y = 1848.6/1.5 = 1232.4km
Approximately to 3.s.f = 1230km
Direction of flight Y
Using sine rule
sin110/ 1848.6 = sinx/1250
X = 39.5°
From the starting point the direction of flight will be 50° + 39.5° = 89.5°
To the nearest degree = 90°
So what is your question? Do you know what bearing means (000 is North, then090 is East)?
Very detailed but the diagram is very important .... hope we can get it
The angle between them is59
the question asked the average speed, sorry but you've not found the speed I'm not seeing the answer to the average speed there.
The average speed of Y is 1230km/hr
Contact me via WhatsApp to get the solutions... 09056199196
Draw a diagram, as usual.
using the law of cosines, the distance z from Kano to Airstrip is
z^2 = 1000^2 + 1500^2 - 2(1000)(1500) cos110°
z = 2068
(a) 2068/1.5
(b) using the law of sines, the angle between X's starting heading and Y's heading is
sinθ/1250 = sin110°/2068