The date is December 13 2005 Air traffic controllers in a co

The date is December 13, 2005. Air traffic controllers in a control tower in central Ulanbangor have been tracking an unidentified aircraft (UAC) on their radar screen for the past 10 minutes. The UAC has entered Ulanbangor\'s eastern border and is flying through its air space heading west-north-west (a bearing of 290 degrees east of north) at an altitude of 35,000 ft and travelling at 560 km/h. This aircraft will not respond to the operators of the control tower and so they receive permission to intercept the UAC to try to determine its identity and purpose and to shoot it down if necessary They call on a fighter jet that can travel twice the speed of the UAC. Twenty minutes after the first sighting of the UAC, at precisely 23:00 hours (Ulanbangor Standard Time (UST)) the fighter jet gets into position directly over the tower at an altitude of 35,000 ft. At that moment the UAC is 200 km north and 52 km east of the tower. What direct bearing (east of north) should the fighter jet take to intercept the UAC. Where will the point of interception be relative to the tower? What will be the time of the interception?

Solution

Take the tower as the origin

==> UAC position at 23:00 hours is 52i+200j

Velocity vector = 560(-icos20+jsin20) = -563.23i+191.53j

Fighter jet\'s starting point be 0i+0j

==> velocity vector = 1120(i cos theta +j sin theta)

==> interception pint will be where

52-526.23t = 1120t cos theta and 200+191.53t = 1120t sin theta

solving the two equations

52/t = 1120cos theta +526.23 ==> 1/t = (1120cos theta +563.23)/52

200/t = 1120t sin theta-191.53==> 1/t = (1120sin theta-191.53)/200

==> (1120cos theta+563.23)/52 = (1120sin theta -191.53)/200

==> 4307.7cos theta + 2023.96+191.53 = 1120sqrt(1-cos^2 theta)

Squaring the equation on both sides gives us

15.793cos^2 theta+15.2163cos theta+2.912=0

==> cos theta = -0.263 and -0.699

earlier interception would be at -0.263 ==> theta = 105.27 degrees N of E ==> direct bearing of 354.72 degrees east of north

c). 52-526.23t = 1120cos theta

take cos theta = -0.263

==> t =0.65 hours or 39.5minutes

b). fighter jet flies at 0.65 hours at 1120km/hr on a bearing of 354.72degrees arriving at

783.79 km(i cos 105.27 +j sin 105.27)

teh interception happens 739.79 km away from the tower or 194.7km west of the tower