3. The date is December 13, 2005. Air traffic controllers in a control tower in

Question

3. 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. (a) What direct bearing (east of north) should the fighter jet take to intercept the UAC? (b) Where will the point of interception be relative to the tower? (c) What will be the time of the interception?

Explanation / Answer

Taking the tower as the origin of a coordinate system,
the UAC position at 23:00 hrs is
52 i + 200 j
and its velocity vector is
(560 km/hr) [ -i cos 20 + j sin 20 ]
= -526.23 i + 191.53 j
The fighter jet's starting position is 0 i + 0 j,
and its velocity vector is
(1120 km/hr) [ i cos theta + j sin theta ]
where theta is an angle NORTH of EAST .
The interception position will be where
52 - 526.23 t = 1120 t cos theta
AND
200 + 191.53 t = 1120 t sin theta.

To see whether cos theta should be positive or negative,
let us first find out how long it will take before the UAC is directly north of the tower:
52 - 526.23 T = 0, this "T" is not the interception time but the time for the
UAC to reach the meridian of the tower:
T = 0.098816 hrs = 5.93 minutes.
In that time, the fighter jet if it flew due north would be 110.67 km north of the tower,
but the UAC would be much farther north; this alone suffices to tell us that the
fighter jet will have to fly somewhere west of north to catch the UAC,
so cos(theta) in my equations above should be a negative number.

Now I return to the pair of simultaneous equations:
52 - 526.23 t = 1120 t cos theta
200 + 191.53 t = 1120 t sin theta
From the first equation,
52/t = 1120 cos theta + 526.23
(1/t) = (1120 cos theta + 526.23)/52
The second equation then becomes:
200(1120 cos theta + 526.23)/52 + 191.53 = 1120 sin(theta)
4307.7 cos theta + 2023.96 + 191.53 = 1120 sqrt(1 - cos^2 theta)
3.84616 cos theta + 1.97812 = sqrt(1 - cos^2 theta)
14.793 cos^2 theta + 15.2163 cos theta + 3.91294 = 1 - cos^2 theta
15.793 cos^2 theta + 15.2163 cos theta + 2.91294 = 0
Using the quadratic formula, we obtain
cos(theta) = -0.263498 or -0.699985
It is conceivable that both of these results will produce an interception,
but if so, the earlier interception would be the one closer to due north,
i.e., cos(theta) = -0.263498
and theta is 105.278 degrees N of E,
which translates into a "direct bearing" of 354.722 degrees east of north.
That answers part(a).

(c) 52 - 526.23 t = 1120 cos theta
= 1120 (-0.263498) = -295.12
347.12 = 526.23 t
t = 0.6596 hours, or 39.58 minutes.

(b): The fighter jet flies 0.6596 hrs at 1120 km/hr
on a bearing of 354.722 degrees,
arriving at
738.79 km (i cos 105.278 + j sin 105.278).
The interception happens 739 km away from the tower,
or 194.7 km west of the tower and 712.7 km north of the tower

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