Date | May Specimen paper | Marks available | 1 | Reference code | SPM.2.AHL.TZ0.4 |
Level | Additional Higher Level | Paper | Paper 2 | Time zone | Time zone 0 |
Command term | State | Question number | 4 | Adapted from | N/A |
Question
An aircraft’s position is given by the coordinates (, , ), where and are the aircraft’s displacement east and north of an airport, and is the height of the aircraft above the ground. All displacements are given in kilometres.
The velocity of the aircraft is given as .
At 13:00 it is detected at a position 30 km east and 10 km north of the airport, and at a height of 5 km. Let be the length of time in hours from 13:00.
If the aircraft continued to fly with the velocity given
When the aircraft is 4 km above the ground it continues to fly on the same bearing but adjusts the angle of its descent so that it will land at the point (0, 0, 0).
Write down a vector equation for the displacement, r of the aircraft in terms of .
verify that it would pass directly over the airport.
state the height of the aircraft at this point.
find the time at which it would fly directly over the airport.
Find the time at which the aircraft is 4 km above the ground.
Find the direct distance of the aircraft from the airport at this point.
Given that the velocity of the aircraft, after the adjustment of the angle of descent, is , find the value of .
Markscheme
r A1A1
[2 marks]
when , M1
EITHER
when , A1
since the two values of are equal the aircraft passes directly over the airport
OR
, A1
[2 marks]
height = 5 − 0.2 × 20 = 1 km A1
[1 mark]
time 13:12 A1
[1 mark]
(3 minutes) (M1)
time 13:03 A1
[2 marks]
displacement is A1
distance is (M1)
= 24.1 km A1
[3 marks]
METHOD 1
time until landing is minutes M1
height to descend =
M1
A1
METHOD 2
M1
M1
A1
[3 marks]