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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 ( x , y , z ), where x and y are the aircraft’s displacement east and north of an airport, and z is the height of the aircraft above the ground. All displacements are given in kilometres.

The velocity of the aircraft is given as  ( 150 50 20 ) km h 1 .

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 t 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 t .

[2]
a.

verify that it would pass directly over the airport.

[2]
b.i.

state the height of the aircraft at this point.

[1]
b.ii.

find the time at which it would fly directly over the airport.

[1]
b.iii.

Find the time at which the aircraft is 4 km above the ground.

[2]
c.i.

Find the direct distance of the aircraft from the airport at this point.

[3]
c.ii.

Given that the velocity of the aircraft, after the adjustment of the angle of descent, is  ( 150 50 a ) km h 1 , find the value of a .

[3]
d.

Markscheme

= ( 30 10 5 ) + t ( 150 50 20 )       A1A1

[2 marks]

a.

when  x = 0 t = 30 150 = 0.2         M1

EITHER

when  y = 0 t = 10 150 = 0.2     A1

since the two values of t are equal the aircraft passes directly over the airport

OR

t = 0.2 y = 0    A1

[2 marks]

b.i.

height = 5 − 0.2 × 20 = 1 km     A1

[1 mark]

b.ii.

time 13:12    A1

[1 mark]

b.iii.

5 20 t = 4 t = 1 20 (3 minutes)     (M1)

time 13:03      A1

[2 marks]

c.i.

displacement is ( 22.5 7.5 4 )      A1

distance is  22.5 2 + 7.5 2 + 4 2     (M1)

= 24.1 km   A1

[3 marks]

c.ii.

METHOD 1

time until landing is  12 3 = 9 minutes        M1

height to descend =  4 km

a = 4 9 60         M1

= 26.7       A1

 

METHOD 2

( 150 50 a ) = s ( 22.5 7.5 4 )         M1

150 = 22.5 s s = 20 3         M1

a = 20 3 × 4

= 26.7       A1

[3 marks]

d.

Examiners report

[N/A]
a.
[N/A]
b.i.
[N/A]
b.ii.
[N/A]
b.iii.
[N/A]
c.i.
[N/A]
c.ii.
[N/A]
d.

Syllabus sections

Topic 3—Geometry and trigonometry » AHL 3.11—Vector equation of a line in 2d and 3d
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Topic 3—Geometry and trigonometry

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