Solved: An aircraft’s position is given by the coordinates ( x, y, z), where x and... [IB Mathematics Applications & Interpretation (Math AI)]

An aircraft’s position is given by the coordinates ( x, y, z), where x and y are the aircraft’sdisplacement 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 heightof 5 km. Let t be the length of time in hours from 13:00.If the aircraft continued to fly with the velocity givenWhen the aircraft is 4 km above the ground it continues to fly on the same bearing but adjuststhe angle of its descent so that it will land at the point (0, 0, 0).

Question

HLPaper 2

An aircraft’s position is given by the coordinates ( $x$ , $y$ , $z$ ), where $x$ and $y$ are the aircraft’sdisplacement 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 $\left( {\begin{array}{*{20}{c}} { - 150} \\ { - 50} \\ { - 20} \end{array}} \right)\,{\text{km}}\,{{\text{h}}^{ - 1}}$ .

At 13:00 it is detected at a position 30 km east and 10 km north of the airport, and at a heightof 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 adjuststhe 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]

Verified

Solution

r $= \left( {\begin{array}{*{20}{c}} {30} \\ {10} \\ 5 \end{array}} \right) + t\left( {\begin{array}{*{20}{c}} { - 150} \\ { - 50} \\ { - 20} \end{array}} \right)\,$ A1A1

[2 marks]

verify that it would pass directly over the airport.

[2]

Verified

Solution

when $x = 0$ , $t = \frac{{30}}{{150}} = 0.2$ M1

EITHER

when $y = 0$ , $t = \frac{{10}}{{150}} = 0.2$ A1

since the two values of $t$ are equal the aircraft passes directlyover the airport

$t = 0.2$ , $y= 0$ A1

[2 marks]

state the height of the aircraft at this point.

[1]

Verified

Solution

height= 5 − 0.2 × 20 = 1 km A1

[1 mark]

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

[1]

Verified

Solution

time 13:12 A1

[1 mark]

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

[2]

Verified

Solution

$5 - 20t = 4 \Rightarrow t = \frac{1}{{20}}$ (3 minutes) (M1)

time 13:03 A1

[2 marks]

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

[3]

Verified

Solution

displacement is $\left( {\begin{array}{*{20}{c}} {22.5} \\ {7.5} \\ 4 \end{array}} \right)$ A1

distance is $\sqrt {{{22.5}^2} + {{7.5}^2} + {4^2}}$ (M1)

= 24.1 km A1

[3 marks]

Given that the velocity of the aircraft, after the adjustment of the angle of descent, is $\left( {\begin{array}{*{20}{c}} { - 150} \\ { - 50} \\ a \end{array}} \right){\text{km}}\,{{\text{h}}^{ - 1}}$ ,find the value of $a$ .

[3]

Verified

Solution

METHOD 1

time until landing is $12 - 3 = 9$ minutes M1

height to descend = $4\,{\text{km}}$

$a = \frac{{ - 4}}{{\frac{9}{{60}}}}$ M1

$= - 26.7$ A1

METHOD 2

$\left( {\begin{array}{*{20}{c}} { - 150} \\ { - 50} \\ a \end{array}} \right) = s\left( {\begin{array}{*{20}{c}} {22.5} \\ {7.5} \\ 4 \end{array}} \right)$ M1

$- 150 = 22.5\,s \Rightarrow s = - \frac{{20}}{3}$ M1

$a = - \frac{{20}}{3} \times 4$

$= - 26.7$ A1

[3 marks]

Question

HLPaper 2

The velocity of the aircraft is given as $\left( {\begin{array}{*{20}{c}} { - 150} \\ { - 50} \\ { - 20} \end{array}} \right)\,{\text{km}}\,{{\text{h}}^{ - 1}}$ .

At 13:00 it is detected at a position 30 km east and 10 km north of the airport, and at a heightof 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 adjuststhe 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]

Verified

Solution

r $= \left( {\begin{array}{*{20}{c}} {30} \\ {10} \\ 5 \end{array}} \right) + t\left( {\begin{array}{*{20}{c}} { - 150} \\ { - 50} \\ { - 20} \end{array}} \right)\,$ A1A1

[2 marks]

verify that it would pass directly over the airport.

[2]

Verified

Solution

when $x = 0$ , $t = \frac{{30}}{{150}} = 0.2$ M1

EITHER

when $y = 0$ , $t = \frac{{10}}{{150}} = 0.2$ A1

since the two values of $t$ are equal the aircraft passes directlyover the airport

$t = 0.2$ , $y= 0$ A1

[2 marks]

state the height of the aircraft at this point.

[1]

Verified

Solution

height= 5 − 0.2 × 20 = 1 km A1

[1 mark]

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

[1]

Verified

Solution

time 13:12 A1

[1 mark]

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

[2]

Verified

Solution

$5 - 20t = 4 \Rightarrow t = \frac{1}{{20}}$ (3 minutes) (M1)

time 13:03 A1

[2 marks]

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

[3]

Verified

Solution

displacement is $\left( {\begin{array}{*{20}{c}} {22.5} \\ {7.5} \\ 4 \end{array}} \right)$ A1

distance is $\sqrt {{{22.5}^2} + {{7.5}^2} + {4^2}}$ (M1)

= 24.1 km A1

[3 marks]

[3]

Verified

Solution

METHOD 1

time until landing is $12 - 3 = 9$ minutes M1

height to descend = $4\,{\text{km}}$

$a = \frac{{ - 4}}{{\frac{9}{{60}}}}$ M1

$= - 26.7$ A1

METHOD 2

$\left( {\begin{array}{*{20}{c}} { - 150} \\ { - 50} \\ a \end{array}} \right) = s\left( {\begin{array}{*{20}{c}} {22.5} \\ {7.5} \\ 4 \end{array}} \right)$ M1

$- 150 = 22.5\,s \Rightarrow s = - \frac{{20}}{3}$ M1

$a = - \frac{{20}}{3} \times 4$

$= - 26.7$ A1

[3 marks]

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