Question

FUS After takeoff, an aircraft enters a straight-line flight path with constant speed at point P(1.5|9|0.5). speed. In one minute it travels a distance vector v=(-1 5 0.2) back. Information in km. a) (1) Determine the speed of the aircraft in km/h. (2) After 4 minutes the plane reaches a big city. Calculate the altitude at this time. b) At an altitude of 2500 m the pilot switches to autopilot. Calculate the coordinates of the point where the aircraft is at this moment and state how much time has passed since the aircraft was at point P. c) Between the points A(-5|38|0.1), B(-4|40|0.11) and C (-7|39|0.1) there is a small natural protected area in the shape of a triangle. (1) Examine triangle ABC for special features. (2) Check whether the aircraft is flying over route AB and, if so, at what distance

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Answer to a math question FUS After takeoff, an aircraft enters a straight-line flight path with constant speed at point P(1.5|9|0.5). speed. In one minute it travels a distance vector v=(-1 5 0.2) back. Information in km. a) (1) Determine the speed of the aircraft in km/h. (2) After 4 minutes the plane reaches a big city. Calculate the altitude at this time. b) At an altitude of 2500 m the pilot switches to autopilot. Calculate the coordinates of the point where the aircraft is at this moment and state how much time has passed since the aircraft was at point P. c) Between the points A(-5|38|0.1), B(-4|40|0.11) and C (-7|39|0.1) there is a small natural protected area in the shape of a triangle. (1) Examine triangle ABC for special features. (2) Check whether the aircraft is flying over route AB and, if so, at what distance

Expert avatar
Santino
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73 Answers
a)
(1) Zur Bestimmung der Geschwindigkeit des Flugzeugs in km/h teilen wir die zurückgelegte Strecke durch die Zeit:
\text{Geschwindigkeit } = \frac{\text{Strecke}}{\text{Zeit}} = \frac{\|\textbf{v}\|}{\text{Zeit}}

Gegeben: $\textbf{v} = (-1, 5, 0.2)$ km und Zeit = 1 Minute = 1/60 Stunde.

Berechnung der Geschwindigkeit:
\|\textbf{v}\| = \sqrt{(-1)^2 + 5^2 + 0.2^2} = \sqrt{1 + 25 + 0.04} = \sqrt{26.04} \approx 5.10 \text{ km}
\text{Geschwindigkeit} = \frac{5.10}{1/60} = 5.1 \times 60 = 306 \text{ km/h}

(2) Die Flughöhe nach 4 Minuten beträgt:
Höhe = 9 + (4 * 0.2) = 9.8 km

b)
Bei einer Höhe von 2500 m = 2.5 km schalten wir auf Autopilot um.

Die Koordinaten des Flugzeugs ergeben sich durch Addition des Vektors $\textbf{v}$ zum Punkt P:
(1 + (-1), 5 + 5, 0.5 + 0.2) = (0, 10, 0.7)

Die Zeit, die seit dem Punkt P vergangen ist, beträgt:
\frac{\sqrt{(0-1)^2 + (10-5)^2 + (0.7-0.5)^2}}{5.1} = \frac{\sqrt{1 + 25 + 0.04}}{5.1} \approx \frac{\sqrt{26.04}}{5.1} \approx \frac{5.1}{5.1} \approx 1 \text{ Stunde}

c)
(1)
Das Dreieck ABC ist ein gleichschenkliges Dreieck, da AB und BC jeweils die gleiche Länge haben.

(2)
Um zu prüfen, ob das Flugzeug die Strecke AB überfliegt, betrachten wir den Abstand des Flugzeugs von der Linie, die durch die Punkte A und B verläuft. Der Abstand kann durch den Normalenvektor auf die Ebene von A und B bestimmt werden.

Den Normalenvektor der Ebene bestimmen:
\textbf{n} = (A-B) \times (A-C) = \begin{pmatrix} -5+4 \ 38-40 \ 0.1-0.11 \end{pmatrix} \times \begin{pmatrix} -5+7 \ 38-39 \ 0.1-0.1 \end{pmatrix} = \begin{pmatrix} -1 \ -2 \ -0.01 \end{pmatrix} \times \begin{pmatrix} 2 \ -1 \ 0 \end{pmatrix}
= \begin{pmatrix} 0.01 \ -0.02 \ -3 \end{pmatrix}

Der Abstand des Flugzeugs von der Ebene beträgt:
\frac{| \textbf{n} \cdot (P - A) |}{\|\textbf{n}\|} = \frac{| \begin{pmatrix} 0.01 \ -0.02 \ -3 \end{pmatrix} \cdot \begin{pmatrix} 1 \ 5 \ 0.5 \end{pmatrix} - \begin{pmatrix} 0.01 \ -0.02 \ -3 \end{pmatrix} \cdot \begin{pmatrix} -5 \ 38 \ 0.1 \end{pmatrix} |}{\sqrt{0.01^2 + (-0.02)^2 + (-3)^2}}

= \frac{| (0.01 \cdot 1 + (-0.02) \cdot 5 + (-3) \cdot 0.5) - (0.01 \cdot (-5) + (-0.02) \cdot 38 + (-3) \cdot 0.1) |}{\sqrt{0.01^2 + (-0.02)^2 + (-3)^2}}

= \frac{| 0.01 - 0.1 + 1.5 - 0.76 |}{\sqrt{0.01^2 + (-0.02)^2 + (-3)^2}} = \frac{| 2.67 |}{\sqrt{9.05}} \approx \frac{2.67}{3} \approx 0.89 \text{ km}

Antwort: a) (1) Die Geschwindigkeit des Flugzeugs beträgt 306 km/h.
b) Nach 4 Minuten beträgt die Flughöhe 9.8 km. Die Koordinaten des Flugzeugs sind (0, 10, 0.7) und es ist 1 Stunde vergangen, seit es sich im Punkt P befand.
c) (1) Das Dreieck ABC ist gleichschenklig.
(2) Das Flugzeug überfliegt die Strecke AB in einem Abstand von etwa 0.89 km.

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