Zadanie 9g, Paper 2, maj 2024, TZ2, Matematyka AA IB SL
Wzorowane na: Matematyka AA IB SL — maj 2024, TZ2, Paper 2 (zadania wzorowane), pytanie 9(g). Treść, dane i kontekst są zmienione; sprawdzane umiejętności, format i punktacja jak w oryginale.
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Zadanie
Wspólne informacje do wiązki zadań
A clockwise rotating scanner at S can reach points up to 24 m away. A straight guide rail runs north–south, 15 m east of S. It intersects the reach circle at A (north) and B (south). The scanner completes one revolution in 18 s. At t=0 its ray points to A; while it sweeps the segment AB, it meets the rail at D, d metres south of A. Define α=angle ASD and β=angle SAB, in radians. A robot moves south along the rail, at distance g(t) = 0.06t2 + 0.8t + 20 metres south of A at time t≥0. For the first sweep, 0≤t≤T, where T is the time the ray takes to move from A to B.
Polecenie
For the first sweep, use . (i) Express the distance w between the robot and D in terms of t. (ii) Find when and where the scanner first reaches the robot.
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Odpowiedź z klucza
(i) w = ∣0.06t2 + 0.8t + 20 − d(t)∣. Before the first meeting the expression inside the absolute value is positive; a signed separation g−d is sufficient for locating the first zero. (ii) Solve g(t)=d(t) on 0≤t≤5.131781…: the first root is t=3.4405869742… s, and g(t)=23.4627279029… m. Thus the first detection is at 3.44 s, 23.5 m south of A. The position lies within AB.
Schemat punktowania
(i) A1: correct difference of g and d; absolute value for distance throughout the sweep, or a clearly stated pre-meeting/signed separation. (ii) M1: solve the zero-separation equation on the physical sweep interval. A1: first time 3.44 s. A1: position 23.5 m south of A.
Komentarz. Paper 2: calculator and AA SL formula booklet allowed. Retain unrounded intermediate values. Give three significant figures unless otherwise specified.