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Date May 2018 Marks available 2 Reference code 18M.2.sl.TZ2.6
Level SL only Paper 2 Time zone TZ2
Command term Find Question number 6 Adapted from N/A

Question

At an amusement park, a Ferris wheel with diameter 111 metres rotates at a constant speed. The bottom of the wheel is k metres above the ground. A seat starts at the bottom of the wheel.

The wheel completes one revolution in 16 minutes.

After t minutes, the height of the seat above ground is given by \(h\left( t \right) = 61.5 + a\,{\text{cos}}\left( {\frac{\pi }{8}t} \right)\), for 0 ≤ t ≤ 32.

After 8 minutes, the seat is 117 m above the ground. Find k.

[2]
a.

Find the value of a.

[3]
b.

Find when the seat is 30 m above the ground for the third time.

[3]
c.

Markscheme

valid approach to find k      (M1)

eg   8 minutes is half a turn, + diameter, + 111 = 117

k = 6      A1 N2

[2 marks]

a.

METHOD 1

valid approach      (M1)
eg  \(\frac{{{\text{max}}\,\, - \,\,{\text{min}}}}{2}\) a = radius

\(\left| a \right| = \frac{{117 - 6}}{2},\,\,55.5\)     (A1)

a = −55.5      A1 N2

 

METHOD 2

attempt to substitute valid point into equation for f      (M1)
eg  h(0) = 6, h(8) = 117

correct equation      (A1)
eg   \(6 = 61.5 + a\,{\text{cos}}\left( {\frac{\pi }{8} \times 0} \right),\,\,117 = 61.5 + a\,{\text{cos}}\left( {\frac{\pi }{8} \times 8} \right),\,\,6 = 61.5 + a\)

a = −55.5      A1 N2

[3 marks]

b.

valid approach      (M1)
eg   sketch of h and \(y = 30,\,\,h = 30,\,\,61.5 - 55.5\,{\text{cos}}\left( {\frac{\pi }{8}t} \right) = 30,\,\,t = 2.46307,\,\,t = 13.5369\)

18.4630

t = 18.5 (minutes)      A1 N3

[3 marks]

c.

Examiners report

[N/A]
a.
[N/A]
b.
[N/A]
c.

Syllabus sections

Topic 3 - Circular functions and trigonometry » 3.5 » Solving trigonometric equations in a finite interval, both graphically and analytically.
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