| 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.
Find the value of a.
Find when the seat is 30 m above the ground for the third time.
Markscheme
valid approach to find k (M1)
eg 8 minutes is half a turn, k + diameter, k + 111 = 117
k = 6 A1 N2
[2 marks]
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]
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]
