Date | May 2018 | Marks available | 3 | 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(t)=61.5+acos(π8t)h(t)=61.5+acos(π8t), 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 max−min2max−min2 a = radius
|a|=117−62,55.5|a|=117−62,55.5 (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+acos(π8×0),117=61.5+acos(π8×8),6=61.5+a6=61.5+acos(π8×0),117=61.5+acos(π8×8),6=61.5+a
a = −55.5 A1 N2
[3 marks]
valid approach (M1)
eg sketch of h and y=30,h=30,61.5−55.5cos(π8t)=30,t=2.46307,t=13.5369y=30,h=30,61.5−55.5cos(π8t)=30,t=2.46307,t=13.5369
18.4630
t = 18.5 (minutes) A1 N3
[3 marks]