June 2023 Paper 1 Q12
12 One of the rides at a theme park is a room where the floor and ceiling both move up and down for \(10\pi\) seconds.
At time \(t\) seconds after the ride begins, the distance \(f\) metres of the floor above the ground is
\[f = 1 - \cos t\]At time \(t\) seconds after the ride begins, the distance \(c\) metres of the ceiling above the ground is
\[c = 8 - 4\sin t\]The ride is shown in the diagram below.
(a) Show that the initial distance between the floor and ceiling is 8 metres. [1 mark]
(b) Show that the distance \(d\) metres between the floor and ceiling at time \(t\) is given by\[d = 7 + R\cos(t + \alpha)\]where \(R\) and \(\alpha\) are positive constants to be found. [5 marks]
(c) Hence, find the minimum distance between the ceiling and the floor.
Give your answer to the nearest centimetre. [2 marks]
| Scheme | Marks | AO |
|---|---|---|
| Substitutes t = 0 into both models and obtains the distance. Condone missing units. | B1 | 3.4 |
| (1) |
Typical solution
\[t = 0\]\[8 - 4\sin 0 - (1 - \cos 0) = 8 \text{ metres}\]| Scheme | Marks | AO |
|---|---|---|
| Models the distance between the ceiling and the floor as \(c - f\) Condone a sign error when expressions for \(c\) and \(f\) are substituted. | M1 | 3.3 |
| Uses a compound angle formula to obtain \(R\cos\alpha = \pm 1 \text{ or } \pm 4\) or \(R\sin\alpha = \pm 4 \text{ or } \pm 1\) or \(\tan\alpha = \pm 4 \text{ or } \pm\dfrac{1}{4}\) PI by \(R = \sqrt{17} \approx 4.1\) or AWRT \(\alpha = 1.33^c\) or AWRT \(\alpha = 76^\circ\) | M1 | 3.1a |
| Obtains \(R = \sqrt{17} \approx 4.1\) Condone correct answer from \(\sqrt{(\pm 1)^2 + (\pm 4)^2}\) Note: M0 M1 A1 is possible | A1 | 1.1b |
| Obtains AWRT \(\alpha = 1.33^c\) or AWRT \(\alpha = 76^\circ\) No incorrect working seen in finding \(\alpha\) Accept other valid values of \(\alpha\) \(\alpha = 1.33^c + 2n\pi\) OE in degrees | A1 | 1.1b |
| Completes argument to obtain \(d = 7 + \sqrt{17}\cos(t + 1.33)\) Accept AWRT 4.1 in place of \(\sqrt{17}\) and \(\alpha = 1.33^c + 2n\pi\) Do not award this mark if \(\sin\alpha = \pm 4 \text{ or } \pm 1\) or \(\cos\alpha = \pm 1 \text{ or } \pm 4\) is used leading to a value of \(\tan\alpha\) | R1 | 2.1 |
| (5) |
Typical solution
\[d = c - f\]\[= 8 - 4\sin t - (1 - \cos t)\]\[= 7 + \cos t - 4\sin t\]\[R = \sqrt{17}\]\[R\cos\alpha = 1\]\[R\sin\alpha = 4\]\[\tan\alpha = 4,\ \alpha = 1.33\]\[d = 7 + \sqrt{17}\cos(t + 1.33)\]| Scheme | Marks | AO |
|---|---|---|
| Subtracts their R from 7 provided their R < 7 | M1 | 1.1a |
| Obtains 2.88 metres or 288 cm Correct units must be seen. | A1 | 3.2a |
| (2) | ||
| (8 marks) |