## Question 15:

### Part (a):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $t = 1 \Rightarrow H = 4 + 1.5\sin\left(\frac{\pi}{6}\right) = 4.75\text{ (m)}$ | B1* | Given answer; score for sight of $H = 4 + 1.5\sin\left(\frac{\pi}{6}\right) = 4.75\text{ (m)}$; alternatively work backwards: $H = 4.75 \Rightarrow \frac{\pi t}{6} = \frac{\pi}{6} \Rightarrow t = 1 \Rightarrow$ time $= 1$ a.m. |

**(1 mark)**

### Part (b):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $t = 14 \Rightarrow H = 4 + 1.5\sin\left(\frac{14\pi}{6}\right) =$ awrt $5.3\text{ (m)}$ | M1A1 | M1 for substituting $t = 14$ into $H$ and attempting to calculate; score for $4 + 1.5\sin\left(\frac{14\pi}{6}\right) = \ldots$; A1 for awrt $5.3$ (m) or $\frac{16+3\sqrt{3}}{4}$ m following correct work. Note: $4 + 1.5\sin\left(\frac{2\pi}{6}\right) = 5.30$ is M0A0 unless explanation given that period is 12 hours |

**(2 marks)**

### Part (c):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $H = 3 \Rightarrow 3 = 4 + 1.5\sin\left(\frac{\pi t}{6}\right)$ | M1 A1 | M1 for substituting $H = 3$ into $H = 4 + 1.5\sin\left(\frac{\pi t}{6}\right)$ WITH some attempt to make $\sin\left(\frac{\pi t}{6}\right)$ the subject; $\left(\frac{\pi t}{6}\right)$ may be replaced by another variable; A1 for $\sin\left(\frac{\pi t}{6}\right) = -\frac{2}{3}$ (condone awrt $-0.67$) |
| $\sin\left(\frac{\pi t}{6}\right) = -\frac{2}{3} \Rightarrow t = \frac{6\times\left(\pi + \arcsin\left(\frac{2}{3}\right)\right)}{\pi} =$ awrt $7.4$ | M1A1 | M1 for correct attempt to find one of the first two positive values of $t$ using their $-0.67$; score for either $t = \frac{6\times\left(2\pi - \arcsin\left(\frac{2}{3}\right)\right)}{\pi}$ or $\frac{6\times\left(\pi + \arcsin\left(\frac{2}{3}\right)\right)}{\pi}$; A1 for one of $t$ awrt $7.4$ or awrt $10.6$ (implies previous M mark) |
| $t = \frac{6\times\left(2\pi - \arcsin\left(\frac{2}{3}\right)\right)}{\pi} =$ awrt $10.6$ | M1 | For correct attempt to find both of the first two positive values of $t$ using their $-0.67$ |
| Times are $7{:}23/7{:}24$ am and $10{:}36/10{:}37$ am | A1 | Times required for this mark only; accept both $7{:}23/7{:}24$ am and $10{:}36/10{:}37$ am in 12-hour times; allow $0723/0724$ and $1036/1037$ in 24-hour times; 0 must be present in $07{:}23/07{:}24$ |

**(6 marks) — Total: 9 marks**

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