# Question 9:

## Part (a) – Way 1:

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\{d = kV^n \Rightarrow\}\ \log_{10} d = \log_{10} k + n\log_{10} V$ or $\log_{10} d = m\log_{10} V + c$ or $\log_{10} d = m\log_{10} V - 1.77$ seen or used as part of argument | M1 | AO 2.1 |
| Alludes to $d = kV^n$ and gives full explanation by comparing result with linear model e.g. $Y = MX + C$ | A1 | AO 2.4 |
| $\{k =\}\ 10^{-1.77} = 0.017$ or $\log 0.017 = -1.77$ linked together in same part of question | B1* | AO 1.1b |

## Part (a) – Way 2:

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\log_{10} d = m\log_{10} V + c$ or $\log_{10} d = m\log_{10} V - 1.77$ or $\log_{10} d = \log_{10} k + n\log_{10} V$ seen or used | M1 | AO 2.1 |
| $\{d = kV^n \Rightarrow\}\ \log_{10} d = \log_{10}(kV^n) \Rightarrow \log_{10} d = \log_{10} k + \log_{10} V^n \Rightarrow \log_{10} d = \log_{10} k + n\log_{10} V$ | A1 | AO 2.4 |
| $\{k =\}\ 10^{-1.77} = 0.017$ or $\log 0.017 = -1.77$ linked together | B1* | AO 1.1b |

## Part (a) – Way 3:

| Answer/Working | Mark | Guidance |
|---|---|---|
| Starts from $\log_{10} d = m\log_{10} V + c$ or $\log_{10} d = m\log_{10} V - 1.77$ | M1 | AO 2.1 |
| $\log_{10} d = m\log_{10} V + c \Rightarrow d = 10^{m\log_{10} V + c} \Rightarrow d = 10^c V^m \Rightarrow d = kV^n$ or $\log_{10} d = m\log_{10} V - 1.77 \Rightarrow d = 10^{m\log_{10} V - 1.77} \Rightarrow d = 10^{-1.77}V^m \Rightarrow d = kV^n$ | A1 | AO 2.4 – with no errors seen in working |
| $\{k =\}\ 10^{-1.77} = 0.017$ or $\log 0.017 = -1.77$ linked together | B1* | AO 1.1b |

**Note:** Allow B1 for $\log_{10} 0.017 = -1.77$ or $\log 0.017 = -1.77$; Give B0 for $10^{-1.77} = 0.01698...$ without reference to 0.017 in same part

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## Part (b):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\{d = 20, V = 30 \Rightarrow\}\ 20 = k(30)^n$ or $\log_{10} 20 = \log_{10} k + n\log_{10} 30$ | M1 | Applies $V = 30$ and $d = 20$ to their model **(correct way round)**; AO 3.4 |
| $20 = k(30)^n \Rightarrow \log 20 = \log k + n\log 30 \Rightarrow n = \frac{\log 20 - \log k}{\log 30} \Rightarrow n = ...$ or $\log_{10} 20 = \log_{10} k + n\log_{10} 30 \Rightarrow n = \frac{\log_{10} 20 - \log_{10} k}{\log_{10} 30} \Rightarrow n = ...$ | M1 | Applies logarithms correctly leading to $n = ...$; AO 1.1b |
| $\{n =$ awrt $2.08 \Rightarrow\}\ d = (0.017)V^{2.08}$ **or** $\log_{10} d = -1.77 + 2.08\log_{10} V$ | A1 | AO 1.1b |

**Note:** Allow $k =$ awrt $0.017$ and/or $n =$ awrt $2.08$ in their final model equation; M0 M1 A0 is a possible score for (b)

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## Part (c):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $d = (0.017)(60)^{2.08}$ | M1 | Applies $V = 60$ to their exponential or logarithmic model; AO 3.4 |
| $13.333...+ 84.918... = 98.251... \Rightarrow$ Sean stops in time | M1 | Uses model in correct problem-solving process — adding "thinking distance" to value of $d$ to find overall stopping distance; AO 3.1b |
| $100 - 13.333... = 86.666...$ & $d = 84.918 \Rightarrow$ Sean stops in time | A1ft | AO 3.2a |

**Note:** $\frac{1}{75}$ or 48 are examples of acceptable thinking distances; The thinking distance must be dimensionally correct for M1 mark (i.e. $0.8\times$ their velocity); A thinking distance of awrt 13 and value of $d$ in range [81.5, 88.5] required for A1ft; Allow "Sean stops in time" or "Yes he stops in time" or "he misses the puddle" as relevant conclusions; M0 M1 A0 is possible in (c)