# Question 8:

## Part 8(a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $s = \frac{1}{2} \times 3 \times 4^2$ **OR** $s = \frac{1}{2} \times 4 \times 12$ | M1 | Complete method to find distance in first 4 s; must be area of a triangle from a $v$-$t$ graph |
| $= 24$ (m) | A1 | Correct answer |

## Part 8(b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $12$ (m s$^{-1}$); $42$ (m s$^{-1}$) | B1 | Both speeds seen anywhere e.g. on a diagram or in part (a) |
| $12 \times 20 + \frac{1}{2} \times 1.5 \times 20^2\ (= 540)$ **OR** $\left(\frac{12+42}{2}\right) \times 20$ | M1 A1ft | Complete method to find total distance in next 20 s; must be area of trapezium from $v$-$t$ graph; correct unsimplified distance, ft on their 12 |
| $42 \times 2 + \frac{1}{2} \times (-4) \times 2^2\ (= 76)$ **OR** $\left(\frac{42+34}{2}\right) \times 2$ | M1 A1ft | Complete method to find total distance in next 2 s; must be area of trapezium from $v$-$t$ graph; correct unsimplified distance, ft on their 42 |
| Total $= 640$ (m) | A1 cao | Correct total distance |

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