**(i)**

$\frac{-20}{2} = -10$ | M1 | Use of a suitable triangle to attempt at $\Delta v / \Delta t$ for suitable interval. Accept wrong sign.

$= -10 \text{ m s}^{-2}$ | A1 | cao. Allow both marks if correct answer seen.

| | 2 |

**(ii) (A)**

Signed area under graph | M1 | Using the relevant area or other complete method

$\frac{1}{2} \times 2 \times 20 = 20$ | A1 |

**(B)**

**either** using areas

Signed area $2 \leq t \leq 5$ is $\frac{1}{2}((5-2)+(4.5-2.4)) \times (-4) = -10.2$ | B1 | Allow $\pm 10.2$.

Signed area $5 \leq t \leq 6$ is $\frac{1}{2} \times 1 \times 8 = 4$ | B1 |

Total displacement is $13.8$ m | B1 | cao but FT from their 20 in part (A)

**or using suvat**

From $t = 0$ to $t = 2.4$: 19.2 | B1 |

From $t = 4.5$ to $t = 6$: 3.0 | B1 |

From $t = 2.4$ to $t = 4.5$: $-8.4$ | B1 | Both required and both must be correct.

Total | : 13.8 |

| | 5 |

**(iii)**

$a = 4t - 14$ | M1 | Differentiate. Do not award for division by $t$.

$a(0.5) = -12$ so $-12 \text{ m s}^{-2}$ | A1 |

| A1 |

| | 3 |

**(iv)**

Model A gives $-4 \text{ m s}^{-1}$ | B1 | May be implied by other working

For model B we need $v$ when $a = 0$ | M1 | Using (iii) or an argument based on symmetry or sketch graph that $a = 0$ when $t = 3.5$

$v(\frac{7}{2}) = -4.5$ | A1 |

so model B is $0.5 \text{ m s}^{-1}$ less | F1 | Accept values without more or less

| | 4 |

**(v)**

Displacement is $\int_0^6 (2t^2 - 14t + 20) dt$ | M1 | Limits not required.

$= \left[\frac{2t^3}{3} - 7t^2 + 20t\right]_0^6$ | A1 | Limits not required. Accept 2 terms correct.

$= 12$ so $12$ m. | M1 | Substitute limits

| A1 | cao. Accept bottom limit not substituted.

| | 4 |

| | 18 |

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