## Question 4:

### Part (i)
$\dfrac{-20}{2} = -10$, so $-10\ \text{m s}^{-2}$ | M1, A1 | Use of suitable triangle to attempt $\Delta v / \Delta t$ for suitable interval; accept wrong sign; allow both marks if correct answer seen

### Part (ii)(A)
Signed area under graph: $\frac{1}{2} \times 2 \times 20 = 20$ | M1, A1 | Using the relevant area or other complete method

### Part (ii)(B)
**Either** using areas:

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

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

Total displacement is $13.8\ \text{m}$ | 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 $t=4.5$: $-8.4$; Total: $13.8$ | B1 | Both required and both must be correct

### Part (iii)
$a = 4t - 14$ | M1, A1 |  Differentiate; do not award for division by $t$

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

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

For model B need $v$ when $a = 0$: $v\!\left(\frac{7}{2}\right) = -4.5$ | M1, A1 | Using (iii) or argument based on symmetry; $a=0$ when $t=3.5$

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

### Part (v)
Displacement is $\displaystyle\int_0^6 (2t^2 - 14t + 20)\, dt$ | M1 | Limits not required

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

Substitute limits | M1 | cao; accept bottom limit not substituted

$= 12$, so $12\ \text{m}$ | A1 |

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