# Question 7:

## Part (i)
| For using the idea that area of the quadrilateral represents distance | M1 | |
| $\frac{1}{2}\times200\times16 + 300\times\frac{1}{2}(16+25)$ | A1 | |
| $+\ \frac{1}{2}\times100\times25\ (=1600+6150+1250)$ | A1 | |
| Distance is $9000$ m | **Total: 3** | |

## Part (ii)
| $a = (0-25)/(600-500)$ | M1 | For using the idea that gradient $(= \text{vel}\div\text{time})$ represents acceleration; or for using $v = u + at$ |
| Deceleration is $0.25$ ms$^{-2}$ | A1 | Allow acceleration $= -0.25$ ms$^{-2}$ | **Total: 2** |

## Part (iii)
| Acceleration is $(1200t - 3t^2)\times10^{-6}$ | M1 | For using $a(t) = \dot{v}(t)$ |
| | A1 | **Total: 2** |

## Part (iv)
| $0.25 - 0.2475$ | M1 | For using 'ans(ii) $- |a_Q(550)|$'; ft ans(ii) only |
| Amount is $\pm0.0025$ ms$^{-2}$ | A1ft | **Total: 2** |

## Part (v)
| $1200t - 3t^2 = 0$ | M1 | For solving $a_Q(t)=0$ or for finding $a_Q(400)$ |
| $t = (0 \text{ or})\ 400$ | A1 | Or for obtaining $a_Q(400) = 0$ | **AG** | **Total: 2** |

## Part (vi)
| $\frac{1}{2}\times200\times16 + 200\times\frac{1}{2}(16+22)$ | M1 | For correct method for $s_P(400)$ |
| | A1 | |
| $s_Q(t) = (200t^3 - t^4/4)\times10^{-6}\ (+C)$ | M1 | For using $s_Q(t) = \int v_Q\, dt$ |
| $6400 - 5400$ | A1 | For using correct limits and finding $|s_Q(400) - s_P(400)|$ |
| | M1 | |
| Distance is $1000$ m | A1 | **Total: 6** |