# Question 6:

## Part (i)
| Answer/Working | Mark | Guidance |
|---|---|---|
| For sketch of single valued, continuous graph consisting of 3 straight line segments with $+^{ve}$, then $-^{ve}$, then $+^{ve}$ slope | B1 | |
| Sketch appears to show $v(0) = 0$ and $v(8) > v(26) > v(20)$ | B1 [2] | |

## Part (ii)
| Answer/Working | Mark | Guidance |
|---|---|---|
| For shading the triangle from $t = 0$ to $t = 8$, the trapezium from $t = 8$ to $t = 20$ and the trapezium from $t = 20$ to a value of $t$ seen to be between 20 and 26 | B1 [1] | |

## Part (iii) — Main Scheme
| Answer/Working | Mark | Guidance |
|---|---|---|
| | M1 | For using area property to find $s(20)$ |
| $s(20) = \frac{1}{2}(8 \times 8) + \frac{1}{2}(8 + 2) \times 12 \; (= 92)$ | A1 | |
| | M1 | For using the gradient property to find acceleration in 3rd phase |
| $a = (6.5 - 2)/6 \; (= 0.75)$ | A1 | |
| $[s(t) = 92 + 2(t - 20) + 0.375(t-20)^2]$ | M1 | |
| Displacement is $0.375t^2 - 13t + 202$ metres | A1 [6] | |

## Part (iii) — Alternative Marking Scheme
| Answer/Working | Mark | Guidance |
|---|---|---|
| $[v(t) = 2 + 0.75(t-20)$; $s(t) = 0.375t^2 - 13t + A$ where $92 = 0.375 \times 400 - 13 \times 20 + A]$ | M1 | For finding $v(t)$, integrating and using $s(20) = 92$ |
| Displacement is $0.375t^2 - 13t + 202$ metres | A1 | |

## Part (iii) — First Alternative Marking Scheme
| Answer/Working | Mark | Guidance |
|---|---|---|
| $a = (6.5 - 2)/(26 - 20) = 0.75$ | B1 | |
| $v = 0.75t \; (+C1)$ | M1 | Integrating |
| $v = 0.75t - 13$ | A1 | Using $v(20) = 2$ or $v(26) = 6.5$ |

## Question 6 (iii) - First Alternative:

| Answer/Working | Mark | Guidance |
|---|---|---|
| $s(20) = 92$ or $s(26) = 117.5$ | B1 | Using area in diagram |
| $s = 0.375t^2 - 13t (+ C_2)$ | M1 | Integrating |
| $s = 0.375t^2 - 13t + 202$ | A1 [6] | Using $s(20)$ or $s(26)$ to find $C_2 = 202$ |

## Question 6 (iii) - Second Alternative:

| Answer/Working | Mark | Guidance |
|---|---|---|
| $s = 0.375t^2 - 13t + 202$ | | Given |
| $v = 0.75t - 13$ | M1 | Differentiating |
| $a = 0.75$ | M1 | Differentiating |
| $a = (6.5-2)/(26-20) = 0.75$ | B1 | Check agreement from graph |
| $v(20) = 0.75(20) - 13 = 2$ or $v(26) = 0.75(26) - 13 = 6.5$ | B1 | Check $v$ agrees at a point between $t = 20$ and $t = 26$ |
| Show $s(20) = 92$ or $s(26) = 117.5$ | B1 | Using area under graph |
| $s(20) = 0.375(20)^2 - 13(20) + 202 = 92$ or $s(26) = 0.375(26)^2 - 13(26) + 202 = 117.5$ | B1 | Check $s$ agrees at a point between $t = 20$ and $t = 26$ |

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