## Question 8:

### Part (a):
| Answer | Mark | Guidance |
|--------|------|----------|
| Straight line segment with positive gradient from origin | B1 | Straight line segment with positive gradient from the origin |
| Subsequent line segment horizontal and above the given line | B1 | Subsequent line segment horizontal and above the given line |
| Gradient change in graph labelled with 5 and 6 (may be on axes) | B1 | Gradient change labelled with 5 and 6 |

### Part (b):
| Answer | Mark | Guidance |
|--------|------|----------|
| Velocity of boy $[t < 5\text{s}]$: $v = 1.2t = 5.4$ | M1 | Equates an expression for boy's velocity to 5.4 |
| Giving $t = 4.5\text{ s}$ | A1 | Mark final answer |

## Question 8(c):

| Answer | Marks | Guidance |
|--------|-------|----------|
| Max distance when $t = 4.5$ | M1 | Recognises that max distance occurs when speeds are equal. Allow for their $t < 5$ from (b) used. Allow for $u = 0$, $v = 5.4$ and $a = 1.2$ in $v^2 = u^2 + 2as$ |
| Boy travels $\frac{1}{2} \times 1.2 \times 4.5^2 = 12.15$ | M1 | Attempt to use suvat or area under the graph and their time $t \leq 5$ to find distance travelled by boy |
| Bus travels $4.5 \times 5.4 = 24.3$ m; Max distance is $24.3 - 12.15 = 12.15$ m | A1 | Cao |

**Alternative method 1:**
| Answer | Marks | Guidance |
|--------|-------|----------|
| Max distance gained by the bus ahead of the boy is represented by the area of the triangle on velocity-time graph | M1 | Recognises that max distance occurs when the speeds are equal. Allow for their time from (b) used |
| Distance $= \frac{1}{2} \times 5.4 \times 4.5 = 12.15$ m | M1, A1 | Attempt to find area of the triangle; Cao |

**Alternative method 2:**
| Answer | Marks | Guidance |
|--------|-------|----------|
| Distance $S$ at time $t$ between $[t < 5]$: $S = 5.4t - \frac{1}{2} \times 1.2t^2$ | M1 | Combines expressions from suvat equations to find expression for the distance between |
| Max occurs when $\frac{dS}{dt} = 5.4 - 1.2t = 0$ | M1 | Equates the derivative of their expression to zero leading to a value for $t$ |
| When $t = 4.5$, max distance is $12.15$ m | A1 | Cao |

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## Question 8(d):

| Answer | Marks | Guidance |
|--------|-------|----------|
| Boy's distance is area of trapezium: $\frac{1}{2} \times 6(T + T - 5) [= 6T - 15]$ | B1 | Expression for the total distance for the boy (area method), e.g. sum of two distances $\frac{1}{2} \times 5 \times 6 + (T-5) \times 6$ |
| Bus's distance $5.4T$; equate distances $5.4T = 6T - 15$ | M1 | Equates their expression for distance to the distance travelled by the bus and attempt to solve for $T$ |
| Distance travelled in $25$ s is $135$ m | A1 | Cao. The value for $T$ need not be seen explicitly |

**Alternative method (relative speed):**
| Answer | Marks | Guidance |
|--------|-------|----------|
| At $t = 5$ the boy has travelled $\frac{1}{2} \times 5 \times 6 = 15$ m and the bus $5.4 \times 5 = 27$ m; Boy needs to catch up $12$ m | B1 | $12$ m seen if clear that it is a distance between the boy and the bus |
| Boy catches up $12$ m at $0.6$ ms$^{-1}$; So time is $\frac{12}{0.6} [= 20$ s$]$ | M1 | Uses relative speed to find the time to catch up |
| Total time $25$ s gives distance $135$ m | A1 | Cao. $25$ s need not be seen explicitly |

**Alternative (numerical) method:**
| Answer | Marks | Guidance |
|--------|-------|----------|
| Finds at least one correct distance for boy for $t > 5$ | B1 | |
| Working towards the time and distance at which the distances are equal | M1 | May be awarded for incorrect time and distances e.g. $0.6t^2$ used for the boy |
| For $25$ s distance travelled $135$ m | A1 | $25$ s must be seen as well as $135$ m |
| No method seen | | SC2 for $135$ m www where $t = 25$ not seen |

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