# Question 6:

## Part (i)
- $v$ is single valued, continuous and positive for $0 < t < 1000$
- $1^{st}$ segment has $+$ve slope
- And two or more of: $v(0) = 0$, $v(1000) = 0$, $2^{nd}$ segment has zero slope, $3^{rd}$ segment has $-$ve slope | M1 | For sketching a graph consisting of 3 straight line segments

Correct sketch | A1 | [2]

## Part (ii)
| M1 | For using 'area under graph' represents distance of $20000$ m
$\frac{1}{2}(600 + 1000)V = 20\,000$ | A1 |
$V = 25$ | A1 | [3]

**SR for candidates who assume $1^{st}$ and $3^{rd}$ time intervals are each 200 s (max 2/3):**
$\frac{1}{2}V\times200 + V\times600 + \frac{1}{2}V\times200 = 20000$ | B1 |
$V = 25$ | B1 |

## Question 6(iii):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $[V/t_1 = 0.15 \Rightarrow t_1 = 166.6...]$ | M1 | For using the gradient property for acceleration (or $v = 0 + at$) to find $t_1$ |
| $t_3 = 233.3...$ | A1ft | ft $400 - V/0.15$ |
| $[s_3 = \frac{1}{2} \times 233.3... \times 25]$ | DM1 | For using the area property for distance or $(u+v)/2 = s/t$. Depends on previous M1 |
| Distance is 2920m | A1 | [4] |

**Alternatively:**

| Answer/Working | Mark | Guidance |
|---|---|---|
| For using $V^2 = 2 \times 0.15s_1$ | M1 | |
| $[\Rightarrow s_1 = 2083.3...]$ | | |
| $s_2 = 15000$ (ft 600V) | B1ft | |
| For $s_3 = 20000 - s_1 - s_2$ | DM1 | |
| Distance is 2920m | A1 | |

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