## Question 4(i):

| Answer | Mark | Guidance |
|--------|------|----------|
| $v^2 - u^2 = 2as$ | | |
| $31^2 - 12^2 = 2 \times 215 \times a$ | M1 | Selection and use of appropriate equation(s) |
| $a = 1.9$ so $1.9 \text{ ms}^{-2}$ | A1 | |

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## Question 4(ii):

| Answer | Mark | Guidance |
|--------|------|----------|
| $v = u + at$ | | |
| $31 = 12 + 1.9t$ | M1 | Selection and use of appropriate equation(s) |
| $t = 10$ s | A1 | FT from their value of $a$ from part (i). |

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## Question 4(iii):

| Answer | Mark | Guidance |
|--------|------|----------|
| $\frac{215}{2} = 12t + \frac{1}{2} \times 1.9 \times t^2$ | M1 | Selection and use of $s = ut + \frac{1}{2}at^2$. Correct elements but condone minor arithmetic errors. |
| $t = \frac{-12 \pm \sqrt{12^2 + 4 \times 0.95 \times 107.5}}{1.9}$ | M1 | Use of quadratic formula (may be implied by answer). |
| $t = 6.055$ (or $-18.69$) | A1 | FT their $a$ only. |

**Alternative (2-stage method):**

| Answer | Mark | Guidance |
|--------|------|----------|
| $v^2 - u^2 = 2as$ and $s = \frac{(u+v)}{2}t$ giving $v = \pm\sqrt{12^2 + 2\times1.9\times107.5} = (\pm)23.505...$ | M1 | Selection and use of a complete valid 2-stage method |
| $s = \frac{(u+v)}{2}t \Rightarrow t = \frac{2\times107.5}{(12+23.505...)}$ | M1 | Using the output from the first stage to find $t$ |
| $t = 6.055$ (or $18.69$) | A1 | FT their $a$ only. |

# Question 4:

## Part (iv):
| Because it is accelerating, it travels less fast in the first half of the distance and so takes more time | B1 | Answer must refer to the two parts of the distance (or "the same distance") — no credit for answers like "Because it is accelerating" and "Because its speed is not uniform". Answers referring to times to cover AM and MB acceptable (may be implicit). e.g. "It is travelling faster between M and B than it is between A and M". Note: uniformity of acceleration is irrelevant. |

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