# Question 2:

## Part 2(a):

| Answer/Working | Marks | Guidance |
|---|---|---|
| $220 = (28 \times 10) - \frac{1}{2}a \times 10^2$ | M1 A1 | Complete method to find equation in $a$ only (note $u=16$). Allow $220=(28\times10)+\frac{1}{2}a\times10^2$ for 'reverse' motion leading to $a=-1.2$: M1A0A0, but if they then change $a$ to $1.2$: M1A1A1 retrospectively. M0 if $u=0$ assumed |
| Other valid equations: $28 = u+10a$; $220=\frac{(u+28)}{2}\times10$; $220=10u+\frac{1}{2}a\times10^2$; $28^2=u^2+2a\times220$ | | Any 2 of these could be used to obtain equation in $a$ only |
| $a = 1.2\ (\text{m s}^{-2})$ | A1 | cao |

## Part 2(b):

| Answer/Working | Marks | Guidance |
|---|---|---|
| $28 = u_4 + 1.2\times6 \Rightarrow u_4 = 20.8$ **or** $28 = u_5 + 1.2\times5 \Rightarrow u_5 = 22$ **or** $s_4 = 16\times4+\frac{1}{2}\times1.2\times4^2 = 73.6$ **or** $s_5 = 16\times5+\frac{1}{2}\times1.2\times5^2=95$ | M1A1ft | Complete method to find speed at $t=4$ or $5$, or a distance at $t=4$ or $5$. M0 if $u=0$ assumed. A1ft: correct speed or distance, follow through on their $a$ and $u$, but only if $u$ has been **used** to find $a$ in (a) |
| Allow distances from $Q$: $s_6 = 28\times6-\frac{1}{2}\times1.2\times6^2=146.4$; $s_5=28\times5-\frac{1}{2}\times1.2\times5^2=125$ | | |
| $s=20.8\times1+\frac{1}{2}\times1.2\times1^2$ **OR** $s=22\times1-\frac{1}{2}\times1.2\times1^2$ **OR** $s=95-73.6$ **OR** $22^2=20.8^2+2\times1.2s$ **OR** $s=146.4-125$ | M1 | Complete method to find the required distance |
| $21.4\ \text{(m)}$, allow $21\ \text{(m)}$ | A1 | cao |

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