## Question 4:

**Part (a)**
| Answer | Mark | Guidance |
|--------|------|----------|
| $\mathbf{v}_B = (-16\mathbf{i} - 3\mathbf{j}) + 5(2.4\mathbf{i} + \mathbf{j})$ | M1 | |
| $\mathbf{v}_B = (-4\mathbf{i} + 2\mathbf{j})$ | A1 | |
| $\sqrt{(-4)^2 + 2^2}$ | M1 | |
| $\sqrt{20} = 2\sqrt{5}$, 4.5 or better (m s$^{-1}$) | A1 | |

**Part (b)**
| Answer | Mark | Guidance |
|--------|------|----------|
| $(4\mathbf{i} + c\mathbf{j}) = (-16\mathbf{i} - 3\mathbf{j})T + \frac{1}{2}(2.4\mathbf{i} + \mathbf{j})T^2 + (44\mathbf{i} - 10\mathbf{j})$ | M1 | Equating **i**-components to give a **quadratic** in $T$ only. Allow $t$ instead of $T$. Allow omission of 44 for M mark. Also allow $\pm 4$ but M0 if 4 not used at all. |
| $4 = -16T + \frac{1}{2} \times 2.4T^2 + 44$ | A1 | |
| $T = 10$ | A1 | |

**Alternative (using B as initial position):**
| Answer | Mark | Guidance |
|--------|------|----------|
| $(4\mathbf{i} + c\mathbf{j}) = (-4\mathbf{i} + 2\mathbf{j})t + \frac{1}{2}(2.4\mathbf{i} + \mathbf{j})t^2 + (-6\mathbf{i} - 12.5\mathbf{j})$ | M1 | Equating **i**-components to give quadratic in $t$. Position vector of $B$ should be $-6\mathbf{i} - 12.5\mathbf{j}$ but no credit for finding this. |

# Question 4:

## Part 4(b):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $4 = -4t + \frac{1}{2} \times 2.4t^2 - 6$ | A1 | 1.1b |
| $t = 5$ so $(T =) 10$ | A1 | 1.1b |
| **(3 marks total)** | | |

## Part 4(c):

| Answer/Working | Mark | Guidance |
|---|---|---|
| Equating **j**-components, with their value of $T$ or $t$ substituted, to give an equation with a square term, in $c$ only. Allow $\pm c$ in equation. Allow omission of $-10$ or $-12.5$ for M mark | M1 | 2.1 |
| If using $A$: $c = (-3\times10) + \frac{1}{2}\times1\times10^2$ scores M1M0A0 | | |
| If using $B$: $c = (2\times5) + \frac{1}{2}\times1\times5^2$ scores M1M0A0 | | |
| If using $A$: $c = (-3\times10) + \frac{1}{2}\times1\times10^2 + (-10)$. Allow $\pm c$ and/or $\pm(-10)$ | M1 | 1.1b |
| If using $B$: $c = (2\times5) + \frac{1}{2}\times1\times5^2 + (-12.5)$. Allow $\pm c$ and/or $\pm(-12.5)$ | | |
| $c = 10$ | A1 | 1.1b |
| **(3 marks total)** | | |

**Notes 4a–4c:**

| Note | Mark | Guidance |
|---|---|---|
| Use of $\mathbf{v} = \mathbf{u} + \mathbf{a}t$ with $t=5$ to give unsimplified $\mathbf{v}_B$. M0 if $\mathbf{u}=\mathbf{0}$ | M1 | |
| If using integration, must find constant and put $t=5$. M0 if constant omitted altogether | | |
| Correct $\mathbf{v}_B$ with **i**'s and **j**'s collected | A1 | |
| Use of Pythagoras on their $\mathbf{v}_B$ to give magnitude (need root). Must be positive | M1, A1 | |
| Equating **i**-components to give equation in $T$ or $t$ only. M0 if $\mathbf{u}=\mathbf{0}$ | M1 (4b) | |
| A correct equation in $T$ or $t$ only (could be in $(T-5)$ if using $B$) | A1 | |
| $T = 10$ | A1 | |
| Equating **j**-components to give equation in $c$ only; allow omission of initial position | M1 (4c) | |
| With their $T$ or $t$, **must include $t=0$ position** (should be $-10$ using $A$, or $-12.5$ using $B$) | M1 | |
| cao | A1 | |

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