## Question 6(a):

| Answer | Marks | Guidance |
|--------|-------|----------|
| For an attempt at differentiation | **\*M1** | Decrease power by 1 and a change in coefficient in at least one term (which must be the same term). Therefore, must see $\frac{d}{dt}(kt^{\frac{1}{2}}) = ct^{-\frac{1}{2}}$ with $c \neq k$. |
| $a = \frac{1}{2}kt^{\frac{1}{2}-1} - 2t^{1-1} + 0 = \frac{1}{2}kt^{-\frac{1}{2}} - 2$ | **A1** | May be unsimplified. |
| $a = 0 \Rightarrow \frac{1}{2}kt^{-\frac{1}{2}} - 2 = 0 \left[t = \frac{k^2}{16}\right]$ or $k = 4t^{\frac{1}{2}}$ | **DM1** | Equate $a$ to 0 and attempt to solve for $t$ or $k$. Must be of the correct form, e.g. $t = \lambda k^2$ or $k = \delta t^{\frac{1}{2}}$ (possibly implied by forming a correct equation in $k$). |
| $(v=)k\left(\frac{k^2}{16}\right)^{\frac{1}{2}} - 2\frac{k^2}{16} - 8 = 4.5$ | **M1** | Or complete method to find $k$ or an equation involving $k$, e.g. $4t^{\frac{1}{2}} \times t^{\frac{1}{2}} - 2t - 8 = 4.5 \Rightarrow t = 6.25$ therefore $k = 4(6.25)^{\frac{1}{2}}$. Dependent on both previous M marks. |
| $\Rightarrow \frac{k^2}{8} = 12.5 \Rightarrow k = 10$ | **A1** | AG - Allow verification. Any errors seen is A0. |

**Alternative method for last two marks of Question 6(a): Completing the square method**

| Answer | Marks | Guidance |
|--------|-------|----------|
| (Let $x = t^{\frac{1}{2}}$) $v = -2\left(x^2 - \frac{k}{2}x + 4\right) = -2\left(x^2 - \frac{k}{2}x + \frac{k^2}{16}\right) + \frac{k^2}{8} - 8$ | **(M1)** | Dependent on both previous M marks. |
| so $\frac{k^2}{8} - 8 = 12.5$, so $k = 10$ | **(A1)** | AG |
| | **5** | |

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## Question 6(b)(i):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $10 \times 1^{\frac{1}{2}} - 2 \times 1 - 8 = 0$ and $10 \times 16^{\frac{1}{2}} - 2 \times 16 - 8 = 0$. Or equate $v=0$ and solve for $t$ to get $t=1$ and $t=16$. | **B1** | AG. Any errors seen is B0. |
| | **1** | |

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## Question 6(b)(ii):

| Answer | Marks | Guidance |
|--------|-------|----------|
| For integration (do not penalise missing $c$) | **\*M1** | Increase power by 1 and a change in coefficient in at least one term (which must be the same term). |
| $\left(s =\right)\frac{10}{\frac{3}{2}}t^{\frac{3}{2}} - \frac{2t^2}{2} - 8t\ [+c] = \frac{20}{3}t^{\frac{3}{2}} - t^2 - 8t\ [+c]$ or $\left(s =\right)\frac{k}{\frac{3}{2}}t^{\frac{3}{2}} - \frac{2t^2}{2} - 8t\ [+c] = \frac{2k}{3}t^{\frac{3}{2}} - t^2 - 8t\ [+c]$ | **A1** | Allow unsimplified; allow with $k$ not substituted. |
| $\left(\frac{20}{3}1^{\frac{3}{2}} - 1^2 - 8\times 1\right) - 0 \left[= -\frac{7}{3}\right]$ and $\left(\frac{20}{3}16^{\frac{3}{2}} - 16^2 - 8\times 16\right) - \left(\frac{20}{3}1^{\frac{3}{2}} - 1^2 - 8\times 1\right) = \left[\frac{128}{3} - -\frac{7}{3} = 45\right]$ | **DM1** | Attempt at correct use of limits 0 and 1 or 1 and 16 – allow a single slip. |
| For both sets of limits applied correctly | **DM1** | Allow at most one slip in both. |
| $\frac{142}{3}$ m | **A1** | Allow 47.3 or better (47.3333…). **SC** if no integration shown: **B1** for $\pm\frac{7}{3}$ or 45, **B1** for $\frac{142}{3}$ (so 2 marks max). |
| | **5** | |

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