## Question 4:

### Part (i)

| Answer/Working | Mark | Guidance |
|---|---|---|
| (A) $4$ m | B1 | |
| (B) $12 - (-4) = 16$ m | M1 | Looking for distance. Need evidence of taking account of $+$ve and $-$ve displacements |
| $16$ m (correct) | A1 | |
| (C) $1 < t < 3.5$ | B1 | The values 1 and 3.5 |
| $1 < t < 3.5$ (strict inequality) | B1 | Strict inequality |
| (D) $t = 1,\ t = 3.5$ | B1 | Do not award if extra values given |

**Subtotal: 6 marks**

### Part (ii)

| Answer/Working | Mark | Guidance |
|---|---|---|
| $v = -8t + 8$ | M1 | Differentiating |
| $v = -8t + 8$ (correct) | A1 | |
| $a = -8$ | F1 | |

**Subtotal: 3 marks**

### Part (iii)

| Answer/Working | Mark | Guidance |
|---|---|---|
| $8t + 8 = 4$ so $t = 0.5$, so $0.5$ s | B1 | FT **their** $v$ |
| $-8t + 8 = -4$ so $t = 1.5$, so $1.5$ s | B1 | FT **their** $v$ |

**Subtotal: 2 marks**

### Part (iv)

**Method 1:**

| Answer/Working | Mark | Guidance |
|---|---|---|
| $v(3) = -8 \times 3 + 8 = -16$ | B1 | FT **their** $v$ from (ii) |
| $v = \int 32\,dt = 32t + C$ | M1 | Accept $32t + C$ or $32t$. SC1 if $\int_3^4 32\,dt$ attempted |
| $v = -16$ when $t=3$ gives $v = 32t - 112$ | A1 | Use of **their** $-16$ from an attempt at $v$ when $t=3$ |
| $y = \int(32t - 112)\,dt = 16t^2 - 112t + D$ | M1 | FT **their** $v$ of the form $pt + q$ with $p \neq 0$ and $q \neq 0$. Accept if at least 1 term correct. Accept no $D$ |
| $y = 0$ when $t = 3$ gives $y = 16t^2 - 112t + 192$ | A1 | cao |

**Method 1 (alternative $s = ut + \frac{1}{2}at^2$):**

| Answer/Working | Mark | Guidance |
|---|---|---|
| $y = -16(t-3) + \frac{1}{2} \times 32 \times (t-3)^2$ | M1 | Use of $s = ut + \frac{1}{2}at^2$ |
| | A1 | Use of **their** $-16$ (not 0) from attempt at $v$ when $t=3$ and 32. Condone use of just $t$ |
| | M1 | Use of $t \pm 3$ |
| (so $y = 16t^2 - 112t + 192$) | A1 | cao |

**Method 2:**

| Answer/Working | Mark | Guidance |
|---|---|---|
| $y = k(t-3)(t-4)$ | M1 | Use of a quadratic function (condone no $k$) |
| | A1 | Correct use of roots |
| | B1 | $k$ present |
| When $t = 3.5,\ y = -4$, so $-4 = k \times \frac{1}{2} \times (-\frac{1}{2})$ | M1 | Or consider velocity at $t = 3$ |
| so $k = 16$ (and $y = 16t^2 - 112t + 192$) | A1 | cao. Accept $k$ without $y$ simplified |

**Subtotal: 5 marks**

**Total: 16 marks**

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