# Question 14:

## Part (i):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $u = 5,\ v = 11.4,\ t = 4$; $a = \frac{v-u}{t} = \frac{11.4 - 5}{4} = 1.6$ | **M1** (AO 3.1b) | Using suvat equation(s) leading to value for $a$ |
| $a = 1.6$ | **A1** (AO 1.1b) | Any form |
| $v = 5 + 1.6t$ | **A1** [3] (AO 3.3) | FT their $a$ |

## Part (ii):

| Answer | Marks | Guidance |
|--------|-------|----------|
| The car would not be able to accelerate indefinitely — the velocity would become too large | **E1** [1] (AO 3.5b) | |

## Part (iii):

| Answer | Marks | Guidance |
|--------|-------|----------|
| When $v = 17.8$: $t = \frac{17.8 - 5}{1.6} = 8$ | **B1** (AO 1.1a) | Calculation or point on graph labelled at $t = 8$ |
| Graph: two line segments with one horizontal | **G1** (AO 1.1a) | Two line segments with one horizontal |
| Axes labelled; $(0, 5)$ and constant speed $17.8$ clear on vertical scale | **G1** [3] (AO 3.5c) | Mark intent for $17.8$ — allow for a linear scale beyond $17.8$ |

## Part (iv):

| Answer | Marks | Guidance |
|--------|-------|----------|
| Dividing area into sections; Area under trapezium $= \frac{1}{2}(5 + 17.8) \times 8 = 91.2$ | **M1** (AO 3.1b) | FT their graph if linear for M1 A0 for a triangle or trapezium area |
| $91.2$ | **A1** (AO 1.1a) | May be found as sum of areas. May be implied by correct total |
| Area rectangle $12 \times 17.8 = 213.6$; Total displacement $= 304.8$ m | **A1** [3] (AO 1.1b) | FT their distance found for first 8s. $213.6$ must be added to another distance |

## Part (v):

| Answer | Marks | Guidance |
|--------|-------|----------|
| When $t = 4$: $v = 5 + 0.3 \times 4^2 - 0.05 \times 4^3 = 11.4\ \text{ms}^{-1}$, which matches the given value | **B1** [1] (AO 3.4) | Allow without comment |

## Part (vi):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{dv}{dt} = 0.6 \times 2t - 0.05 \times 3t^2\ \left[= 1.2t - 0.15t^2\right]$ | **M1** (AO 1.1a) | Need not be simplified |
| When $t = 8$: $v = 1.2 \times 8 - 0.15 \times 64 = 0$; Acceleration is zero at $t = 8$ | **A1** (AO 3.2a) | Must mention acceleration |
| Which means that the car reaches its maximum speed without the sudden change in acceleration in model A | **E1** [3] (AO 3.2a) | Must compare with model A |

## Part (vii):

| Answer | Marks | Guidance |
|--------|-------|----------|
| **EITHER:** $\int_0^8 (5 + 0.6t^2 - 0.05t^3)\, dt = \left[5t + 0.2t^3 - 0.0125t^4\right]_0^8 = 91.2$ m | **M1** (AO 2.1) | Allow for correct definite integral stated and calculator used |
| $= 91.2$ m | **A1** (AO 1.1b) **BC** | Also allow M1A1 for $5\times8 - 0.2\times8^3 - 0.0125\times8^4$ seen |
| Which is same as model A for the first 8s; Distance is the same for the remainder of the time; So this is the same as model A at $t = 20$ | **E1** [3] (AO 2.1) | Must consider to $t = 20$ |
| **OR:** $\int_0^8 (5 + 0.6t^2 - 0.05t^3)\, dt = \left[5t + 0.2t^3 - 0.0125t^4\right]_0^8 = 91.2$ m | **M1** | Allow for correct definite integral stated and calculator used |
| $= 91.2$ m; Distance at $17.8\ \text{ms}^{-1}$: $213.6$; Total distance $304.8$ m [which is the same as model A] | **A1**, **A1** | Must consider to $t = 20$ |