| Exam Board | OCR MEI |
|---|---|
| Module | AS Paper 1 (AS Paper 1) |
| Year | 2022 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Travel graphs |
| Type | Model refinement or criticism |
| Difficulty | Moderate -0.8 This is a straightforward multi-part SUVAT question requiring only standard techniques: constant acceleration formula (part a), substituting into an equation to find k (part b), differentiating to find acceleration (part c), and integrating to compare distances (part d). All steps are routine applications of AS-level mechanics and calculus with no problem-solving insight required. |
| Spec | 1.08f Area between two curves: using integration3.02d Constant acceleration: SUVAT formulae |
| Answer | Marks | Guidance |
|---|---|---|
| Using \(s = \frac{1}{2}(u+v)t\) with \(u=0,\ v=9,\ t=5\) | M1 | Allow for any sequence of suvat equations leading to a value for \(s\) (\(a = 1.8\) may be seen) |
| \(s = \frac{1}{2}(0+9)5 = 22.5\ \text{m}\) | A1* |
| Answer | Marks | Guidance |
|---|---|---|
| \(9 = 0.05 \times 5^3 + 5k\) | M1 | Uses \(t=5,\ v=9\) to form an equation for \(k\) |
| \(k = 0.55\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(a = \frac{\mathrm{d}v}{\mathrm{d}t} = 0.15t^2 + k\) | M1 | Differentiation of their model B |
| When \(t=5,\ a = 0.15 \times 5^2 + k = 4.3\ \text{ms}^{-2}\) | A1 | FT their \(k\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(\int_0^5 v\, \mathrm{d}t = \int_0^5 (0.05t^3 + 0.55t)\, \mathrm{d}t\) | M1 | Definite integral; Do not allow \(s = 0.05\frac{t^4}{4} + 0.55\frac{t^2}{2} + c\) unless limits used or attempt to evaluate \(c\) |
| Distance \(14.6875\ \text{m}\) \((14.7\ \text{m})\) | A1* | FT their \(k\); BC acceptable |
| Model B distance 14.7 m is closer to 16 m than model A (22.5 m) so B models this better | E1 (dep) | Comment which quotes at least two values. Dependent on both A marks indicated * |
# Question 11:
## Part (a):
Using $s = \frac{1}{2}(u+v)t$ with $u=0,\ v=9,\ t=5$ | **M1** | Allow for any sequence of suvat equations leading to a value for $s$ ($a = 1.8$ may be seen)
$s = \frac{1}{2}(0+9)5 = 22.5\ \text{m}$ | **A1*** |
## Part (b):
Substituting $t=5,\ v=9$ into $v = 0.05t^3 + kt$
$9 = 0.05 \times 5^3 + 5k$ | **M1** | Uses $t=5,\ v=9$ to form an equation for $k$
$k = 0.55$ | **A1** |
## Part (c):
$a = \frac{\mathrm{d}v}{\mathrm{d}t} = 0.15t^2 + k$ | **M1** | Differentiation of their model B
When $t=5,\ a = 0.15 \times 5^2 + k = 4.3\ \text{ms}^{-2}$ | **A1** | FT their $k$
## Part (d):
$\int_0^5 v\, \mathrm{d}t = \int_0^5 (0.05t^3 + 0.55t)\, \mathrm{d}t$ | **M1** | Definite integral; Do not allow $s = 0.05\frac{t^4}{4} + 0.55\frac{t^2}{2} + c$ unless limits used or attempt to evaluate $c$
Distance $14.6875\ \text{m}$ $(14.7\ \text{m})$ | **A1*** | FT their $k$; BC acceptable
Model B distance 14.7 m is closer to 16 m than model A (22.5 m) so B models this better | **E1 (dep)** | Comment which quotes at least two values. Dependent on both A marks indicated *
---11 A sports car accelerates along a straight road from rest. After 5 s its velocity is $9 \mathrm {~ms} ^ { - 1 }$.
In model A, the acceleration is assumed to be constant.
\begin{enumerate}[label=(\alph*)]
\item Calculate the distance travelled by the car in the first 5 seconds according to model A .
In model B , the velocity $v$ in $\mathrm { ms } ^ { - 1 }$ is given by $\mathrm { v } = 0.05 \mathrm { t } ^ { 3 } + \mathrm { kt }$, where $t$ is the time in seconds after the start and $k$ is a constant.
\item Find the value of $k$ which gives the correct value of $v$ when $t = 5$.
\item Using this value of $k$ in model B , calculate the acceleration of the car when $t = 5$.
The car travels 16 m in the first 5 seconds.
\item Show that model B, with the value of $k$ found in part (b), better fits this information than model A does.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI AS Paper 1 2022 Q11 [9]}}