| Exam Board | OCR MEI |
|---|---|
| Module | AS Paper 1 (AS Paper 1) |
| Year | 2019 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable acceleration (1D) |
| Type | Finding constants from motion conditions |
| Difficulty | Moderate -0.3 This is a straightforward mechanics question requiring substitution to find a constant, use of v=u+at, and integration/comparison of displacements. All steps are routine applications of standard techniques with no conceptual challenges or novel problem-solving required, making it slightly easier than average. |
| Spec | 3.02d Constant acceleration: SUVAT formulae3.02f Non-uniform acceleration: using differentiation and integration |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Using given values in \(v = 1.2t^2 - kt^3\): \(25.6 = 1.2 \times 8^2 - k \times 8^3\) | M1 | 3.3: method must be clear; allow M1 for substitution of \(t=8, k=0.1\) to obtain \(v=25.6\) |
| \(k = \frac{25.6 - 76.8}{-8^3} = 0.1\) | A1 [2] | 2.1: AG; A1 for comment made that \(k=0.1\) is correct value |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Uses \(v = u + at\), \(u=0\), \(v=25.6\), \(t=8\) \(\Rightarrow 25.6 = 0 + 8a\) | M1 | 3.3 |
| \(a = 3.2 \text{ ms}^{-2}\) | A1 [2] | 1.1: cao |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(s = \int_0^8 (1.2t^2 - 0.1t^3)\,dt = \left[0.4t^3 - 0.025t^4\right]_0^8\) | M1 | 3.4: Attempt to integrate – expression for indefinite integral must be seen; allow SC1 for \(\int_0^8(1.2t^2-0.1t^3)\,dt = 102.4\) |
| \((0.4\times8^3 - 0.025\times8^4) - (0.4\times0 - 0.025\times0)\) | M1 | 1.1: Both limits seen or establishes that \(c=0\) if indefinite integral used |
| \(= 102.4\) m | A1 | 1.1: Allow following first M mark |
| Second model: use *suvat* with \(u=0\), \(v=25.6\), \(t=8\), \(a=\) their value: \(s = 0\times8 + \frac{1}{2}\times3.2\times8^2\) or \(s=\frac{1}{2}(0+25.6)8\) | M1 | 3.4: Method must be clear |
| \(= 102.4\) m | A1 [5] | 1.1: Allow without comment |
## Question 9(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Using given values in $v = 1.2t^2 - kt^3$: $25.6 = 1.2 \times 8^2 - k \times 8^3$ | M1 | 3.3: method must be clear; allow M1 for substitution of $t=8, k=0.1$ to obtain $v=25.6$ |
| $k = \frac{25.6 - 76.8}{-8^3} = 0.1$ | A1 [2] | 2.1: AG; A1 for comment made that $k=0.1$ is correct value |
---
## Question 9(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Uses $v = u + at$, $u=0$, $v=25.6$, $t=8$ $\Rightarrow 25.6 = 0 + 8a$ | M1 | 3.3 |
| $a = 3.2 \text{ ms}^{-2}$ | A1 [2] | 1.1: cao |
---
## Question 9(c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $s = \int_0^8 (1.2t^2 - 0.1t^3)\,dt = \left[0.4t^3 - 0.025t^4\right]_0^8$ | M1 | 3.4: Attempt to integrate – expression for indefinite integral must be seen; allow SC1 for $\int_0^8(1.2t^2-0.1t^3)\,dt = 102.4$ |
| $(0.4\times8^3 - 0.025\times8^4) - (0.4\times0 - 0.025\times0)$ | M1 | 1.1: Both limits seen or establishes that $c=0$ if indefinite integral used |
| $= 102.4$ m | A1 | 1.1: Allow following first M mark |
| **Second model:** use *suvat* with $u=0$, $v=25.6$, $t=8$, $a=$ their value: $s = 0\times8 + \frac{1}{2}\times3.2\times8^2$ or $s=\frac{1}{2}(0+25.6)8$ | M1 | 3.4: Method must be clear |
| $= 102.4$ m | A1 [5] | 1.1: Allow without comment |
---9 In this question you must show detailed reasoning.
A car accelerates from rest along a straight level road. The velocity of the car after 8 s is $25.6 \mathrm {~ms} ^ { - 1 }$.\\
In one model for the motion, the velocity $v \mathrm {~ms} ^ { - 1 }$ at time $t$ seconds is given by $v = 1.2 t ^ { 2 } - k t ^ { 3 }$, where $k$ is a constant and $0 \leqslant t \leqslant 8$.
\begin{enumerate}[label=(\alph*)]
\item The model gives the correct velocity of $25.6 \mathrm {~ms} ^ { - 1 }$ at time 8 s . Show that $k = 0.1$.
A second model for the motion uses constant acceleration.
\item Find the value of the acceleration which gives the correct velocity of $25.6 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ at time 8 s .
\item Show that these two models give the same value for the displacement in the first 8 s .
\end{enumerate}
\hfill \mbox{\textit{OCR MEI AS Paper 1 2019 Q9 [9]}}