| Exam Board | OCR MEI |
|---|---|
| Module | AS Paper 1 (AS Paper 1) |
| Year | 2024 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable acceleration (1D) |
| Type | Total distance with direction changes |
| Difficulty | Moderate -0.3 This is a straightforward kinematics question requiring standard techniques: solving a quadratic equation for v=0, differentiating to find acceleration, and integrating velocity (with attention to sign changes) to find distance. While part (c) requires care with the direction change, these are all routine A-level mechanics procedures with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.08e Area between curve and x-axis: using definite integrals3.02f Non-uniform acceleration: using differentiation and integration |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(v = 0.6t^2 - 2.1t + 1.5 = 0\) | M1 | Form quadratic equation and attempt to solve by any method |
| So stationary when \(t = 1, 2.5\) | A1 [2] | cao. May be implied by correct times |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(a = 1.2t - 2.1\) | M1 | Attempt to differentiate \(v\). Do not allow if \(0.6t\) seen |
| When \(t = 1\), \(a = -0.9\) m s\(^{-2}\) | A1FT [2] | FT their (a) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\int_1^{2.5}(0.6t^2 - 2.1t + 1.5)\,\mathrm{d}t = \left[-\frac{27}{80}\right]\) | M1 | Attempt to evaluate definite integral – must be seen but may be evaluated BC. FT their limits for M mark only |
| So distance is \(\frac{27}{80} = 0.3375\) m | A1 [2] | Do not allow if given as the value of the definite integral |
## Question 7(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $v = 0.6t^2 - 2.1t + 1.5 = 0$ | M1 | Form quadratic equation and attempt to solve by any method |
| So stationary when $t = 1, 2.5$ | A1 [2] | cao. May be implied by correct times |
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## Question 7(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $a = 1.2t - 2.1$ | M1 | Attempt to differentiate $v$. Do not allow if $0.6t$ seen |
| When $t = 1$, $a = -0.9$ m s$^{-2}$ | A1FT [2] | FT their (a) |
---
## Question 7(c):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\int_1^{2.5}(0.6t^2 - 2.1t + 1.5)\,\mathrm{d}t = \left[-\frac{27}{80}\right]$ | M1 | Attempt to evaluate definite integral – must be seen but may be evaluated BC. FT their limits for M mark only |
| So distance is $\frac{27}{80} = 0.3375$ m | A1 [2] | Do not allow if given as the value of the definite integral |7 The velocity of a particle moving in a straight line is modelled by $\mathbf { v } = 0.6 \mathbf { t } ^ { 2 } - 2.1 \mathbf { t } + 1.5$ where $v$ is the velocity in metres per second and $t$ is the time in seconds.
\begin{enumerate}[label=(\alph*)]
\item Determine the times at which the particle is stationary.
\item Find the acceleration of the particle at the first of the times at which it is stationary.
\item Find the distance travelled by the particle between the times at which it is stationary.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI AS Paper 1 2024 Q7 [6]}}