| Exam Board | OCR MEI |
|---|---|
| Module | AS Paper 1 (AS Paper 1) |
| Year | 2019 |
| Session | June |
| Marks | 7 |
| 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 question requiring basic algebraic manipulation and SUVAT application. Part (a) is 'show that' with the answer given, (b) uses v=u+at directly, and (c) involves comparing two simple displacement calculations. All steps are routine with no problem-solving insight required. |
| Spec | 3.02d Constant acceleration: SUVAT formulae3.02f Non-uniform acceleration: using differentiation and integration |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(y = x^2 - 4x^{\frac{1}{2}} \Rightarrow \frac{dy}{dx} = 2x - 4 \times \frac{1}{2}x^{-\frac{1}{2}}\) | M1 | 2.1: Uses fractional power in an attempt to differentiate |
| \(\frac{dy}{dx} = 0\) gives \(x^{\frac{3}{2}} = 1 \Rightarrow x = 1\) | M1 | 3.1a: Attempt to solve their \(\frac{dy}{dx} = 0\); allow SC1 for verifying \(x=1\) gives \(\frac{dy}{dx}=0\) |
| There is only one solution so there is only one stationary point | A1 | 2.2a: Must obtain \(x=1\) from correct working and indicate this is the only stationary point |
| When \(x=1\), \(y = 1 - 4 = -3\) | B1 | 2.1: From correct working seen (AG) |
| \(\frac{d^2y}{dx^2} = 2 - 2\left(-\frac{1}{2}\right)x^{-\frac{3}{2}}\) | M1 | 1.1a: Attempt to find second derivative |
| When \(x=1\), \(\frac{d^2y}{dx^2} = 3 > 0\) | M1 | 1.1a: Substituting into their expression |
| So the stationary point is a minimum point | A1 [7] | 2.2a: Conclusion from correct working (AG) |
| Alternative for final three marks: Attempt to evaluate \(\frac{dy}{dx}\) at a point \(x \neq 1\) | M1 | As there is only one stationary point, allow for similarly evaluating \(y\) and comparing with \(-3\) |
| Attempt to evaluate \(\frac{dy}{dx}\) at a point the other side of \(x=1\) | M1 | |
| Correct conclusion from correct values | A1 |
## Question 8:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $y = x^2 - 4x^{\frac{1}{2}} \Rightarrow \frac{dy}{dx} = 2x - 4 \times \frac{1}{2}x^{-\frac{1}{2}}$ | M1 | 2.1: Uses fractional power in an attempt to differentiate |
| $\frac{dy}{dx} = 0$ gives $x^{\frac{3}{2}} = 1 \Rightarrow x = 1$ | M1 | 3.1a: Attempt to solve their $\frac{dy}{dx} = 0$; allow SC1 for verifying $x=1$ gives $\frac{dy}{dx}=0$ |
| There is only one solution so there is only one stationary point | A1 | 2.2a: Must obtain $x=1$ from correct working and indicate this is the only stationary point |
| When $x=1$, $y = 1 - 4 = -3$ | B1 | 2.1: From correct working seen (AG) |
| $\frac{d^2y}{dx^2} = 2 - 2\left(-\frac{1}{2}\right)x^{-\frac{3}{2}}$ | M1 | 1.1a: Attempt to find second derivative |
| When $x=1$, $\frac{d^2y}{dx^2} = 3 > 0$ | M1 | 1.1a: Substituting into their expression |
| So the stationary point is a minimum point | A1 [7] | 2.2a: Conclusion from correct working (AG) |
| **Alternative for final three marks:** Attempt to evaluate $\frac{dy}{dx}$ at a point $x \neq 1$ | M1 | As there is only one stationary point, allow for similarly evaluating $y$ and comparing with $-3$ |
| Attempt to evaluate $\frac{dy}{dx}$ at a point the other side of $x=1$ | M1 | |
| Correct conclusion from correct values | A1 | |
---\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 Q8 [7]}}