| Exam Board | OCR MEI |
|---|---|
| Module | AS Paper 1 (AS Paper 1) |
| Year | 2020 |
| Session | November |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Travel graphs |
| Type | Velocity-time graph sketching |
| Difficulty | Moderate -0.3 This is a straightforward SUVAT problem requiring a standard velocity-time graph sketch and solving two simultaneous equations from area (distance) and gradient (acceleration). The setup is clear, the methods are routine, and it's slightly easier than average due to being a standard textbook-style question with no conceptual surprises. |
| Spec | 3.02b Kinematic graphs: displacement-time and velocity-time3.02c Interpret kinematic graphs: gradient and area3.02d Constant acceleration: SUVAT formulae |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Two line segments starting and ending on the \(x\)-axis | B1 | Two line segments starting and ending on the \(x\)-axis |
| \(T\) and \(3T\) seen on \(t\)-axis | B1 [2] | Instead of \(3T\), allow \(2T\) shown on horizontal axis for second phase |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Acceleration phase \(\frac{V}{T} = 2.5\) | B1 | soi |
| Area under graph \(\frac{1}{2} \times V \times 3T = 240\) | B1 | May be sum of two areas soi; \(\frac{1}{2}VT + VT = 240\) |
| Solving simultaneously: \(\frac{3}{2}(2.5T)T = 240 \Rightarrow T^2 = 64\) | M1 | Attempt to eliminate one variable |
| \(T = 8\) and \(V = 20\) | A1 [4] | Correct pair of answers |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(V = 2.5 \times T\) | B1 | May be final step using \(T=8\), award if seen |
| First phase has distance \(240 \div 3 = 80\) m | M1 | |
| Using \(s = 80\), \(u = 0\), \(a = 2.5\): \(80 = \frac{1}{2} \times 2.5T^2\) | B1 | Using suvat equation(s) leading to a value for \(t\) with \(s = 80\) |
| \(T = 8\) and \(V = 20\) | A1 | Correct pair of answers |
## Question 9:
**Part (a):**
| Answer | Marks | Guidance |
|--------|-------|----------|
| Two line segments starting and ending on the $x$-axis | B1 | Two line segments starting and ending on the $x$-axis |
| $T$ and $3T$ seen on $t$-axis | B1 [2] | Instead of $3T$, allow $2T$ shown on horizontal axis for second phase |
**Part (b):**
**EITHER method:**
| Answer | Marks | Guidance |
|--------|-------|----------|
| Acceleration phase $\frac{V}{T} = 2.5$ | B1 | soi |
| Area under graph $\frac{1}{2} \times V \times 3T = 240$ | B1 | May be sum of two areas soi; $\frac{1}{2}VT + VT = 240$ |
| Solving simultaneously: $\frac{3}{2}(2.5T)T = 240 \Rightarrow T^2 = 64$ | M1 | Attempt to eliminate one variable |
| $T = 8$ and $V = 20$ | A1 [4] | Correct pair of answers |
**OR method:**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $V = 2.5 \times T$ | B1 | May be final step using $T=8$, award if seen |
| First phase has distance $240 \div 3 = 80$ m | M1 | |
| Using $s = 80$, $u = 0$, $a = 2.5$: $80 = \frac{1}{2} \times 2.5T^2$ | B1 | Using suvat equation(s) leading to a value for $t$ with $s = 80$ |
| $T = 8$ and $V = 20$ | A1 | Correct pair of answers |
---9 A car travelling in a straight line accelerates uniformly from rest to $V \mathrm {~ms} ^ { - 1 }$ in $T \mathrm {~s}$. It then slows down uniformly, coming to rest after a further $2 T$ s.
\begin{enumerate}[label=(\alph*)]
\item Sketch a velocity-time graph for the car.
The acceleration in the first stage of the motion is $2.5 \mathrm {~ms} ^ { - 2 }$ and the total distance travelled is 240 m .
\item Calculate the values of $V$ and $T$.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI AS Paper 1 2020 Q9 [6]}}