| Exam Board | OCR MEI |
|---|---|
| Module | AS Paper 1 (AS Paper 1) |
| Year | 2021 |
| Session | November |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Travel graphs |
| Type | Velocity from displacement function using calculus |
| Difficulty | Moderate -0.3 This is a straightforward multi-part question on displacement-time graphs requiring standard techniques: understanding displacement vs distance (conceptual but commonly taught), differentiating a quadratic for velocity, and substituting values. All parts are routine AS-level mechanics with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.07i Differentiate x^n: for rational n and sums1.08d Evaluate definite integrals: between limits3.02a Kinematics language: position, displacement, velocity, acceleration3.02b Kinematic graphs: displacement-time and velocity-time |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \((2, 11)\) on the graph tells Kai that the displacement from the origin at that time is \(11\) m and not the distance travelled. [The particle starts \(7\) m from the origin, so actual distance travelled is \(11 - 7 = 4\) m] | E1 [1] | Draws a clear distinction between displacement from the origin at \(t = 2\) and the distance travelled. Allow for a statement that the displacement during the first 2 s is 4 m |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| When \(t = 10\), \(s = -53\) | M1 | Evaluating displacement or distance from the origin at \(t = 10\). May be implied by substitution seen or by \(-53\) seen |
| Distance travelled is \(53 + 11 + 4 = 68\) m | M1 | Adds at least one distance to their 53m |
| \(= 68\) m | A1 [3] | — |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(v = \frac{ds}{dt} = 4 - 2t\) | M1, A1 [2] | Attempt to differentiate |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| When \(t = 10\), \(v = 4 - 2 \times 10\ [-16]\), therefore the speed is \(16\ \text{m s}^{-1}\) | M1, A1 [2] | Substitution of \(t = 10\). Allow for speed \(= 16\ \text{m s}^{-1}\) seen. Allow M1A0 for speed \(-16\ \text{m s}^{-1}\) seen or for \(4 - 2 \times 10 = 16\) |
# Question 6(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $(2, 11)$ on the graph tells Kai that the displacement from the origin at that time is $11$ m and not the distance travelled. [The particle starts $7$ m from the origin, so actual distance travelled is $11 - 7 = 4$ m] | E1 [1] | Draws a clear distinction between displacement from the origin at $t = 2$ and the distance travelled. Allow for a statement that the displacement during the first 2 s is 4 m |
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# Question 6(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| When $t = 10$, $s = -53$ | M1 | Evaluating displacement or distance from the origin at $t = 10$. May be implied by substitution seen or by $-53$ seen |
| Distance travelled is $53 + 11 + 4 = 68$ m | M1 | Adds at least one distance to their 53m |
| $= 68$ m | A1 [3] | — |
---
# Question 6(c):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $v = \frac{ds}{dt} = 4 - 2t$ | M1, A1 [2] | Attempt to differentiate |
---
# Question 6(d):
| Answer | Marks | Guidance |
|--------|-------|----------|
| When $t = 10$, $v = 4 - 2 \times 10\ [-16]$, therefore the speed is $16\ \text{m s}^{-1}$ | M1, A1 [2] | Substitution of $t = 10$. Allow for speed $= 16\ \text{m s}^{-1}$ seen. Allow M1A0 for speed $-16\ \text{m s}^{-1}$ seen or for $4 - 2 \times 10 = 16$ |
---6 The displacement of a particle is modelled by the equation $\mathrm { s } = 7 + 4 \mathrm { t } - \mathrm { t } ^ { 2 }$, where $s$ metres is the displacement from the origin at time $t$ seconds. The diagram shows part of the displacement-time graph for the particle. The point $( 2,11 )$ is the maximum point on the graph.\\
\includegraphics[max width=\textwidth, alt={}, center]{5428eabf-431d-4db1-8c25-1f2b9570d9aa-4_513_1381_422_255}
\begin{enumerate}[label=(\alph*)]
\item Kai argues that the point $( 2,11 )$ is on the graph, so the particle has travelled a distance of 11 metres in the first 2 seconds.
Comment on the validity of Kai's argument.
\item Determine the total distance the particle travels in the first 10 seconds.
\item Find an expression for the velocity of the particle at time $t$.
\item Find the speed of the particle when $t = 10$.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI AS Paper 1 2021 Q6 [8]}}