| Exam Board | OCR MEI |
|---|---|
| Module | M1 (Mechanics 1) |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable acceleration (1D) |
| Type | Displacement from velocity by integration |
| Difficulty | Moderate -0.8 This is a straightforward mechanics question requiring basic differentiation of a quadratic to find acceleration, interpretation of zero acceleration, and integration of a polynomial for distance. All techniques are routine A-level procedures with no problem-solving insight needed, making it easier than average but not trivial due to the multi-part structure. |
| Spec | 1.07i Differentiate x^n: for rational n and sums1.08d Evaluate definite integrals: between limits3.02b Kinematic graphs: displacement-time and velocity-time3.02c Interpret kinematic graphs: gradient and area3.02f Non-uniform acceleration: using differentiation and integration |
| Answer | Marks |
|---|---|
| \(4\ \text{m}\) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(12-(-4) = 16\ \text{m}\) | M1, A1 | Looking for distance; need evidence of taking account of \(+\)ve and \(-\)ve displacements |
| Answer | Marks | Guidance |
|---|---|---|
| \(1 < t < 3.5\) | B1, B1 | The values 1 and 3.5; strict inequality |
| Answer | Marks | Guidance |
|---|---|---|
| \(t=1,\ t=3.5\) | B1 | Do not award if extra values given |
| Answer | Marks | Guidance |
|---|---|---|
| \(v = -8t + 8\) | M1, A1 | Differentiating |
| \(a = -8\) | F1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(8t + 8 = 4\) so \(t = 0.5\), so \(0.5\ \text{s}\) | B1 | FT their \(v\) |
| \(-8t + 8 = -4\) so \(t = 1.5\), so \(1.5\ \text{s}\) | B1 | FT their \(v\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(v(3) = -8\times3+8 = -16\) | B1 | FT their \(v\) from (ii) |
| \(v = \int 32\, dt = 32t + C\) | M1 | Accept \(32t+C\) or \(32t\); SC1 if \(\int_3^4 32\,dt\) attempted |
| \(v=-16\) when \(t=3\) gives \(v = 32t - 112\) | A1 | Use of their \(-16\) from attempt at \(v\) when \(t=3\) |
| \(y = \int(32t-112)\,dt = 16t^2 - 112t + D\) | M1 | FT their \(v\) of form \(pt+q\) with \(p\neq0\), \(q\neq0\); accept if at least 1 term correct |
| \(y=0\) when \(t=3\) gives \(y = 16t^2 - 112t + 192\) | A1 | cao |
| Or: \(y = -16(t-3)+\frac{1}{2}\times32\times(t-3)^2\) | M1, A1, M1, A1 | Use of \(s=ut+\frac{1}{2}at^2\); use of their \(-16\); condone use of just \(t\); use of \(t\pm3\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(y = k(t-3)(t-4)\) | M1, A1 | Use of quadratic function |
| \(k\) present | B1 | |
| When \(t=3.5\), \(y=-4\): \(-4 = k\times\frac{1}{2}\times(-\frac{1}{2})\), so \(k=16\) | M1, A1 | Or consider velocity at \(t=3\); cao; accept \(k\) without \(y\) simplified |
## Question 5:
### Part (i)(A)
$4\ \text{m}$ | B1 |
### Part (i)(B)
$12-(-4) = 16\ \text{m}$ | M1, A1 | Looking for distance; need evidence of taking account of $+$ve and $-$ve displacements
### Part (i)(C)
$1 < t < 3.5$ | B1, B1 | The values 1 and 3.5; strict inequality
### Part (i)(D)
$t=1,\ t=3.5$ | B1 | Do not award if extra values given
### Part (ii)
$v = -8t + 8$ | M1, A1 | Differentiating
$a = -8$ | F1 |
### Part (iii)
$8t + 8 = 4$ so $t = 0.5$, so $0.5\ \text{s}$ | B1 | FT their $v$
$-8t + 8 = -4$ so $t = 1.5$, so $1.5\ \text{s}$ | B1 | FT their $v$
### Part (iv)
**Method 1:**
$v(3) = -8\times3+8 = -16$ | B1 | FT their $v$ from (ii)
$v = \int 32\, dt = 32t + C$ | M1 | Accept $32t+C$ or $32t$; SC1 if $\int_3^4 32\,dt$ attempted
$v=-16$ when $t=3$ gives $v = 32t - 112$ | A1 | Use of their $-16$ from attempt at $v$ when $t=3$
$y = \int(32t-112)\,dt = 16t^2 - 112t + D$ | M1 | FT their $v$ of form $pt+q$ with $p\neq0$, $q\neq0$; accept if at least 1 term correct
$y=0$ when $t=3$ gives $y = 16t^2 - 112t + 192$ | A1 | cao
**Or:** $y = -16(t-3)+\frac{1}{2}\times32\times(t-3)^2$ | M1, A1, M1, A1 | Use of $s=ut+\frac{1}{2}at^2$; use of their $-16$; condone use of just $t$; use of $t\pm3$
(so $y = 16t^2 - 112t + 192$)
**Method 2:**
Since acceleration is constant, $y$ is quadratic; $y=0$ at $t=3$ and $t=4$:
$y = k(t-3)(t-4)$ | M1, A1 | Use of quadratic function
$k$ present | B1 |
When $t=3.5$, $y=-4$: $-4 = k\times\frac{1}{2}\times(-\frac{1}{2})$, so $k=16$ | M1, A1 | Or consider velocity at $t=3$; cao; accept $k$ without $y$ simplified
(so $y = 16t^2 - 112t + 192$)5 Fig. 3 is a sketch of the velocity-time graph modelling the velocity of a sprinter at the start of a race.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{569e7c0e-7c33-47c9-b986-8587ea239f0a-5_575_1086_482_551}
\captionsetup{labelformat=empty}
\caption{Fig. 3}
\end{center}
\end{figure}
(i) How can you tell from the sketch that the acceleration is not modelled as being constant for $0 \leqslant t \leqslant 4$ ?
The velocity of the sprinter, $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$, for the time interval $0 \leqslant t \leqslant 4$ is modelled by the expression
$$v = 3 t - \frac { 3 } { 8 } t ^ { 2 } .$$
(ii) Find the acceleration that the model predicts for $t = 4$ and comment on what this suggests about the running of the sprinter.\\
(iii) Calculate the distance run by the sprinter from $t = 1$ to $t = 4$.
\hfill \mbox{\textit{OCR MEI M1 Q5 [8]}}