| Exam Board | OCR MEI |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2010 |
| Session | June |
| Marks | 16 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Constant acceleration (SUVAT) |
| Type | Read and interpret velocity-time graph |
| Difficulty | Moderate -0.3 This is a multi-part mechanics question requiring standard techniques: reading graphs, differentiating a quadratic to find velocity/acceleration, solving simple equations, and using SUVAT with integration. All parts are routine applications with no novel problem-solving required, making it slightly easier than average for A-level. |
| Spec | 1.07b Gradient as rate of change: dy/dx notation1.08a Fundamental theorem of calculus: integration as reverse of differentiation3.02b Kinematic graphs: displacement-time and velocity-time3.02c Interpret kinematic graphs: gradient and area |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| (A) \(4 \text{ m}\) | B1 | |
| (B) \(12-(-4) = 16 \text{ m}\) | M1 | Looking for distance. Need evidence of taking account of +ve and \(-\)ve displacements. |
| A1 | ||
| (C) \(1 < t < 3.5\) | B1 | The values 1 and 3.5 |
| B1 | Strict inequality | |
| (D) \(t = 1\), \(t = 3.5\) | B1 | Do not award if extra values given. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(v = -8t + 8\) | M1 | Differentiating |
| A1 | ||
| \(a = -8\) | F1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | 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 |
|---|---|---|
| Answer/Working | Mark | Guidance |
| method 1: Need velocity at \(t = 3\): \(v(3) = -8\times3+8 = -16\) | B1 | FT their \(v\) from (ii) |
| \(v = \int 32 \, dt = 32t + C\) | M1 | Accept \(32t + C\) or \(32t\). SC1 if \(\int_3^4 32dt\) attempted. |
| \(v = -16\) when \(t=3\) gives \(v = 32t - 112\) | A1 | Use of their \(-16\) from an attempt at \(v\) when \(t=3\) |
| \(y = \int(32t - 112)dt = 16t^2 - 112t + D\) | M1 | FT their \(v\) of the form \(pt + q\) with \(p \neq 0\) and \(q \neq 0\). Accept if at least 1 term correct. Accept no \(D\). |
| \(y = 0\) 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 | Use of \(s = ut + \frac{1}{2}at^2\) |
| A1 | Use of their \(-16\) (not 0) from an attempt at \(v\) when \(t=3\) and 32. Condone use of just \(t\) | |
| M1 | Use of \(t \pm 3\) | |
| (so \(y = 16t^2 - 112t + 192\)) | A1 | cao |
| method 2: \(y = k(t-3)(t-4)\) | M1 | Use of a quadratic function (condone no \(k\)) |
| A1 | Correct use of roots | |
| B1 | \(k\) present | |
| When \(t = 3.5\), \(y = -4\): so \(-4 = k\times\frac{1}{2}\times-\frac{1}{2}\) | M1 | Or consider velocity at \(t = 3\) |
| so \(k = 16\) (and \(y = 16t^2 - 112t + 192\)) | A1 | cao. Accept \(k\) without \(y\) simplified. |
# Question 7:
## Part (i)
| Answer/Working | Mark | Guidance |
|---|---|---|
| (A) $4 \text{ m}$ | B1 | |
| (B) $12-(-4) = 16 \text{ m}$ | M1 | Looking for distance. Need evidence of taking account of +ve and $-$ve displacements. |
| | A1 | |
| (C) $1 < t < 3.5$ | B1 | The values 1 and 3.5 |
| | B1 | Strict inequality |
| (D) $t = 1$, $t = 3.5$ | B1 | Do not award if extra values given. |
## Part (ii)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $v = -8t + 8$ | M1 | Differentiating |
| | A1 | |
| $a = -8$ | F1 | |
## Part (iii)
| Answer/Working | Mark | 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** $v$. |
## Part (iv)
| Answer/Working | Mark | Guidance |
|---|---|---|
| **method 1**: Need velocity at $t = 3$: $v(3) = -8\times3+8 = -16$ | B1 | FT **their** $v$ from (ii) |
| $v = \int 32 \, dt = 32t + C$ | M1 | Accept $32t + C$ or $32t$. SC1 if $\int_3^4 32dt$ attempted. |
| $v = -16$ when $t=3$ gives $v = 32t - 112$ | A1 | Use of **their** $-16$ from an attempt at $v$ when $t=3$ |
| $y = \int(32t - 112)dt = 16t^2 - 112t + D$ | M1 | FT **their** $v$ of the form $pt + q$ with $p \neq 0$ and $q \neq 0$. Accept if at least 1 term correct. Accept no $D$. |
| $y = 0$ 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 | Use of $s = ut + \frac{1}{2}at^2$ |
| | A1 | Use of **their** $-16$ (not 0) from an attempt at $v$ when $t=3$ and 32. Condone use of just $t$ |
| | M1 | Use of $t \pm 3$ |
| (so $y = 16t^2 - 112t + 192$) | A1 | cao |
| **method 2**: $y = k(t-3)(t-4)$ | M1 | Use of a quadratic function (condone no $k$) |
| | A1 | Correct use of roots |
| | B1 | $k$ present |
| When $t = 3.5$, $y = -4$: so $-4 = k\times\frac{1}{2}\times-\frac{1}{2}$ | M1 | Or consider velocity at $t = 3$ |
| so $k = 16$ (and $y = 16t^2 - 112t + 192$) | A1 | cao. Accept $k$ without $y$ simplified. |
---7 A point P on a piece of machinery is moving in a vertical straight line. The displacement of P above ground level at time $t$ seconds is $y$ metres. The displacement-time graph for the motion during the time interval $0 \leqslant t \leqslant 4$ is shown in Fig. 7 .
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{6cca1e5e-82b0-487d-8048-b9db7745dea6-4_1026_1339_516_404}
\captionsetup{labelformat=empty}
\caption{Fig. 7}
\end{center}
\end{figure}
\begin{enumerate}[label=(\roman*)]
\item Using the graph, determine for the time interval $0 \leqslant t \leqslant 4$\\
(A) the greatest displacement of P above its position when $t = 0$,\\
(B) the greatest distance of P from its position when $t = 0$,\\
(C) the time interval in which P is moving downwards,\\
(D) the times when P is instantaneously at rest.
The displacement of P in the time interval $0 \leqslant t \leqslant 3$ is given by $y = - 4 t ^ { 2 } + 8 t + 12$.
\item Use calculus to find expressions in terms of $t$ for the velocity and for the acceleration of P in the interval $0 \leqslant t \leqslant 3$.
\item At what times does P have a speed of $4 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ in the interval $0 \leqslant t \leqslant 3$ ?
In the time interval $3 \leqslant t \leqslant 4 , \mathrm { P }$ has a constant acceleration of $32 \mathrm {~m} \mathrm {~s} ^ { - 2 }$. There is no sudden change in velocity when $t = 3$.
\item Find an expression in terms of $t$ for the displacement of P in the interval $3 \leqslant t \leqslant 4$.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI M1 2010 Q7 [16]}}