| Exam Board | OCR MEI |
|---|---|
| Module | M1 (Mechanics 1) |
| 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 graph reading, basic differentiation (v = dy/dt, a = dv/dt), solving quadratic equations, and using SUVAT with integration. While it has many parts (7 marks typical), each individual step uses standard M1 techniques with no novel problem-solving required. Slightly easier than average due to straightforward calculus and routine application of kinematic principles. |
| Spec | 1.07i Differentiate x^n: for rational n and sums3.02b Kinematic graphs: displacement-time and velocity-time3.02c Interpret kinematic graphs: gradient and area3.02d Constant acceleration: SUVAT formulae3.02f Non-uniform acceleration: using differentiation and integration |
| Answer | Marks | Guidance |
|---|---|---|
| False; this is a speed-time graph not one for displacement-time | M1, A1 | Accept: "no information about direction of travel"; "distance is given by area under graph but this is not the same as displacement"; "speed is not a vector"; Do not accept: "the particle moves only in the positive direction" |
| Answer | Marks | Guidance |
|---|---|---|
| True | B1 | Ignore subsequent working |
| Answer | Marks | Guidance |
|---|---|---|
| True | B1 | Ignore subsequent working |
| Answer | Marks | Guidance |
|---|---|---|
| False; the area under the graph is 420 not 400 | M1, A1 | Accept area up to time 55 s is 400 m; the calculation in the false example must be correct |
## Question 3:
### Part (A)
False; this is a speed-time graph not one for displacement-time | M1, A1 | Accept: "no information about direction of travel"; "distance is given by area under graph but this is not the same as displacement"; "speed is not a vector"; Do not accept: "the particle moves only in the positive direction"
### Part (B)
True | B1 | Ignore subsequent working
### Part (C)
True | B1 | Ignore subsequent working
### Part (D)
False; the area under the graph is 420 not 400 | M1, A1 | Accept area up to time 55 s is 400 m; the calculation in the false example must be correct
---3 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]{569e7c0e-7c33-47c9-b986-8587ea239f0a-3_1020_1333_352_439}
\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 Q3 [16]}}