| Exam Board | OCR MEI |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2007 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Constant acceleration (SUVAT) |
| Type | Multi-phase journey: find total distance |
| Difficulty | Moderate -0.8 This is a straightforward multi-stage SUVAT problem with all parameters explicitly given. Students must sketch a speed-time graph (routine), calculate distance using trapezium areas (standard technique), and solve for deceleration time using given total distance. All steps are mechanical applications of basic kinematics formulas with no problem-solving insight required. |
| 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 |
| Correct axes labelled with \(t\) and \(v\) | B1 | |
| Three correct line segments with values: \((0,10), (10,30), (25,40), (45,40)\) | B1 | Correct shape |
| All values correct | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Area under graph: \(\frac{1}{2}(10+30)\times10 + \frac{1}{2}(30+40)\times15 + 40\times20\) | M1 | Attempt at area of trapezium/triangles |
| \(= 200 + 525 + 800\) | A1 | At least two correct |
| \(= 1525\) m | A1 | cao |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(1700 - 1525 = 175\) m remaining after B | M1 | |
| \(175 = \frac{1}{2}\times40\times T \Rightarrow T = 8.75\) s | A1 | cao |
# Question 2:
## Part (i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Correct axes labelled with $t$ and $v$ | B1 | |
| Three correct line segments with values: $(0,10), (10,30), (25,40), (45,40)$ | B1 | Correct shape |
| All values correct | B1 | |
## Part (ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Area under graph: $\frac{1}{2}(10+30)\times10 + \frac{1}{2}(30+40)\times15 + 40\times20$ | M1 | Attempt at area of trapezium/triangles |
| $= 200 + 525 + 800$ | A1 | At least two correct |
| $= 1525$ m | A1 | cao |
## Part (iii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $1700 - 1525 = 175$ m remaining after B | M1 | |
| $175 = \frac{1}{2}\times40\times T \Rightarrow T = 8.75$ s | A1 | cao |
---2 A car passes a point A travelling at $10 \mathrm {~m} \mathrm {~s} ^ { - 1 }$. Its motion over the next 45 seconds is modelled as follows.
\begin{itemize}
\item The car's speed increases uniformly from $10 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ to $30 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ over the first 10 s .
\item Its speed then increases uniformly to $40 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ over the next 15 s .
\item The car then maintains this speed for a further 20 s at which time it reaches the point B .\\
(i) Sketch a speed-time graph to represent this motion.\\
(ii) Calculate the distance from A to B .\\
(iii) When it reaches the point B , the car is brought uniformly to rest in $T$ seconds. The total distance from A is now 1700 m . Calculate the value of $T$.
\end{itemize}
\hfill \mbox{\textit{OCR MEI M1 2007 Q2 [8]}}