| Exam Board | OCR MEI |
|---|---|
| Module | M1 (Mechanics 1) |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Constant acceleration (SUVAT) |
| Type | Multi-phase journey: find total distance |
| Difficulty | Moderate -0.3 This is a straightforward multi-stage SUVAT problem with all parameters explicitly given. Part (i) is routine graph sketching, part (ii) requires calculating trapezoid areas under the velocity-time graph, and part (iii) involves one additional calculation using the deceleration phase. All steps are standard M1 techniques with no problem-solving insight required, making it slightly easier than average. |
| 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 |
|---|---|---|
| \(0.5 \times 8 \times 10 = 40\) m | M1, A1 (sub: 2) | Attempt to find whole area or... If *suvat* used in 2 parts, accept any \(t\) value \(0 \leq t \leq 8\) for max. |
| Answer | Marks | Guidance |
|---|---|---|
| \(0.5 \times 5(T-8) = 10\) | A1 | cao |
| M1 | \(0.5 \times 5 \times k = 10\) seen. Accept \(\pm 5\) and \(\pm 10\) only. If *suvat* used need whole area; if in 2 parts, accept any \(t\) value \(8 \leq t \leq T\) for min. | |
| \(T = 12\) | B1 | Attempt to use \(k = T - 8\) |
| A1 (sub: 3) | cao. [Award 3 if \(T = 12\) seen] |
| Answer | Marks | Guidance |
|---|---|---|
| \(40 - 10 = 30\) m | B1 (sub: 1) | FT their 40. |
## Question 6:
**Part (i):**
$0.5 \times 8 \times 10 = 40$ m | M1, A1 (sub: 2) | Attempt to find whole area or... If *suvat* used in 2 parts, accept any $t$ value $0 \leq t \leq 8$ for max.
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**Part (ii):**
$0.5 \times 5(T-8) = 10$ | A1 | cao
| M1 | $0.5 \times 5 \times k = 10$ seen. Accept $\pm 5$ and $\pm 10$ only. If *suvat* used need whole area; if in 2 parts, accept any $t$ value $8 \leq t \leq T$ for min.
$T = 12$ | B1 | Attempt to use $k = T - 8$
| A1 (sub: 3) | cao. [Award 3 if $T = 12$ seen]
---
**Part (iii):**
$40 - 10 = 30$ m | B1 (sub: 1) | FT **their** 40.
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**Total: 6 marks**6 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 $10 \mathrm {~ms} { } ^ { 1 }$ to $30 \mathrm {~ms} { } ^ { 1 }$ 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 Q6 [8]}}