| Exam Board | Edexcel |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2017 |
| Session | January |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Constant acceleration (SUVAT) |
| Type | Multi-phase journey: find total distance |
| Difficulty | Easy -1.2 This is a straightforward three-stage SUVAT problem where all parameters are explicitly given. Students simply apply v=u+at twice to find missing times, then calculate distances using standard formulae, and finally compute average speed. No problem-solving insight required—pure mechanical application of memorized equations. |
| Spec | 3.02d Constant acceleration: SUVAT formulae |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Use of \(v = u + at\) to find \(t_1\) or \(t_2\) | M1 | Use of \(v = u + at\) or gradient or any other complete method to find \(t_1\) or \(t_2\) (condone sign errors) |
| \(t_1 = 15 \div 0.5 = 30\) (s) OR \(t_2 = 15 \div 0.25 = 60\) | A1 | First A1 for either 30 or 60 (A0 if negative) |
| Total time \(= 30 + 200 + 60 = 290\) (s) | A1 cso | Second A1 for 290 with no errors seen |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Use area/suvat to find distance: \(\text{distance} = \frac{1}{2} \times 30 \times 15 + 200 \times 15 + \frac{1}{2} \times 60 \times 15\) | M1A2 ft | M1 for complete method (must have \(\frac{1}{2}\)) either by trapezium rule or 2 triangles and rectangle. Follow their \(t_1\) and \(t_2\). A2 ft on their \(t_1\) & \(t_2\) (-1 each error) |
| \(= 3675\) (m) (3.675 km) | A1 | A1 for 3675 (m) or 3.675 km |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Ave. speed \(= \dfrac{\text{their(b)}}{\text{their(a)}}\) | M1 | M1 for \(= \dfrac{\text{their(b)}}{\text{their(a)}}\) |
| \(= \dfrac{3675}{290}\) oe (m s\(^{-1}\)) (12.6724...) | A1 | A1 for 13 or better |
# Question 1:
## Part 1a:
| Answer/Working | Mark | Guidance |
|---|---|---|
| Use of $v = u + at$ to find $t_1$ or $t_2$ | M1 | Use of $v = u + at$ or gradient or any other complete method to find $t_1$ or $t_2$ (condone sign errors) |
| $t_1 = 15 \div 0.5 = 30$ (s) **OR** $t_2 = 15 \div 0.25 = 60$ | A1 | First A1 for either 30 or 60 (A0 if negative) |
| Total time $= 30 + 200 + 60 = 290$ (s) | A1 cso | Second A1 for 290 with no errors seen |
## Part 1b:
| Answer/Working | Mark | Guidance |
|---|---|---|
| Use area/suvat to find distance: $\text{distance} = \frac{1}{2} \times 30 \times 15 + 200 \times 15 + \frac{1}{2} \times 60 \times 15$ | M1A2 ft | M1 for complete method (must have $\frac{1}{2}$) either by trapezium rule or 2 triangles and rectangle. Follow their $t_1$ and $t_2$. A2 ft on their $t_1$ & $t_2$ (-1 each error) |
| $= 3675$ (m) (3.675 km) | A1 | A1 for 3675 (m) or 3.675 km |
## Part 1c:
| Answer/Working | Mark | Guidance |
|---|---|---|
| Ave. speed $= \dfrac{\text{their(b)}}{\text{their(a)}}$ | M1 | M1 for $= \dfrac{\text{their(b)}}{\text{their(a)}}$ |
| $= \dfrac{3675}{290}$ oe (m s$^{-1}$) (12.6724...) | A1 | A1 for 13 or better |
---\begin{enumerate}
\item A train moves along a straight horizontal track between two stations $R$ and $S$. Initially the train is at rest at $R$. The train accelerates uniformly at $\frac { 1 } { 2 } \mathrm {~m} \mathrm {~s} ^ { - 2 }$ from rest at $R$ until it is moving with speed $15 \mathrm {~m} \mathrm {~s} ^ { - 1 }$. For the next 200 seconds the train maintains a constant speed of $15 \mathrm {~m} \mathrm {~s} ^ { - 1 }$. The train then decelerates uniformly at $\frac { 1 } { 4 } \mathrm {~m} \mathrm {~s} ^ { - 2 }$ until it comes to rest at $S$.
\end{enumerate}
Find\\
(a) the time taken by the train to travel from $R$ to $S$,\\
(b) the distance from $R$ to $S$,\\
(c) the average speed of the train during the journey from $R$ to $S$.\\
\hfill \mbox{\textit{Edexcel M1 2017 Q1 [9]}}