| Exam Board | Edexcel |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2023 |
| Session | January |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Constant acceleration (SUVAT) |
| Type | Multi-phase journey: find unknown speed or time |
| Difficulty | Moderate -0.3 This is a standard M1 SUVAT question requiring a speed-time graph sketch and solving for unknowns using area under the graph equals distance. The multi-stage setup and algebraic relationship (acceleration = 2×deceleration) add mild complexity, but the solution follows a routine template with straightforward simultaneous equations. Slightly easier than average due to predictable structure and clear given information. |
| Spec | 3.02b Kinematic graphs: displacement-time and velocity-time3.02d Constant acceleration: SUVAT formulae |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Trapezoid/triangle shape with acceleration phase steeper than deceleration phase | B1 | Correct shape with acceleration steeper than deceleration (ignore entries on the axes) |
| Maximum speed = 20 (m s\(^{-1}\)) marked on vertical axis | B1 | Correct vertical label |
| \(T\), \(T+180\), \(3T+180\) marked on horizontal axis | B1 | Correct horizontal labels. Accept use of their \(T\) or appropriately labelled delineators |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(T\) and \(2T\) seen or implied for acceleration and deceleration times respectively | B1 | Correct relationship seen or implied between time accelerating and time decelerating |
| Any of: \(4800 = \left(\frac{20 \times T}{2}\right) + (180 \times 20) + \left(\frac{20 \times 2T}{2}\right)\) or \(4800 = \left(\frac{20 \times T}{2}\right) + \frac{1}{2} \times 20(180 + (180+2T))\) or \(4800 = \frac{1}{2} \times 20(180+T+180) + \left(\frac{20 \times 2T}{2}\right)\) or \(4800 = \frac{1}{2} \times 20(180+3T+180)\) or \(4800 = 20 \times (180+3T) - \left(\frac{20 \times T}{2}\right) - \left(\frac{20 \times 2T}{2}\right)\) | M1 | A clear attempt to use total area under graph (or use suvat formulae) and equate to 4800 to form equation in \(T\) only. Must involve triangle or trapezium (M0 if single suvat formula used for whole motion) |
| Equation with at most one error | A1 | Use of 3 instead of 180 is one error. Having \(T\) and \(2T\) the wrong way round treated as one error |
| Fully correct equation | A1 | |
| \(T = 40\) (allow \(t\)) | A1 | cao. N.B. \(\frac{1}{2} \times 20(x+180) = 4800 \Rightarrow x = 300\) ONLY scores B0M1A1A0A0 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(20 = a \times T\) (using their \(T\)) | M1 | Correct equation in \(a\), using their \(T\) |
| Acceleration \(= \frac{1}{2}\) (m s\(^{-2}\)) | A1ft | Correct answer, follow through on their '40' |
# Question 1:
## Part (a)
| Answer/Working | Mark | Guidance |
|---|---|---|
| Trapezoid/triangle shape with acceleration phase steeper than deceleration phase | B1 | Correct shape with acceleration steeper than deceleration (ignore entries on the axes) |
| Maximum speed = 20 (m s$^{-1}$) marked on vertical axis | B1 | Correct vertical label |
| $T$, $T+180$, $3T+180$ marked on horizontal axis | B1 | Correct horizontal labels. Accept use of their $T$ or appropriately labelled delineators |
## Part (b)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $T$ and $2T$ seen or implied for acceleration and deceleration times respectively | B1 | Correct relationship seen or implied between time accelerating and time decelerating |
| Any of: $4800 = \left(\frac{20 \times T}{2}\right) + (180 \times 20) + \left(\frac{20 \times 2T}{2}\right)$ or $4800 = \left(\frac{20 \times T}{2}\right) + \frac{1}{2} \times 20(180 + (180+2T))$ or $4800 = \frac{1}{2} \times 20(180+T+180) + \left(\frac{20 \times 2T}{2}\right)$ or $4800 = \frac{1}{2} \times 20(180+3T+180)$ or $4800 = 20 \times (180+3T) - \left(\frac{20 \times T}{2}\right) - \left(\frac{20 \times 2T}{2}\right)$ | M1 | A clear attempt to use total area under graph (or use suvat formulae) and equate to 4800 to form equation in $T$ only. Must involve triangle or trapezium (M0 if single suvat formula used for whole motion) |
| Equation with at most one error | A1 | Use of 3 instead of 180 is one error. Having $T$ and $2T$ the wrong way round treated as one error |
| Fully correct equation | A1 | |
| $T = 40$ (allow $t$) | A1 | cao. N.B. $\frac{1}{2} \times 20(x+180) = 4800 \Rightarrow x = 300$ ONLY scores B0M1A1A0A0 |
## Part (c)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $20 = a \times T$ (using their $T$) | M1 | Correct equation in $a$, using their $T$ |
| Acceleration $= \frac{1}{2}$ (m s$^{-2}$) | A1ft | Correct answer, follow through on their '40' |
---\begin{enumerate}
\item A train travels along a straight horizontal track between two stations $A$ and $B$.
\end{enumerate}
The train starts from rest at station $A$ and accelerates uniformly for $T$ seconds until it reaches a speed of $20 \mathrm {~ms} ^ { - 1 }$
The train then travels at a constant speed of $20 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ for 3 minutes before decelerating uniformly until it comes to rest at station $B$.
The magnitude of the acceleration of the train is twice the magnitude of the deceleration.\\
(a) On the axes below, sketch a speed-time graph to illustrate the motion of the train as it moves from station $A$ to station $B$.\\
\includegraphics[max width=\textwidth, alt={}, center]{84c0eead-0a87-4d87-b33d-794a94bb466c-02_670_1422_813_312}
If you need to redraw your graph, use the axes on page 3
Stations $A$ and $B$ are 4.8 km apart.\\
(b) Find the value of $T$\\
(c) Find the acceleration of the train during the first $T$ seconds of its motion.
Only use these axes if you need to redraw your graph. ${ } _ { O } ^ { \substack { \text { speed } \\ \left( \mathrm { ms } ^ { - 1 } \right) } }$\\
\hfill \mbox{\textit{Edexcel M1 2023 Q1 [10]}}