| Exam Board | Edexcel |
|---|---|
| Module | AS Paper 2 (AS Paper 2) |
| Year | 2022 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Constant acceleration (SUVAT) |
| Type | Multi-phase journey: find unknown speed or time |
| Difficulty | Standard +0.3 This is a standard SUVAT multi-stage motion problem requiring a speed-time graph sketch, algebraic setup with one unknown, and straightforward calculations. While it has multiple parts (5 marks total likely 8-10), each step follows directly from the previous with no novel insight required—students set up equations from the trapezium area and given constraints, then solve linearly. Slightly above average difficulty due to the algebraic manipulation and multi-stage nature, but well within typical AS mechanics scope. |
| 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/Working | Mark | Guidance |
| Trapezium shape starting at origin, peak at \((0, 25)\), ending at \((700, 0)\), with deceleration phase longer than acceleration phase | B1 | AO 1.1b; overall shape correct; ignore incorrect figs on axes; deceleration phase longer than acceleration phase if nothing on \(t\)-axis; allow if \(t\) (or 40) and \(4t\) (or 160) clearly and correctly marked |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Using total area \(= 15000\) to set up equation in one unknown, or use suvat on sections (considering all sections) | M1 | AO 3.4; need all sections with correct structure for each section, with \(\frac{1}{2}\)'s where appropriate; allow \(= 15\) or \(150\) or \(1500\) etc instead of \(15000\) |
| \(\frac{1}{2}(700 + 700 - t - 4t) \times 25 = 15000\) OR \(\frac{1}{2} \times 25 \times t + 25(700 - t - 4t) + \frac{1}{2} \times 25 \times 4t = 15000\) | A1 | AO 1.1b; correct equation in \(t\) only, seen or implied (or with \(t = 40\) for verification) |
| \(t = 40\) (s) | A1* | AO 1.1b; cso; at least one line of working with brackets removed and \(t\)'s collected |
| Verification: \(4 \times 40 = 160\); \(700 - 40 - 160 = 500\); \(\frac{(700+500)}{2} \times 25 = 15000 = 15\text{ km}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(0.63\) or \(0.625\) or \(\frac{5}{8}\) oe \((\text{m s}^{-2})\) | B1 | AO 1.1b / 2.2a; cao |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Complete method to find speed at \(t = 572\): e.g. \(\pm\left(25 - \left(32 \times \frac{5}{32}\right)\right)\) or \(\pm\left(128 \times \frac{5}{32}\right)\) | M1 | AO 3.1b; any complete method; correct figures; condone sign errors |
| \(20\ (\text{m s}^{-1})\) | A1 | AO 1.1b; cao; must be positive and exact |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| e.g. train cannot instantaneously change acceleration; train won't move with constant acceleration; train won't move with constant speed | B1 | AO 3.5b; any appropriate limitation of the model; B0 if any incorrect extra answers given; B0 if friction or air resistance or size of train mentioned |
# Question 2:
## Part 2(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Trapezium shape starting at origin, peak at $(0, 25)$, ending at $(700, 0)$, with deceleration phase longer than acceleration phase | B1 | AO 1.1b; overall shape correct; ignore incorrect figs on axes; deceleration phase longer than acceleration phase if nothing on $t$-axis; allow if $t$ (or 40) and $4t$ (or 160) clearly and correctly marked |
## Part 2(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Using total area $= 15000$ to set up equation in one unknown, or use suvat on sections (considering all sections) | M1 | AO 3.4; need all sections with correct structure for each section, with $\frac{1}{2}$'s where appropriate; allow $= 15$ or $150$ or $1500$ etc instead of $15000$ |
| $\frac{1}{2}(700 + 700 - t - 4t) \times 25 = 15000$ **OR** $\frac{1}{2} \times 25 \times t + 25(700 - t - 4t) + \frac{1}{2} \times 25 \times 4t = 15000$ | A1 | AO 1.1b; correct equation in $t$ only, seen or implied (or with $t = 40$ for verification) |
| $t = 40$ (s) | A1* | AO 1.1b; cso; at least one line of working with brackets removed and $t$'s collected |
| Verification: $4 \times 40 = 160$; $700 - 40 - 160 = 500$; $\frac{(700+500)}{2} \times 25 = 15000 = 15\text{ km}$ | | |
## Part 2(c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $0.63$ or $0.625$ or $\frac{5}{8}$ oe $(\text{m s}^{-2})$ | B1 | AO 1.1b / 2.2a; cao |
## Part 2(d):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Complete method to find speed at $t = 572$: e.g. $\pm\left(25 - \left(32 \times \frac{5}{32}\right)\right)$ or $\pm\left(128 \times \frac{5}{32}\right)$ | M1 | AO 3.1b; any complete method; correct figures; condone sign errors |
| $20\ (\text{m s}^{-1})$ | A1 | AO 1.1b; cao; must be positive and exact |
## Part 2(e):
| Answer/Working | Mark | Guidance |
|---|---|---|
| e.g. train cannot instantaneously change acceleration; train won't move with constant acceleration; train won't move with constant speed | B1 | AO 3.5b; any appropriate limitation of the model; B0 if any incorrect extra answers given; B0 if friction or air resistance or size of train mentioned |
---\begin{enumerate}
\item A train travels along a straight horizontal track from station $P$ to station $Q$.
\end{enumerate}
In a model of the motion of the train, at time $t = 0$ the train starts from rest at $P$, and moves with constant acceleration until it reaches its maximum speed of $25 \mathrm {~m} \mathrm {~s} ^ { - 1 }$
The train then travels at this constant speed of $25 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ before finally moving with constant deceleration until it comes to rest at $Q$.
The time spent decelerating is four times the time spent accelerating.\\
The journey from $P$ to $Q$ takes 700 s .\\
Using the model,\\
(a) sketch a speed-time graph for the motion of the train between the two stations $P$ and $Q$.
The distance between the two stations is 15 km .\\
Using the model,\\
(b) show that the time spent accelerating by the train is 40 s ,\\
(c) find the acceleration, in $\mathrm { m } \mathrm { s } ^ { - 2 }$, of the train,\\
(d) find the speed of the train 572s after leaving $P$.\\
(e) State one limitation of the model which could affect your answers to parts (b) and (c).
\hfill \mbox{\textit{Edexcel AS Paper 2 2022 Q2 [8]}}