| Exam Board | Edexcel |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2023 |
| Session | October |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Travel graphs |
| Type | Acceleration-time graph sketching or interpretation |
| Difficulty | Standard +0.2 This is a straightforward M1 question requiring standard interpretation of an acceleration-time graph. Part (a) involves calculating area under the graph to find change in velocity (basic kinematics), part (b) requires sketching the corresponding speed-time graph (routine skill), and part (c) uses area under the speed-time graph for distance. All steps are mechanical applications of standard techniques with no problem-solving insight required, making it 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 |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(v = u + at: \quad w = 8 + (-0.5)(4)\) | M1 | Complete method for finding velocity \(w\) when \(t=4\). M0 if \(u=0\). Note: 6 on its own can imply this mark |
| \(v = u + at: \quad v = w + (1.2)(10)\) | M1 | Method completed to show speed when \(t=20\). M0 if initial speed is not \(w\) |
| \(v = 18\)* | A1* | Fully correct solution leading to given answer |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Correct shape of graph | B1 | No solid vertical lines |
| Time labels 4, 10, 20 | B1 | Correct time labels |
| Speed labels 6, 8, 18 | B1 | Correct speed labels |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Clear attempt to find distance using area under graph from \(t=0\) to \(t=20\) | M1 | Award even if using wrong shapes |
| \(\text{Distance} = \frac{(8+\text{"6"})\times 4}{2} + (6\times\text{"6"}) + \frac{(\text{"6"}+18)\times 10}{2}\) | A1ft A1ft | First A1ft: equation with at most one error, ft their "6". Second A1ft: correct equation, ft their "6" |
| \(= 184\) (m) | A1 | Correct final answer |
## Question 2(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $v = u + at: \quad w = 8 + (-0.5)(4)$ | M1 | Complete method for finding velocity $w$ when $t=4$. M0 if $u=0$. Note: 6 on its own can imply this mark |
| $v = u + at: \quad v = w + (1.2)(10)$ | M1 | Method completed to show speed when $t=20$. M0 if initial speed is not $w$ |
| $v = 18$* | A1* | Fully correct solution leading to given answer |
## Question 2(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Correct shape of graph | B1 | No solid vertical lines |
| Time labels 4, 10, 20 | B1 | Correct time labels |
| Speed labels 6, 8, 18 | B1 | Correct speed labels |
## Question 2(c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Clear attempt to find distance using area under graph from $t=0$ to $t=20$ | M1 | Award even if using wrong shapes |
| $\text{Distance} = \frac{(8+\text{"6"})\times 4}{2} + (6\times\text{"6"}) + \frac{(\text{"6"}+18)\times 10}{2}$ | A1ft A1ft | First A1ft: equation with at most one error, ft their "6". Second A1ft: correct equation, ft their "6" |
| $= 184$ (m) | A1 | Correct final answer |
---2.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{017cc2b0-9ec3-45ff-94c0-9d989badfd5d-04_677_1620_294_169}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\end{center}
\end{figure}
Two fixed points, $A$ and $B$, are on a straight horizontal road.\\
The acceleration-time graph in Figure 2 represents the motion of a car travelling along the road as it moves from $A$ to $B$.
At time $t = 0$, the car passes through $A$ with speed $8 \mathrm {~m} \mathrm {~s} ^ { - 1 }$\\
At time $t = 20 \mathrm {~s}$, the car passes through $B$ with speed $v \mathrm {~ms} ^ { - 1 }$
\begin{enumerate}[label=(\alph*)]
\item Show that $v = 18$
\item Sketch a speed-time graph for the motion of the car from $A$ to $B$.
\item Find the distance $A B$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M1 2023 Q2 [10]}}