| Exam Board | Edexcel |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2013 |
| Session | January |
| Marks | 15 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Travel graphs |
| Type | Multi-stage motion with velocity-time graph given |
| Difficulty | Moderate -0.8 This is a standard M1 kinematics question using velocity-time graphs with straightforward application of SUVAT equations and area-under-graph calculations. Part (a) uses v²=u²+2as directly, part (b) requires finding total time by calculating areas (trapezium), parts (c-d) involve sketching a triangular v-t graph and using equal areas/times. All steps are routine textbook exercises with no problem-solving insight required, making it easier than average but not trivial due to the multi-part structure and arithmetic involved. |
| 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 |
|---|---|---|
| \(30^2 = 2a.300\) | M1 | |
| \(a = 1.5\) | A1 | (2) |
| Answer | Marks | Guidance |
|---|---|---|
| \(0^2 = 30^2 - 2 \times 1.25s\) OR \(0 = 30 - 1.25t_2\) | M1 | |
| \(s = 360\) | \(t_2 = 24\) | A1 |
| \(300 + 30T + 360 = 1500\) | \(\frac{(20 + T + 24 + T)}{2} \times 30 = 1500\) | M1 A1 |
| \(T = 28\) | \(T = 28\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Triangle, drawn on the diagram, with base coinciding with base of trapezium, top vertex above line \(v = 30\) and meeting trapezium at least once | B1 | |
| DB1 | ||
| \(V\) marked correctly | (2) |
| Answer | Marks | Guidance |
|---|---|---|
| \(30 = 1.5t_1 \Rightarrow t_1 = 20\) | M1 | |
| \(30 = 1.25t_2 \Rightarrow t_2 = 24\) | A1 | |
| \(\frac{1}{2}(20 + 28 + 24)V = 1500\) | M1 A1 | |
| \(V = \frac{750}{18} = 41.67\) | A1 | |
| \(= \frac{125}{3}\) (oe) 0r 42 (or better) | (6) |
### Part (a)
$30^2 = 2a.300$ | M1 |
$a = 1.5$ | A1 | (2)
### Part (b)
$0^2 = 30^2 - 2 \times 1.25s$ OR $0 = 30 - 1.25t_2$ | M1 |
$s = 360$ | $t_2 = 24$ | A1 |
$300 + 30T + 360 = 1500$ | $\frac{(20 + T + 24 + T)}{2} \times 30 = 1500$ | M1 A1 |
$T = 28$ | $T = 28$ | A1 | (5)
### Part (c)
Triangle, drawn on the diagram, with base coinciding with base of trapezium, top vertex above line $v = 30$ and meeting trapezium at least once | B1 |
| DB1 |
$V$ marked correctly | | (2)
### Part (d)
$30 = 1.5t_1 \Rightarrow t_1 = 20$ | M1 |
$30 = 1.25t_2 \Rightarrow t_2 = 24$ | A1 |
$\frac{1}{2}(20 + 28 + 24)V = 1500$ | M1 A1 |
$V = \frac{750}{18} = 41.67$ | A1 |
$= \frac{125}{3}$ **(oe) 0r 42 (or better)** | | (6)
**Total for Question 5: 15**
---\includegraphics{figure_4}
The velocity-time graph in Figure 4 represents the journey of a train $P$ travelling along a straight horizontal track between two stations which are 1.5 km apart. The train $P$ leaves the first station, accelerating uniformly from rest for 300 m until it reaches a speed of 30 m s$^{-1}$. The train then maintains this speed for 7 seconds before decelerating uniformly at 1.25 m s$^{-2}$, coming to rest at the next station.
\begin{enumerate}[label=(\alph*)]
\item Find the acceleration of $P$ during the first 300 m of its journey. [2]
\item Find the value of $T$. [5]
\end{enumerate}
A second train $Q$ completes the same journey in the same total time. The train leaves the first station, accelerating uniformly from rest until it reaches a speed of $V$ m s$^{-1}$ and then immediately decelerates uniformly until it comes to rest at the next station.
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{2}
\item Sketch on the diagram above, a velocity-time graph which represents the journey of train $Q$. [2]
\item Find the value of $V$. [6]
\end{enumerate}
\hfill \mbox{\textit{Edexcel M1 2013 Q5 [15]}}