| Exam Board | Edexcel |
|---|---|
| Module | M1 (Mechanics 1) |
| Session | Specimen |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Travel graphs |
| Type | Multi-stage motion with all parameters given |
| Difficulty | Moderate -0.8 This is a straightforward M1 kinematics question involving standard SUVAT calculations and interpreting speed-time graphs. Parts (a) and (b) require routine application of kinematic equations with constant acceleration, part (c) is simple observation, and part (d) uses the area-under-graph method with basic algebra. All techniques are standard textbook exercises 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 |
A train $T_1$ moves from rest at Station $A$ with constant acceleration $2 \text{ m s}^{-2}$ until it reaches a speed of $36 \text{ m s}^{-1}$. In maintains this constant speed for 90 s before the brakes are applied, which produce constant retardation $3 \text{ m s}^{-2}$. The train $T_1$ comes to rest at station $B$.
\begin{enumerate}[label=(\alph*)]
\item Sketch a speed-time graph to illustrate the journey of $T_1$ from $A$ to $B$. [3]
\item Show that the distance between $A$ and $B$ is 3780 m. [5]
\end{enumerate}
\includegraphics{figure_3}
A second train $T_2$ takes 150 s to move form rest at $A$ to rest at $B$. Figure 3 shows the speed-time graph illustrating this journey.
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{2}
\item Explain briefly one way in which $T_1$'s journey differs from $T_2$'s journey. [1]
\item Find the greatest speed, in m s$^{-1}$, attained by $T_2$ during its journey. [3]
\end{enumerate}
\hfill \mbox{\textit{Edexcel M1 Q4 [12]}}