| Exam Board | Edexcel |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2003 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Constant acceleration (SUVAT) |
| Type | Two vehicles: overtaking or meeting (algebraic) |
| Difficulty | Standard +0.3 This is a standard two-particle SUVAT problem requiring students to find acceleration from given data, sketch speed-time graphs, and equate distances using areas under graphs. While it involves multiple stages and careful bookkeeping of time intervals, it follows a well-practiced M1 template with no novel problem-solving required—slightly easier than average due to its routine nature. |
| 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 |
|---|---|---|
| Working | Marks | Notes |
| Shape for \(A\) | B1 | |
| Shape for \(B\) with parallel slope | B1 | |
| Figures (12, 30, 40, 60, \(T\) on axes) | B1 | (3) |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Notes |
| Distance moved by \(A = \frac{1}{2}\times 12\times 30 + 30(T-12)\) | B1, M1 A1 | |
| \(B\) accelerates for 24 s | B1 | |
| Distance moved by \(B = \frac{1}{2}\times 24\times 60 + 60(T-64)\) | B1, M1 A1 | |
| \(\frac{1}{2}\times 12\times 30 + 30(T-12) = \frac{1}{2}\times 24\times 60 + 60(T-64)\) | M1 | |
| \(\Rightarrow T = 98\) s | A1 | (9) |
# Question 7:
## Part (a):
| Working | Marks | Notes |
|---------|-------|-------|
| Shape for $A$ | B1 | |
| Shape for $B$ with parallel slope | B1 | |
| Figures (12, 30, 40, 60, $T$ on axes) | B1 | (3) |
## Part (b):
| Working | Marks | Notes |
|---------|-------|-------|
| Distance moved by $A = \frac{1}{2}\times 12\times 30 + 30(T-12)$ | B1, M1 A1 | |
| $B$ accelerates for 24 s | B1 | |
| Distance moved by $B = \frac{1}{2}\times 24\times 60 + 60(T-64)$ | B1, M1 A1 | |
| $\frac{1}{2}\times 12\times 30 + 30(T-12) = \frac{1}{2}\times 24\times 60 + 60(T-64)$ | M1 | |
| $\Rightarrow T = 98$ s | A1 | (9) |
---7. Two trains $A$ and $B$ run on parallel straight tracks. Initially both are at rest in a station and level with each other. At time $t = 0 , A$ starts to move. It moves with constant acceleration for 12 s up to a speed of $30 \mathrm {~m} \mathrm {~s} ^ { - 1 }$, and then moves at a constant speed of $30 \mathrm {~m} \mathrm {~s} ^ { - 1 }$. Train $B$ starts to move in the same direction as $A$ when $t = 40$, where $t$ is measured in seconds. It accelerates with the same initial acceleration as $A$, up to a speed of $60 \mathrm {~m} \mathrm {~s} ^ { - 1 }$. It then moves at a constant speed of $60 \mathrm {~m} \mathrm {~s} ^ { - 1 }$. Train $B$ overtakes $A$ after both trains have reached their maximum speed. Train $B$ overtakes $A$ when $t = T$.
\begin{enumerate}[label=(\alph*)]
\item Sketch, on the same diagram, the speed-time graphs of both trains for $0 \leq t \leq T$.
\item Find the value of $T$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M1 2003 Q7 [12]}}