Edexcel M1 2004 January — Question 6 14 marks

Exam BoardEdexcel
ModuleM1 (Mechanics 1)
Year2004
SessionJanuary
Marks14
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicTravel graphs
TypeMulti-stage motion with all parameters given
DifficultyModerate -0.8 This is a straightforward kinematics question using constant acceleration equations (suvat) with clearly defined phases. Part (a) is routine calculation, part (b) is basic graph sketching, parts (c) and (d) require applying standard formulas across multiple phases but involve no conceptual challenges or novel problem-solving—purely methodical application of well-practiced techniques.
Spec3.02b Kinematic graphs: displacement-time and velocity-time3.02c Interpret kinematic graphs: gradient and area3.02d Constant acceleration: SUVAT formulae

A train starts from rest at a station \(A\) and moves along a straight horizontal track. For the first 10 s, the train moves with constant acceleration 1.2 m s\(^{-2}\). For the next 24 s it moves at a constant acceleration 0.75 m s\(^{-2}\). It then moves with constant speed for \(T\) seconds. Finally it slows down with constant deceleration 3 m s\(^{-2}\) until it comes to a rest at station \(B\).
  1. Show that, 34 s after leaving \(A\), the speed of the train is 30 m s\(^{-1}\). [3]
  2. Sketch a speed-time graph to illustrate the motion of the train as it moves from \(A\) to \(B\). [3]
  3. Find the distance moved by the train during the first 34 s of its journey from \(A\). [4]
The distance from \(A\) to \(B\) is 3 km.
  1. Find the value of \(T\). [4]

Part (a)
Answer/Working:
- After 10 s, speed \(= 1.2 \times 10 = 12 \text{ m s}^{-1}\)
- After next 24 s, \(v = "u + at" = 12 + 0.75 \times 24 = 30 \text{ m s}^{-1}\)
Marks: B1, M1, A1 (3 marks)
Part (b)
Answer/Working:
- Shape \(0 \leq t \leq 34\): (shown in graph)
- Shape \(t \geq 34\): (shown in graph)
- Figures: 12, 30, 300 at appropriate positions
Marks: B1, B1, B1 (3 marks)
Part (c)
Answer/Working: Distance \(= \frac{1}{2} \times 10 \times 12 + \frac{1}{2}(30+12) \times 24 = 60 + 504 = 564 \text{ m}\)
Marks: B1, M1, A1, A1 (4 marks)
Part (d)
Answer/Working:
- Distance travelled decelerating \(= \frac{1}{2} \times 30 \times 10\)
- \(564 + 307 + \frac{1}{2} \times 30 \times 10 = 3000\)
- \(\Rightarrow T = 76.2 \text{ s}\)
Marks: B1, M1, A1, A1 (4 marks)
## Part (a)
**Answer/Working:**
- After 10 s, speed $= 1.2 \times 10 = 12 \text{ m s}^{-1}$
- After next 24 s, $v = "u + at" = 12 + 0.75 \times 24 = 30 \text{ m s}^{-1}$

**Marks:** B1, M1, A1 (3 marks)

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## Part (b)
**Answer/Working:**
- Shape $0 \leq t \leq 34$: (shown in graph)
- Shape $t \geq 34$: (shown in graph)
- Figures: 12, 30, 300 at appropriate positions

**Marks:** B1, B1, B1 (3 marks)

---

## Part (c)
**Answer/Working:** Distance $= \frac{1}{2} \times 10 \times 12 + \frac{1}{2}(30+12) \times 24 = 60 + 504 = 564 \text{ m}$

**Marks:** B1, M1, A1, A1 (4 marks)

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## Part (d)
**Answer/Working:**
- Distance travelled decelerating $= \frac{1}{2} \times 30 \times 10$
- $564 + 307 + \frac{1}{2} \times 30 \times 10 = 3000$
- $\Rightarrow T = 76.2 \text{ s}$

**Marks:** B1, M1, A1, A1 (4 marks)

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A train starts from rest at a station $A$ and moves along a straight horizontal track. For the first 10 s, the train moves with constant acceleration 1.2 m s$^{-2}$. For the next 24 s it moves at a constant acceleration 0.75 m s$^{-2}$. It then moves with constant speed for $T$ seconds. Finally it slows down with constant deceleration 3 m s$^{-2}$ until it comes to a rest at station $B$.

\begin{enumerate}[label=(\alph*)]
\item Show that, 34 s after leaving $A$, the speed of the train is 30 m s$^{-1}$. [3]
\item Sketch a speed-time graph to illustrate the motion of the train as it moves from $A$ to $B$. [3]
\item Find the distance moved by the train during the first 34 s of its journey from $A$. [4]
\end{enumerate}

The distance from $A$ to $B$ is 3 km.

\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{3}
\item Find the value of $T$. [4]
\end{enumerate}

\hfill \mbox{\textit{Edexcel M1 2004 Q6 [14]}}