Edexcel M1 — Question 2 7 marks

Exam BoardEdexcel
ModuleM1 (Mechanics 1)
Marks7
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicConstant acceleration (SUVAT)
TypeMulti-phase journey: find total distance
DifficultyModerate -0.8 This is a straightforward SUVAT problem with clearly defined stages and all necessary information provided. Students need to recognize that acceleration time equals deceleration time (total 116s - 84s = 32s, so 16s each), then calculate distance using the trapezium area under the velocity-time graph. It requires only routine application of kinematic equations with no problem-solving insight, making it easier than average but not trivial due to the multi-stage setup.
Spec3.02b Kinematic graphs: displacement-time and velocity-time3.02d Constant acceleration: SUVAT formulae

2. An underground train accelerates uniformly from rest at station \(A\) to a velocity of \(24 \mathrm {~ms} ^ { - 1 }\). It maintains this speed for 84 seconds, until it decelerates uniformly to rest at station \(B\). The total journey time is 116 seconds and the magnitudes of the acceleration and deceleration are equal.
  1. Find the time it takes the train to accelerate from rest to \(24 \mathrm {~m} \mathrm {~s} ^ { - 1 }\).
  2. Illustrate this information on a velocity-time graph.
  3. Using your graph, or otherwise, find the distance between the two stations.

AnswerMarks Guidance
(a) \(t = \frac{116-84}{2} = 16\) secondsM1 A1
(b) Velocity-time graph with axes labeled: velocity (m s⁻¹) up to 24, time (seconds) showing points at 0, 16, 100, 116B2
(c) dist. = area under graph = \(\frac{1}{2}(116 + 84)(24) = 2400\) mM2 A1 (7 marks)
**(a)** $t = \frac{116-84}{2} = 16$ seconds | M1 A1 |

**(b)** Velocity-time graph with axes labeled: velocity (m s⁻¹) up to 24, time (seconds) showing points at 0, 16, 100, 116 | B2 |

**(c)** dist. = area under graph = $\frac{1}{2}(116 + 84)(24) = 2400$ m | M2 A1 | (7 marks)
2. An underground train accelerates uniformly from rest at station $A$ to a velocity of $24 \mathrm {~ms} ^ { - 1 }$. It maintains this speed for 84 seconds, until it decelerates uniformly to rest at station $B$. The total journey time is 116 seconds and the magnitudes of the acceleration and deceleration are equal.
\begin{enumerate}[label=(\alph*)]
\item Find the time it takes the train to accelerate from rest to $24 \mathrm {~m} \mathrm {~s} ^ { - 1 }$.
\item Illustrate this information on a velocity-time graph.
\item Using your graph, or otherwise, find the distance between the two stations.
\end{enumerate}

\hfill \mbox{\textit{Edexcel M1  Q2 [7]}}