Edexcel M1 — Question 4 11 marks

Exam BoardEdexcel
ModuleM1 (Mechanics 1)
Marks11
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicTravel graphs
TypeMulti-stage motion with algebraic unknowns
DifficultyModerate -0.8 This is a standard M1 kinematics question involving constant acceleration in three phases. Part (a) requires sketching a trapezoid velocity-time graph with given algebraic expressions. Part (b) uses the area under the graph to form an equation (routine application of distance = area under v-t graph). Part (c) solves a quadratic equation. All steps are textbook-standard with no novel insight required, making it easier than average, though the algebraic manipulation with parameter x adds minor complexity compared to purely numerical questions.
Spec3.02c Interpret kinematic graphs: gradient and area3.02d Constant acceleration: SUVAT formulae

A train starts from rest at a station \(S\) and accelerates at a constant rate for \(2x\) seconds to a speed of \(5x\) ms\(^{-1}\). It maintains this speed until 126 seconds after it left \(S\) and then decelerates at a constant rate until it comes to rest at another station \(T\), \(20x\) seconds after it left \(S\).
  1. Sketch a velocity-time graph for this journey. [4 marks]
Given that the distance between \(S\) and \(T\) is \(5.4\) km,
  1. show that \(x^2 + 7x = 120\). [4 marks]
  2. Find the value of \(x\). [3 marks]

AnswerMarks Guidance
(a) Velocity-time graph with axes labeled, trapezium shape with height \(5x\), horizontal section from \(t=2x\) to \(t=12g\), sloped sections reaching \(t=20x\)B2 graph; B2 labelling
(b) Area = \(\frac{1}{2} \times 5x(20x + 126 - 2x) = 45x^2 + 315x = 5400\) (given)M1 M1 A1
\(\div 45: x^2 + 7x = 120\)A1
(c) \(x^2 + 7x - 120 = 0\) giving \((x-8)(x+15) = 0\) so \(x = 8\)M1 A1 A1 total: 11
(a) Velocity-time graph with axes labeled, trapezium shape with height $5x$, horizontal section from $t=2x$ to $t=12g$, sloped sections reaching $t=20x$ | B2 graph; B2 labelling | 

(b) Area = $\frac{1}{2} \times 5x(20x + 126 - 2x) = 45x^2 + 315x = 5400$ (given) | M1 M1 A1 |

$\div 45: x^2 + 7x = 120$ | A1 | 

(c) $x^2 + 7x - 120 = 0$ giving $(x-8)(x+15) = 0$ so $x = 8$ | M1 A1 A1 | total: 11
A train starts from rest at a station $S$ and accelerates at a constant rate for $2x$ seconds to a speed of $5x$ ms$^{-1}$. It maintains this speed until 126 seconds after it left $S$ and then decelerates at a constant rate until it comes to rest at another station $T$, $20x$ seconds after it left $S$.

\begin{enumerate}[label=(\alph*)]
\item Sketch a velocity-time graph for this journey. [4 marks]
\end{enumerate}

Given that the distance between $S$ and $T$ is $5.4$ km,
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item show that $x^2 + 7x = 120$. [4 marks]
\item Find the value of $x$. [3 marks]
\end{enumerate}

\hfill \mbox{\textit{Edexcel M1  Q4 [11]}}