Edexcel M1 — Question 6 14 marks

Exam BoardEdexcel
ModuleM1 (Mechanics 1)
Marks14
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicTravel graphs
TypeMulti-stage motion with velocity-time graph given
DifficultyStandard +0.3 This is a standard M1 velocity-time graph question requiring area calculations and kinematic equations. Part (a) involves finding areas under trapezoids and solving simultaneous equations (routine). Parts (b) and (c) apply the same techniques to a second scenario. While multi-step with 14 marks total, it requires only straightforward application of well-practiced methods with no novel insight or complex problem-solving.
Spec3.02c Interpret kinematic graphs: gradient and area3.02d Constant acceleration: SUVAT formulae

The diagram shows the velocity-time graph for a cyclist's journey. Each section has constant acceleration or deceleration and the three sections are of equal duration \(x\) seconds each. \includegraphics{figure_6} Given that the total distance travelled is \(792\) m,
  1. find the value of \(x\) and the acceleration for the first section of the journey. [6 marks]
Another cyclist covers the same journey in three sections of equal duration, accelerating at \(\frac{1}{11} \text{ ms}^{-2}\) for the first section, travelling at constant speed for the second section and decelerating at \(\frac{1}{11} \text{ ms}^{-2}\) for the third section.
  1. Find the time taken by this cyclist to complete the journey. [6 marks]
  2. Show that the maximum speeds of both cyclists are the same. [2 marks]

AnswerMarks Guidance
(a) Sum of areas = \(3x + 5.5x + 2.5x = 792\) → \(11x = 792\) → \(x = 72\)M1 A1 M1 A1
Acc. = \(6 + 72 = \frac{1}{12}\) ms\(^{-2}\)M1 A1
(b) Area under new graph = \(\frac{1}{2}(3t)(t) + (11t) = 792\) → \(4t^2 = 22 \times 792\)M1 M1 A1
\(t^2 = 4356\) → \(t = 66\) → Total time = \(3t = 198\) sM1 A1 A1
(c) \(v_{\max} = 66 x \frac{1}{11} = 6\) ms\(^{-1}\), as for first cyclistM1 A1 14 marks
**(a)** Sum of areas = $3x + 5.5x + 2.5x = 792$ → $11x = 792$ → $x = 72$ | M1 A1 M1 A1 |
Acc. = $6 + 72 = \frac{1}{12}$ ms$^{-2}$ | M1 A1 |

**(b)** Area under new graph = $\frac{1}{2}(3t)(t) + (11t) = 792$ → $4t^2 = 22 \times 792$ | M1 M1 A1 |
$t^2 = 4356$ → $t = 66$ → Total time = $3t = 198$ s | M1 A1 A1 |

**(c)** $v_{\max} = 66 x \frac{1}{11} = 6$ ms$^{-1}$, as for first cyclist | M1 A1 | 14 marks |
The diagram shows the velocity-time graph for a cyclist's journey. Each section has constant acceleration or deceleration and the three sections are of equal duration $x$ seconds each.

\includegraphics{figure_6}

Given that the total distance travelled is $792$ m,

\begin{enumerate}[label=(\alph*)]
\item find the value of $x$ and the acceleration for the first section of the journey. [6 marks]
\end{enumerate}

Another cyclist covers the same journey in three sections of equal duration, accelerating at $\frac{1}{11} \text{ ms}^{-2}$ for the first section, travelling at constant speed for the second section and decelerating at $\frac{1}{11} \text{ ms}^{-2}$ for the third section.

\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item Find the time taken by this cyclist to complete the journey. [6 marks]
\item Show that the maximum speeds of both cyclists are the same. [2 marks]
\end{enumerate}

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