Edexcel M2 2003 January — Question 5 11 marks

Exam BoardEdexcel
ModuleM2 (Mechanics 2)
Year2003
SessionJanuary
Marks11
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicVariable acceleration (1D)
TypeVariable acceleration with initial conditions
DifficultyStandard +0.3 This is a standard M2 integration question requiring integration of acceleration to find velocity, then solving a quadratic to find rest points, and finally integrating velocity to find displacement. While it involves multiple steps (11 marks total), each step follows routine procedures with no novel insight required—slightly easier than average due to straightforward setup and clean arithmetic.
Spec1.08d Evaluate definite integrals: between limits3.02a Kinematics language: position, displacement, velocity, acceleration3.02f Non-uniform acceleration: using differentiation and integration

A particle \(P\) moves on the \(x\)-axis. The acceleration of \(P\) at time \(t\) seconds is \((4t - 8)\) m s\(^{-2}\), measured in the direction of \(x\) increasing. The velocity of \(P\) at time \(t\) seconds is \(v\) m s\(^{-1}\). Given that \(v = 6\) when \(t = 0\), find
  1. \(v\) in terms of \(t\), [4]
  2. the distance between the two points where \(P\) is instantaneously at rest. [7]

Part (a)
AnswerMarks Guidance
\(v = \int a \, dt = 2t^2 - 8t \,(+c)\)M1 A1
Using \(v = 6\), \(t = 0\): \(v = 2t^2 - 8t + 6\)M1 A1 (4 marks)
\(v = 0 \Rightarrow 2t^2 - 8t + 6 = 0, \Rightarrow t = 1, 3\)M1 A1
\(S = \int (2t^2 - 8t + 6) \, dt = \left[\frac{2}{3}t^3 - 4t^2 + 6t\right]\)M1 A2, 1, 0
\(= 0 - 2\frac{2}{3}\)M1
Distance is \((\pm)2\frac{2}{3}\) mA1 (7 marks)
(11 marks total)
## Part (a)

$v = \int a \, dt = 2t^2 - 8t \,(+c)$ | M1 A1 |

Using $v = 6$, $t = 0$: $v = 2t^2 - 8t + 6$ | M1 A1 | **(4 marks)**

$v = 0 \Rightarrow 2t^2 - 8t + 6 = 0, \Rightarrow t = 1, 3$ | M1 A1 |

$S = \int (2t^2 - 8t + 6) \, dt = \left[\frac{2}{3}t^3 - 4t^2 + 6t\right]$ | M1 A2, 1, 0 |

$= 0 - 2\frac{2}{3}$ | M1 |

Distance is $(\pm)2\frac{2}{3}$ m | A1 | **(7 marks)**

**(11 marks total)**

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A particle $P$ moves on the $x$-axis. The acceleration of $P$ at time $t$ seconds is $(4t - 8)$ m s$^{-2}$, measured in the direction of $x$ increasing. The velocity of $P$ at time $t$ seconds is $v$ m s$^{-1}$. Given that $v = 6$ when $t = 0$, find

\begin{enumerate}[label=(\alph*)]
\item $v$ in terms of $t$, [4]
\item the distance between the two points where $P$ is instantaneously at rest. [7]
\end{enumerate}

\hfill \mbox{\textit{Edexcel M2 2003 Q5 [11]}}