Edexcel M2 — Question 3 9 marks

Exam BoardEdexcel
ModuleM2 (Mechanics 2)
Marks9
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicVariable acceleration (1D)
TypeTotal distance with direction changes
DifficultyModerate -0.3 This is a straightforward M2 kinematics question requiring basic calculus operations: substituting t=0 for initial velocity, solving a quadratic equation to find when v=0, and integrating velocity to find displacement. All techniques are standard with no conceptual challenges or novel problem-solving required, making it slightly easier than average.
Spec3.02f Non-uniform acceleration: using differentiation and integration

3. A particle moves in a straight horizontal line such that its velocity, \(v \mathrm {~m} \mathrm {~s} ^ { - 1 }\), at time \(t\) seconds is given by \(v = 2 t ^ { 2 } - 9 t + 4\). Initially, the particle has displacement 9 m from a fixed point \(O\) on the line.
  1. Find the initial velocity of the particle.
  2. Show that the particle is at rest when \(t = 4\) and find the other value of \(t\) when it is at rest.
  3. Find the displacement of the particle from \(O\) when \(t = 6\).

Part (a):
AnswerMarks
when \(t = 0\), \(v = 4 \text{ ms}^{-1}\)A1
Part (b):
AnswerMarks
particle at rest when \(2t^2 - 9t + 4 = 0\) i.e. \((2t-1)(t-4) = 0\)M1 A1
\(t = \frac{1}{2}, 4\)A1
Part (c):
AnswerMarks Guidance
\(s = \int v dt = \frac{2}{3}t^3 - \frac{9}{2}t^2 + 4t + c\)M1 A1
when \(t = 0\), \(s = 9\) so \(c = 9\) \(\therefore s = \frac{2}{3}t^3 - \frac{9}{2}t^2 + 4t + 9\)A1
disp. when \(t = 6\) is \(\frac{2}{3}(6)^3 - \frac{9}{2}(6)^2 + 4(6) + 9\)M1
\(= 144 - 162 + 24 + 9 = 15 \text{ m}\)A1 (9)
**Part (a):**
when $t = 0$, $v = 4 \text{ ms}^{-1}$ | A1 |

**Part (b):**
particle at rest when $2t^2 - 9t + 4 = 0$ i.e. $(2t-1)(t-4) = 0$ | M1 A1 |

$t = \frac{1}{2}, 4$ | A1 |

**Part (c):**
$s = \int v dt = \frac{2}{3}t^3 - \frac{9}{2}t^2 + 4t + c$ | M1 A1 |

when $t = 0$, $s = 9$ so $c = 9$ $\therefore s = \frac{2}{3}t^3 - \frac{9}{2}t^2 + 4t + 9$ | A1 |

disp. when $t = 6$ is $\frac{2}{3}(6)^3 - \frac{9}{2}(6)^2 + 4(6) + 9$ | M1 |

$= 144 - 162 + 24 + 9 = 15 \text{ m}$ | A1 | (9)

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3. A particle moves in a straight horizontal line such that its velocity, $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$, at time $t$ seconds is given by $v = 2 t ^ { 2 } - 9 t + 4$. Initially, the particle has displacement 9 m from a fixed point $O$ on the line.
\begin{enumerate}[label=(\alph*)]
\item Find the initial velocity of the particle.
\item Show that the particle is at rest when $t = 4$ and find the other value of $t$ when it is at rest.
\item Find the displacement of the particle from $O$ when $t = 6$.
\end{enumerate}

\hfill \mbox{\textit{Edexcel M2  Q3 [9]}}