Edexcel M2 — Question 1 7 marks

Exam BoardEdexcel
ModuleM2 (Mechanics 2)
Marks7
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicVariable acceleration (1D)
TypeVelocity from acceleration by integration
DifficultyModerate -0.3 This is a straightforward kinematics question requiring integration of acceleration to find velocity, then using given conditions to find the constant of integration. Part (b) involves solving a quadratic equation. While it requires multiple steps, the techniques are standard M2 material with no conceptual challenges or novel problem-solving required, making it slightly easier than average.
Spec3.02f Non-uniform acceleration: using differentiation and integration

The acceleration of a particle \(P\) is \((8t - 18)\) ms\(^{-2}\), where \(t\) seconds is the time that has elapsed since \(P\) passed through a fixed point \(O\) on the straight line on which it is moving. At time \(t = 3\), \(P\) has speed \(2\) ms\(^{-1}\). Find
  1. the velocity of \(P\) at time \(t\), [4 marks]
  2. the values of \(t\) when \(P\) is instantaneously at rest. [3 marks]

AnswerMarks Guidance
(a) \(v = \int a \, dt = 4t^2 - 18t + c\), \(v(3) = 2\): \(c = 20\), \(v = 4t^2 - 18t + 20\)M1 A1 M1 A1
(b) \(v = 0\): \(2(t-2)(2t-5) = 0\), \(t = 2, t = 2.5\)M1 A1 A1 Total: 7 marks
(a) $v = \int a \, dt = 4t^2 - 18t + c$, $v(3) = 2$: $c = 20$, $v = 4t^2 - 18t + 20$ | M1 A1 M1 A1 |

(b) $v = 0$: $2(t-2)(2t-5) = 0$, $t = 2, t = 2.5$ | M1 A1 A1 | Total: 7 marks
The acceleration of a particle $P$ is $(8t - 18)$ ms$^{-2}$, where $t$ seconds is the time that has elapsed since $P$ passed through a fixed point $O$ on the straight line on which it is moving.

At time $t = 3$, $P$ has speed $2$ ms$^{-1}$. Find
\begin{enumerate}[label=(\alph*)]
\item the velocity of $P$ at time $t$, [4 marks]
\item the values of $t$ when $P$ is instantaneously at rest. [3 marks]
\end{enumerate}

\hfill \mbox{\textit{Edexcel M2  Q1 [7]}}