AQA M2 2008 June — Question 1 8 marks

Exam BoardAQA
ModuleM2 (Mechanics 2)
Year2008
SessionJune
Marks8
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicVariable acceleration (1D)
TypeAcceleration from velocity differentiation
DifficultyEasy -1.2 This is a straightforward mechanics question requiring only standard differentiation and integration of polynomials, plus direct application of F=ma. All three parts are routine textbook exercises with no problem-solving or conceptual challenges—simpler than average A-level questions.
Spec1.08b Integrate x^n: where n != -1 and sums3.02f Non-uniform acceleration: using differentiation and integration3.03b Newton's first law: equilibrium

1 A particle moves in a straight line and at time \(t\) seconds has velocity \(v \mathrm {~m} \mathrm {~s} ^ { - 1 }\), where $$v = 6 t ^ { 2 } + 4 t - 7 , \quad t \geqslant 0$$
  1. Find an expression for the acceleration of the particle at time \(t\).
  2. The mass of the particle is 3 kg . Find the resultant force on the particle when \(t = 4\).
  3. When \(t = 0\), the displacement of the particle from the origin is 5 metres. Find an expression for the displacement of the particle from the origin at time \(t\).

Part (a)
AnswerMarks Guidance
\(a = \frac{dv}{dt} = 12t + 4\)M1 A1 2 marks
Part (b)
AnswerMarks Guidance
Using \(F = ma\), Force \(= 3 \times (12t + 4)\). When \(t = 4\), force \(= 3(12 \times 4 + 4) = 156\) NM1 A1 2 marks
Part (c)
AnswerMarks Guidance
\(r = 2t^3 + 2t^2 - 7t + c\). When \(t = 0, r = 5\), so \(c = 5\). Therefore \(r = 2t^3 + 2t^2 - 7t + 5\)M1 A1 M1 A1 4 marks
Total: 8 marks
**Part (a)**
$a = \frac{dv}{dt} = 12t + 4$ | M1 A1 | 2 marks

**Part (b)**
Using $F = ma$, Force $= 3 \times (12t + 4)$. When $t = 4$, force $= 3(12 \times 4 + 4) = 156$ N | M1 A1 | 2 marks

**Part (c)**
$r = 2t^3 + 2t^2 - 7t + c$. When $t = 0, r = 5$, so $c = 5$. Therefore $r = 2t^3 + 2t^2 - 7t + 5$ | M1 A1 M1 A1 | 4 marks | SC3 if no '+c' seen

**Total: 8 marks**

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1 A particle moves in a straight line and at time $t$ seconds has velocity $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$, where

$$v = 6 t ^ { 2 } + 4 t - 7 , \quad t \geqslant 0$$
\begin{enumerate}[label=(\alph*)]
\item Find an expression for the acceleration of the particle at time $t$.
\item The mass of the particle is 3 kg .

Find the resultant force on the particle when $t = 4$.
\item When $t = 0$, the displacement of the particle from the origin is 5 metres.

Find an expression for the displacement of the particle from the origin at time $t$.
\end{enumerate}

\hfill \mbox{\textit{AQA M2 2008 Q1 [8]}}