| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable acceleration (1D) |
| Type | Total distance with direction changes |
| Difficulty | Moderate -0.8 This is a straightforward kinematics question requiring basic differentiation to find acceleration and integration for distance. Part (a) involves finding when v=0 then differentiating; part (b) requires recognizing that the particle changes direction (v is negative initially) and using integration with appropriate limits. While it requires careful attention to the sign of velocity, these are standard M2 techniques with no conceptual challenges beyond routine application. |
| Spec | 1.07b Gradient as rate of change: dy/dx notation1.08d Evaluate definite integrals: between limits3.02f Non-uniform acceleration: using differentiation and integration |
| Answer | Marks | Guidance |
|---|---|---|
| (a) When \(v = 0\), \(4t^2 = 9\), \(t = 1.5\) | \(a = 8t = 12 \text{ ms}^{-2}\) | M1 A1 A1 |
| (b) \(s = \int_0^{1.5} v \, dt = \left[\frac{4}{3}t^3 - 9t\right]_0^{1.5} = 4.5 - 13.5\), so distance = 9 m | M1 M1 A1 A1 | 7 marks |
**(a)** When $v = 0$, $4t^2 = 9$, $t = 1.5$ | $a = 8t = 12 \text{ ms}^{-2}$ | M1 A1 A1 |
**(b)** $s = \int_0^{1.5} v \, dt = \left[\frac{4}{3}t^3 - 9t\right]_0^{1.5} = 4.5 - 13.5$, so distance = 9 m | M1 M1 A1 A1 | 7 marksThe velocity, $v$ ms$^{-1}$, of a particle at time $t$ s is given by $v = 4t^2 - 9$.
\begin{enumerate}[label=(\alph*)]
\item Find the acceleration of the particle when it is instantaneously at rest. [3 marks]
\item Find the distance travelled by the particle from time $t = 0$ until it comes to rest. [4 marks]
\end{enumerate}
\hfill \mbox{\textit{Edexcel M2 Q2 [7]}}