| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2011 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable acceleration (1D) |
| Type | Variable acceleration with initial conditions |
| Difficulty | Moderate -0.3 This is a standard M2 variable acceleration question requiring straightforward integration twice (acceleration to velocity to displacement) and solving a quadratic. While it has multiple parts and requires careful bookkeeping of constants, it follows a completely routine template with no conceptual challenges or novel problem-solving required, making it slightly easier than average. |
| Spec | 3.02f Non-uniform acceleration: using differentiation and integration |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\frac{dv}{dt} = t - 4\) | ||
| \(v = \frac{1}{2}t^2 - 4t(+c)\) | M1 A1 | |
| \(t = 0\) \(v = 6\) \(\Rightarrow c = 6\) | M1 | |
| \(\therefore v = \frac{1}{2}t^2 - 4t + 6\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(v = 0\) \(\Rightarrow 0 = t^2 - 8t + 12\) | M1 | |
| \((t - 6)(t - 2) = 0\) | DM1 | |
| \(t = 6\) \(t = 2\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(x = \frac{t^3}{6} - 2t^2 + 6t + k\) | M1 A1 ft | |
| \(x_6 - x_2 = \frac{6^3}{6} - 2 \times 6^2 + 6 \times 6 + k - \left(\frac{2^3}{6} - 2 \times 2^2 + 6 \times 2 + k\right)\) | DM1 | |
| \(= -5\frac{1}{3}\) | ||
| \(\therefore\) Distance is \(5\frac{1}{3}\) m | A1 |
## Part (a):
| **Answer/Working** | **Marks** | **Guidance** |
|---|---|---|
| $\frac{dv}{dt} = t - 4$ | | |
| $v = \frac{1}{2}t^2 - 4t(+c)$ | M1 A1 | |
| $t = 0$ $v = 6$ $\Rightarrow c = 6$ | M1 | |
| $\therefore v = \frac{1}{2}t^2 - 4t + 6$ | A1 | |
## Part (b):
| **Answer/Working** | **Marks** | **Guidance** |
|---|---|---|
| $v = 0$ $\Rightarrow 0 = t^2 - 8t + 12$ | M1 | |
| $(t - 6)(t - 2) = 0$ | DM1 | |
| $t = 6$ $t = 2$ | A1 | |
## Part (c):
| **Answer/Working** | **Marks** | **Guidance** |
|---|---|---|
| $x = \frac{t^3}{6} - 2t^2 + 6t + k$ | M1 A1 ft | |
| $x_6 - x_2 = \frac{6^3}{6} - 2 \times 6^2 + 6 \times 6 + k - \left(\frac{2^3}{6} - 2 \times 2^2 + 6 \times 2 + k\right)$ | DM1 | |
| $= -5\frac{1}{3}$ | | |
| $\therefore$ Distance is $5\frac{1}{3}$ m | A1 | |A particle $P$ moves on the $x$-axis. The acceleration of $P$ at time $t$ seconds is $(t - 4)$ m s$^{-2}$ in the positive $x$-direction. The velocity of $P$ at time $t$ seconds is $v$ m s$^{-1}$. When $t = 0$, $v = 6$.
Find
\begin{enumerate}[label=(\alph*)]
\item $v$ in terms of $t$, [4]
\item the values of $t$ when $P$ is instantaneously at rest, [3]
\item the distance between the two points at which $P$ is instantaneously at rest. [4]
\end{enumerate}
\hfill \mbox{\textit{Edexcel M2 2011 Q6 [11]}}