| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2009 |
| Session | January |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable acceleration (1D) |
| Type | Piecewise motion functions |
| Difficulty | Standard +0.3 This is a straightforward piecewise motion problem requiring integration of given velocity functions to find displacement. Students must integrate two different expressions and add results, but the integrations are routine (polynomial and power rule) with no conceptual challenges beyond recognizing the piecewise structure. |
| Spec | 3.02a Kinematics language: position, displacement, velocity, acceleration3.02b Kinematic graphs: displacement-time and velocity-time3.02f Non-uniform acceleration: using differentiation and integration |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(v = 10t - 2t^2,\ s = \int v\, dt\) | M1 | |
| \(= 5t^2 - \dfrac{2t^3}{3}\ (+C)\) | A1 | |
| \(t = 6 \Rightarrow s = 180 - 144 = \underline{36}\) (m) | A1 | |
| Total: (3) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(s = \int v\, dt = \dfrac{-432t^{-1}}{-1}(+K) = \dfrac{432}{t}(+K)\) | B1 | |
| \(t = 6,\ s = \text{"36"} \Rightarrow 36 = \dfrac{432}{6} + K\) | M1* | |
| \(\Rightarrow K = -36\) | A1 | |
| At \(t = 10,\ s = \dfrac{432}{10} - 36 = \underline{7.2}\) (m) | d*M1, A1 | |
| Total: [8] |
## Question 4:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $v = 10t - 2t^2,\ s = \int v\, dt$ | M1 | |
| $= 5t^2 - \dfrac{2t^3}{3}\ (+C)$ | A1 | |
| $t = 6 \Rightarrow s = 180 - 144 = \underline{36}$ (m) | A1 | |
| **Total: (3)** | | |
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $s = \int v\, dt = \dfrac{-432t^{-1}}{-1}(+K) = \dfrac{432}{t}(+K)$ | B1 | |
| $t = 6,\ s = \text{"36"} \Rightarrow 36 = \dfrac{432}{6} + K$ | M1* | |
| $\Rightarrow K = -36$ | A1 | |
| At $t = 10,\ s = \dfrac{432}{10} - 36 = \underline{7.2}$ (m) | d*M1, A1 | |
| **Total: [8]** | | |
---4. A particle $P$ moves along the $x$-axis in a straight line so that, at time $t$ seconds, the velocity of $P$ is $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$, where
$$v = \begin{cases} 10 t - 2 t ^ { 2 } , & 0 \leqslant t \leqslant 6 \\ \frac { - 432 } { t ^ { 2 } } , & t > 6 \end{cases}$$
At $t = 0 , P$ is at the origin $O$. Find the displacement of $P$ from $O$ when
\begin{enumerate}[label=(\alph*)]
\item $t = 6$,
\item $t = 10$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M2 2009 Q4 [8]}}