| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2013 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable acceleration (1D) |
| Type | Total distance with direction changes |
| Difficulty | Moderate -0.3 This is a straightforward M2 kinematics question requiring standard techniques: (a) solving a quadratic equation when v=0, and (b) integrating velocity to find displacement, identifying when the particle changes direction, and calculating total distance. While it requires careful attention to when the particle reverses direction, the methods are routine for M2 students with no novel problem-solving required. |
| Spec | 3.02f Non-uniform acceleration: using differentiation and integration |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{1}{2}t^2 - 3t + 4 = 0\) | M1 | Set \(v = 0\) |
| \(t^2 - 6t + 8 = 0\) | ||
| \((t-2)(t-4) = 0\) | DM1 | Solve for v |
| \(t = 2\) s or \(4\) s | A1 A1 | |
| (4) |
| Answer | Marks | Guidance |
|---|---|---|
| \(\int \frac{1}{2}t^2 - 3t + 4dt\) | M1 | Integration – majority of powers increasing |
| \(= \frac{1}{6}t^3 - \frac{3}{2}t^2 + 4t(+C)\) | A1 | Correct (+C not required) |
| \(s = \int_0^2 \frac{1}{2}t^2 - 3t + 4 dt - \int_2^4 \frac{1}{2}t^2 - 3t + 4 dt\) | DM1 | Correct strategy for finding the required distance. Follow their "2". Subtraction/swap limits/modulus signs |
| \(= \left[\frac{1}{6}t^3 - \frac{3}{2}t^2 + 4t\right]_0^2 - \left[\frac{1}{6}t^3 - \frac{3}{2}t^2 + 4t\right]_2^4\) | ||
| \(= \frac{8}{6} - 6 + 8 - (\frac{64}{6} - 24 + 16 - (\frac{8}{6} - 6 + 8))\) | A1 | Correct unsimplified |
| \(= \frac{10}{6} - \frac{8}{3} + \frac{10}{3}\) | ||
| \(= 4\) | A1 | |
| (5) | ||
| [9] |
**Part (a):**
| $\frac{1}{2}t^2 - 3t + 4 = 0$ | M1 | Set $v = 0$ |
| $t^2 - 6t + 8 = 0$ | | |
| $(t-2)(t-4) = 0$ | DM1 | Solve for v |
| $t = 2$ s or $4$ s | A1 A1 | |
| | (4) | |
**Part (b):**
| $\int \frac{1}{2}t^2 - 3t + 4dt$ | M1 | Integration – majority of powers increasing |
| $= \frac{1}{6}t^3 - \frac{3}{2}t^2 + 4t(+C)$ | A1 | Correct (+C not required) |
| $s = \int_0^2 \frac{1}{2}t^2 - 3t + 4 dt - \int_2^4 \frac{1}{2}t^2 - 3t + 4 dt$ | DM1 | Correct strategy for finding the required distance. Follow their "2". Subtraction/swap limits/modulus signs |
| $= \left[\frac{1}{6}t^3 - \frac{3}{2}t^2 + 4t\right]_0^2 - \left[\frac{1}{6}t^3 - \frac{3}{2}t^2 + 4t\right]_2^4$ | | |
| $= \frac{8}{6} - 6 + 8 - (\frac{64}{6} - 24 + 16 - (\frac{8}{6} - 6 + 8))$ | A1 | Correct unsimplified |
| $= \frac{10}{6} - \frac{8}{3} + \frac{10}{3}$ | | |
| $= 4$ | A1 | |
| | (5) |
| | [9] | |
---A particle $P$ moves along a straight line in such a way that at time $t$ seconds its velocity $v$ m s$^{-1}$ is given by
$$v = \frac{1}{2}t^2 - 3t + 4$$
Find
\begin{enumerate}[label=(\alph*)]
\item the times when $P$ is at rest, [4]
\item the total distance travelled by $P$ between $t = 0$ and $t = 4$. [5]
\end{enumerate}
\hfill \mbox{\textit{Edexcel M2 2013 Q3 [9]}}