| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable acceleration (1D) |
| Type | Finding when particle at rest |
| Difficulty | Moderate -0.8 This is a straightforward M2 question requiring only basic differentiation of a polynomial and solving simple equations. Part (a) involves factorising a cubic (which factors easily), and part (b) requires differentiating to find velocity and solving a quadratic. Both are routine procedures with no problem-solving insight needed, making this easier than average. |
| Spec | 3.02f Non-uniform acceleration: using differentiation and integration |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Notes |
| (a) \(s = t(2t^2 - 13t + 20) = t(2t-5)(t-4)\) | M1 A1 | |
| particle at \(O\) when \(s = 0\) \(\therefore\) at \(t = 0,\ \frac{5}{2},\ 4\) seconds | M1 A1 | |
| (b) at rest when \(v = 0\), \(v = \frac{ds}{dt} = 6t^2 - 26t + 20\) | M1 A1 | |
| \(\therefore 3t^2 - 13t + 10 = 0\), \((t-1)(3t-10) = 0\) | M1 | |
| \(t = 1,\ \frac{10}{3}\) seconds | A1 | (8) |
## Question 3:
| Answer/Working | Marks | Notes |
|---|---|---|
| **(a)** $s = t(2t^2 - 13t + 20) = t(2t-5)(t-4)$ | M1 A1 | |
| particle at $O$ when $s = 0$ $\therefore$ at $t = 0,\ \frac{5}{2},\ 4$ seconds | M1 A1 | |
| **(b)** at rest when $v = 0$, $v = \frac{ds}{dt} = 6t^2 - 26t + 20$ | M1 A1 | |
| $\therefore 3t^2 - 13t + 10 = 0$, $(t-1)(3t-10) = 0$ | M1 | |
| $t = 1,\ \frac{10}{3}$ seconds | A1 | **(8)** |
---3. A particle moves along a straight horizontal track such that its displacement, $s$ metres, from a fixed point $O$ on the line after $t$ seconds is given by
$$s = 2 t ^ { 3 } - 13 t ^ { 2 } + 20 t$$
\begin{enumerate}[label=(\alph*)]
\item Find the values of $t$ for which the particle is at $O$.
\item Find the values of $t$ at which the particle comes instantaneously to rest.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M2 Q3 [8]}}