| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2024 |
| Session | January |
| 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 mechanics question requiring standard differentiation techniques. Part (a) involves finding velocity by differentiating and solving a quadratic equation; part (b) requires evaluating displacement at endpoints and checking for direction changes; part (c) is simple second differentiation and substitution. All steps are routine M2 procedures with no problem-solving insight needed, making it easier than average. |
| Spec | 3.02a Kinematics language: position, displacement, velocity, acceleration3.02f Non-uniform acceleration: using differentiation and integration |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Use of \(v = \frac{\mathrm{d}x}{\mathrm{d}t}\) | M1 | At least 2 powers going down by 1. Clear division by \(t\) is M0 |
| \(v = 6t^2 - 42t + 60\) | A1 | Correct only |
| Set \(v = 0\) and correctly solves to obtain 2 values for \(t\) | M1 | Complete method to obtain both values (implied by correct answers seen) \(\left(0 = t^2 - 7t + 10 = (t-2)(t-5)\right)\) |
| Obtain \(t = 2\) and \(t = 5\) | A1 | Correct only. Allow 2.0, 5.0 |
| [4] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Distance \(= \ | x_2 - x_1\ | + \ |
| \(= 11 + 7 = 18\) (m) | A1 | Correct only |
| [2] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Use of \(a = \frac{\mathrm{d}v}{\mathrm{d}t}\) | M1 | Differentiate their \(v\). Clear division by \(t\) is M0. A power going down by 1 \((a = 12t - 42)\) |
| Obtain \(6\left(\mathrm{ms}^{-2}\right)\) | A1 | Must be positive – the Q asks for magnitude |
| [2] |
## Question 1:
**Part 1a:**
| Answer/Working | Mark | Guidance |
|---|---|---|
| Use of $v = \frac{\mathrm{d}x}{\mathrm{d}t}$ | M1 | At least 2 powers going down by 1. Clear division by $t$ is M0 |
| $v = 6t^2 - 42t + 60$ | A1 | Correct only |
| Set $v = 0$ and correctly solves to obtain 2 values for $t$ | M1 | Complete method to obtain both values (implied by correct answers seen) $\left(0 = t^2 - 7t + 10 = (t-2)(t-5)\right)$ |
| Obtain $t = 2$ and $t = 5$ | A1 | Correct only. Allow 2.0, 5.0 |
| **[4]** | | |
**Part 1b:**
| Answer/Working | Mark | Guidance |
|---|---|---|
| Distance $= \|x_2 - x_1\| + \|x_3 - x_2\|$ $\left(= \|45-52\| + \|52-41\|\right)$ | M1 | Correct strategy dependent on their $t$ being in $1 < t < 3$ |
| $= 11 + 7 = 18$ (m) | A1 | Correct only |
| **[2]** | | |
**Part 1c:**
| Answer/Working | Mark | Guidance |
|---|---|---|
| Use of $a = \frac{\mathrm{d}v}{\mathrm{d}t}$ | M1 | Differentiate their $v$. Clear division by $t$ is M0. A power going down by 1 $(a = 12t - 42)$ |
| Obtain $6\left(\mathrm{ms}^{-2}\right)$ | A1 | Must be positive – the Q asks for magnitude |
| **[2]** | | |
---\begin{enumerate}
\item A particle $P$ moves along a straight line. The fixed point $O$ is on the line. At time $t$ seconds, $t > 0$, the displacement of $P$ from $O$ is $x$ metres, where
\end{enumerate}
$$x = 2 t ^ { 3 } - 21 t ^ { 2 } + 60 t$$
Find\\
(a) the values of $t$ for which $P$ is instantaneously at rest\\
(b) the distance travelled by $P$ in the interval $1 \leqslant t \leqslant 3$\\
(c) the magnitude of the acceleration of $P$ at the instant when $t = 3$
\hfill \mbox{\textit{Edexcel M2 2024 Q1 [8]}}