| Exam Board | Edexcel |
|---|---|
| Module | FM2 AS (Further Mechanics 2 AS) |
| Session | Specimen |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable acceleration (1D) |
| Type | Total distance with direction changes |
| Difficulty | Standard +0.3 This is a straightforward Further Maths mechanics question requiring differentiation of velocity for acceleration, integration with attention to sign changes for distance, and analysis of displacement. While it involves non-constant acceleration and requires careful handling of the particle changing direction, the techniques are standard and the algebraic manipulation is routine. Slightly above average difficulty due to being FM content and requiring part (c)'s reasoning, but well within expected FM2 AS level. |
| Spec | 1.08d Evaluate definite integrals: between limits3.02a Kinematics language: position, displacement, velocity, acceleration3.02f Non-uniform acceleration: using differentiation and integration |
| Answer | Marks | Guidance |
|---|---|---|
| Multiply out and differentiate wrt \(t\) | M1 | 1.1b |
| \(v = 3t^2 - 16t + 20 \Rightarrow a = 6t - 16\) | A1 | 1.1b |
| Answer | Marks | Guidance |
|---|---|---|
| Multiply out and integrate wrt \(t\) | M1 | 1.1b |
| \(s = \int 3t^2 - 16t + 20 \, dt = t^3 - 8t^2 + 20t (+C)\) | A1 | 1.1b |
| Answer | Marks | Guidance |
|---|---|---|
| \(t = 2, s = 8 - 32 + 40 = 16\) | A1 | 1.1b |
| Answer | Marks | Guidance |
|---|---|---|
| \(s = 0 \Rightarrow t^3 - 8t^2 + 20t = 0\) and \(t \neq 0 \Rightarrow t^2 - 8t + 20 = 0\) | M1 | 2.1 |
| Explanation to show that \(t^2 - 8t + 20 > 0\) for all \(t\) | M1 | 2.4 |
| So \(s = 0\) has no non-zero solutions, so \(s\) is never zero again, so never returns to \(O\) | A1* | 3.2a |
# Question 1
## 1(a)
Multiply out and differentiate wrt $t$ | M1 | 1.1b
$v = 3t^2 - 16t + 20 \Rightarrow a = 6t - 16$ | A1 | 1.1b
(2)
## 1(b)
Multiply out and integrate wrt $t$ | M1 | 1.1b
$s = \int 3t^2 - 16t + 20 \, dt = t^3 - 8t^2 + 20t (+C)$ | A1 | 1.1b
$t = 0, s = 0 \Rightarrow C = 0$
$t = 2, s = 8 - 32 + 40 = 16$ | A1 | 1.1b
(3)
## 1(c)
$s = 0 \Rightarrow t^3 - 8t^2 + 20t = 0$ and $t \neq 0 \Rightarrow t^2 - 8t + 20 = 0$ | M1 | 2.1
Explanation to show that $t^2 - 8t + 20 > 0$ for all $t$ | M1 | 2.4
So $s = 0$ has no non-zero solutions, so $s$ is never zero again, so never returns to $O$ | A1* | 3.2a
(3)
## Notes
**(a)**
- M1: For multiplying out and differentiating (powers decreasing by 1)
- A1: For a correct expression for $a$
**(b)**
- M1: For multiplying out and integrating (powers increasing by 1)
- A1: For a correct expression for $s$ with or without $C$
- A1: For $C = 0$ and correct final answer
**(c)**
- M1: For equating their $s$ to 0 and producing a quadratic
- M1: For clear explanation that $t^2 - 8t + 20 > 0$ for all $t$ (e.g. completing the square or another complete method)
- A1*: For a correct conclusion in context\begin{enumerate}
\item A particle $P$ moves on the $x$-axis. At time $t$ seconds the velocity of $P$ is $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$ in the direction of $x$ increasing, where
\end{enumerate}
$$v = ( t - 2 ) ( 3 t - 10 ) , \quad t \geqslant 0$$
When $t = 0 , P$ is at the origin $O$.\\
(a) Find the acceleration of $P$ at time $t$ seconds.\\
(b) Find the total distance travelled by $P$ in the first 2 seconds of its motion.\\
(c) Show that $P$ never returns to $O$, explaining your reasoning.\\
\hfill \mbox{\textit{Edexcel FM2 AS Q1 [8]}}