| Exam Board | Edexcel |
|---|---|
| Module | FM2 AS (Further Mechanics 2 AS) |
| Year | 2024 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable acceleration (1D) |
| Type | Acceleration as function of velocity (separation of variables) |
| Difficulty | Standard +0.3 This is a standard Further Maths variable acceleration question requiring separation of variables and integration with initial conditions. The differential equation a = dv/dt = 4 - 3v is straightforward to separate and integrate, and part (b) requires a second integration. While it's FM content, it follows a well-established template with no conceptual surprises, making it slightly easier than average overall. |
| Spec | 1.08d Evaluate definite integrals: between limits1.08k Separable differential equations: dy/dx = f(x)g(y)3.02f Non-uniform acceleration: using differentiation and integration |
| Answer | Marks | Guidance |
|---|---|---|
| \(a = 4 - 3v \Rightarrow \int \frac{1}{4-3v}dv = \int 1dt\) | M1 | 2.1 |
| Integrate both sides of the equation | M1 | 1.1b |
| \(\Rightarrow -\frac{1}{3}\ln\ | 4-3v\ | = t(+C)\) |
| Use \(t=0, v=0\) | M1 | 3.4 |
| \(t = \frac{1}{3}\ln\left(\frac{4}{4-3v}\right)\) | A1 | 1.1b |
| \(\Rightarrow e^{3t} = \frac{4}{4-3v}, \quad 4-3v = 4e^{-3t}\) | M1 | 1.1b |
| \(v = \frac{4}{3}\left(1 - e^{-3t}\right)\) * | A1* | 2.2a |
| (7) |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{dx}{dt} = k\left(1 - e^{-3t}\right) \Rightarrow \int 1dx = \int k\left(1 - e^{-3t}\right)dt\) | M1 | 3.3 |
| \(x = k\left(t + \frac{1}{3}e^{-3t}\right)(+C) = \frac{4}{3}\left(t + \frac{1}{3}e^{-3t}\right)(+C)\) | M1 | 1.1b |
| Use \(t=0, x=0\) | M1 | 3.4 |
| \(x = k\left(t + \frac{1}{3}e^{-3t} - \frac{1}{3}\right) = \frac{4}{3}\left(t + \frac{1}{3}e^{-3t} - \frac{1}{3}\right)\) | A1ft | 1.1b |
| (4) | ||
| (11 marks) |
**3a**
| $a = 4 - 3v \Rightarrow \int \frac{1}{4-3v}dv = \int 1dt$ | M1 | 2.1 |
| Integrate both sides of the equation | M1 | 1.1b |
| $\Rightarrow -\frac{1}{3}\ln\|4-3v\| = t(+C)$ | A1 | 1.1b |
| Use $t=0, v=0$ | M1 | 3.4 |
| $t = \frac{1}{3}\ln\left(\frac{4}{4-3v}\right)$ | A1 | 1.1b |
| $\Rightarrow e^{3t} = \frac{4}{4-3v}, \quad 4-3v = 4e^{-3t}$ | M1 | 1.1b |
| $v = \frac{4}{3}\left(1 - e^{-3t}\right)$ * | A1* | 2.2a |
| | (7) | |
**3b**
| $\frac{dx}{dt} = k\left(1 - e^{-3t}\right) \Rightarrow \int 1dx = \int k\left(1 - e^{-3t}\right)dt$ | M1 | 3.3 |
| $x = k\left(t + \frac{1}{3}e^{-3t}\right)(+C) = \frac{4}{3}\left(t + \frac{1}{3}e^{-3t}\right)(+C)$ | M1 | 1.1b |
| Use $t=0, x=0$ | M1 | 3.4 |
| $x = k\left(t + \frac{1}{3}e^{-3t} - \frac{1}{3}\right) = \frac{4}{3}\left(t + \frac{1}{3}e^{-3t} - \frac{1}{3}\right)$ | A1ft | 1.1b |
| | (4) | |
| | (11 marks) | |
**Notes for 3a:**
- M1: Use $a = \frac{dv}{dt}$ and separate the variables to form integrals in $v$ and $t$
- M1: Integrate to obtain terms $p\ln(4-3v)$ and $qt$
- A1: Or equivalent. Accept with brackets in place of modulus signs. Condone missing constant of integration
- M1: Use boundary conditions in the model to evaluate constant of integration or as limits on a definite integral
- A1: Or equivalent
- M1: Rearrange to obtain expression for $v$ in terms of $t$
- A1*: Obtain given form with $k = \frac{4}{3}$ from correct working
**Notes for 3b:**
- M1: Use $v = \frac{dx}{dt}$ to form integrals in $x$ and $t$
- M1: Integrate to obtain $\lambda t + \mu e^{-3t}(+C)$
- M1: Use boundary conditions in the model to evaluate constant of integration or as limits on a definite integral
- A1 ft: Any equivalent form. Follow their $k$
---\begin{enumerate}
\item A particle $P$ is moving along the $x$-axis. At time $t$ seconds, $P$ has velocity $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$ in the positive $x$ direction and acceleration $a \mathrm {~ms} ^ { - 2 }$ in the positive $x$ direction.
\end{enumerate}
In a model of the motion of $P$
$$a = 4 - 3 v$$
When $t = 0 , v = 0$\\
(a) Use integration to show that $v = k \left( 1 - \mathrm { e } ^ { - 3 t } \right)$, where $k$ is a constant to be found.
When $t = 0 , P$ is at the origin $O$\\
(b) Find, in terms of $t$ only, the distance of $P$ from $O$ at time $t$ seconds.
\hfill \mbox{\textit{Edexcel FM2 AS 2024 Q3 [11]}}