| Exam Board | Edexcel |
|---|---|
| Module | FM2 (Further Mechanics 2) |
| Year | 2024 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable acceleration (1D) |
| Type | Velocity from acceleration by integration |
| Difficulty | Standard +0.3 This is a straightforward Further Maths mechanics question requiring integration of a given acceleration function with initial conditions. Part (a) involves a single integration with substitution (or recognition of chain rule), part (b) is a simple limit, and part (c) requires a second integration. While it's FM2 content, the techniques are routine calculus applications with no problem-solving insight required—slightly easier than an average A-level question overall due to its mechanical nature. |
| Spec | 1.08a Fundamental theorem of calculus: integration as reverse of differentiation1.08d Evaluate definite integrals: between limits3.02f Non-uniform acceleration: using differentiation and integration |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\int \frac{96}{(3t+5)^3} \, dt = \int 1 \, dv \Rightarrow v = \ldots\) | M1 | Form a differential equation in \(v\) and \(t\) and integrate. Must attempt integration of \(\frac{k}{(3t+5)^3}\). RHS can be implied. |
| \(-\frac{96}{2 \times 3 \times (3t+5)^2}(+C) = v\) | A1 | Correct integration. Ignore any limits. Accept without constant of integration. |
| Use limits \(v = 0, t = 0\) | M1 | Use \(v=0, t=0\) as limits in a definite integral or to find the constant of integration. |
| \(v = \frac{96}{6 \times (5)^2} - \frac{96}{6 \times (3t+5)^2} = \frac{16}{25} - \frac{16}{(3t+5)^2}\) | A1* | Obtain given answer in the form \(v = p - \frac{q}{(3t+5)^2}\) from correct working. Accept if correct form given and values of \(p\) and \(q\) stated separately. Must have "\(v =\)". |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(t \to \infty \Rightarrow v \to \frac{16}{25} (= 0.64)\) | B1ft | Follow through their \(p\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\int 1 \, dx = \int \frac{16}{25} - \frac{16}{(3t+5)^2} \, dt \Rightarrow x = rt + s\frac{1}{3t+5}\) | M1 | Form a differential equation in \(x\) and \(t\) and integrate to obtain \(rt + s\frac{1}{3t+5}\) where \(r\) and \(s\) are rational |
| \(x = \frac{16}{25}t + \frac{16}{3(3t+5)}(+D)\) | A1ft | Correct integration. Ignore limits and condone no constant of integration. Follow through their \(p\) and their \(-\frac{q}{3}\) |
| \(x = \left[\frac{16}{25}t + \frac{16}{3(3t+5)}\right]_0^2\) | M1 | Use \(x=0, t=0\) as limits in a definite integral or substituted to find the constant of integration and find \(x\) when \(t=2\) |
| \(x = \left(\frac{32}{25} + \frac{16}{3(11)}\right) - \left(\frac{16}{3(5)}\right) = \left(\frac{192}{275}\right) = 0.70\) or better | A1 | 0.70 or better. \((0.698181\ldots)\) |
# Question 1:
## Part 1a:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\int \frac{96}{(3t+5)^3} \, dt = \int 1 \, dv \Rightarrow v = \ldots$ | M1 | Form a differential equation in $v$ and $t$ and integrate. Must attempt integration of $\frac{k}{(3t+5)^3}$. RHS can be implied. |
| $-\frac{96}{2 \times 3 \times (3t+5)^2}(+C) = v$ | A1 | Correct integration. Ignore any limits. Accept without constant of integration. |
| Use limits $v = 0, t = 0$ | M1 | Use $v=0, t=0$ as limits in a definite integral or to find the constant of integration. |
| $v = \frac{96}{6 \times (5)^2} - \frac{96}{6 \times (3t+5)^2} = \frac{16}{25} - \frac{16}{(3t+5)^2}$ | A1* | Obtain given answer in the form $v = p - \frac{q}{(3t+5)^2}$ from correct working. Accept if correct form given and values of $p$ and $q$ stated separately. Must have "$v =$". |
## Part 1b:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $t \to \infty \Rightarrow v \to \frac{16}{25} (= 0.64)$ | B1ft | Follow through their $p$ |
## Part 1c:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\int 1 \, dx = \int \frac{16}{25} - \frac{16}{(3t+5)^2} \, dt \Rightarrow x = rt + s\frac{1}{3t+5}$ | M1 | Form a differential equation in $x$ and $t$ and integrate to obtain $rt + s\frac{1}{3t+5}$ where $r$ and $s$ are rational |
| $x = \frac{16}{25}t + \frac{16}{3(3t+5)}(+D)$ | A1ft | Correct integration. Ignore limits and condone no constant of integration. Follow through their $p$ and their $-\frac{q}{3}$ |
| $x = \left[\frac{16}{25}t + \frac{16}{3(3t+5)}\right]_0^2$ | M1 | Use $x=0, t=0$ as limits in a definite integral or substituted to find the constant of integration **and** find $x$ when $t=2$ |
| $x = \left(\frac{32}{25} + \frac{16}{3(11)}\right) - \left(\frac{16}{3(5)}\right) = \left(\frac{192}{275}\right) = 0.70$ or better | A1 | 0.70 or better. $(0.698181\ldots)$ |
---\begin{enumerate}
\item In this question you must show all stages of your working. Solutions relying entirely on calculator technology are not acceptable.
\end{enumerate}
A particle $P$ moves along a straight line. Initially $P$ is at rest at the point $O$ on the line.
At time $t$ seconds, where $t \geqslant 0$
\begin{itemize}
\item the displacement of $P$ from $O$ is $x$ metres
\item the velocity of $P$ is $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$ in the positive $x$ direction
\item the acceleration of $P$ is $\frac { 96 } { ( 3 t + 5 ) ^ { 3 } } \mathrm {~ms} ^ { - 2 }$ in the positive $x$ direction\\
(a) Show that, at time $t$ seconds, $v = p - \frac { q } { ( 3 t + 5 ) ^ { 2 } }$, where $p$ and $q$ are constants to be determined.\\
(b) Find the limiting value of $v$ as $t$ increases.\\
(c) Find the value of $x$ when $t = 2$
\end{itemize}
\hfill \mbox{\textit{Edexcel FM2 2024 Q1 [9]}}