| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2012 |
| Session | January |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable acceleration (1D) |
| Type | Acceleration as function of velocity (separation of variables) |
| Difficulty | Standard +0.8 This M3 question requires understanding v as a function of x, using the chain rule to find acceleration (a = v dv/dx), then separating variables to integrate. While the calculus is standard, the conceptual step of relating acceleration to position rather than time, plus the two-part structure requiring both differentiation and integration, makes this moderately challenging for A-level. |
| Spec | 6.06a Variable force: dv/dt or v*dv/dx methods |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(a = v\frac{dv}{dx} = \frac{10}{x+6} \times \frac{-10}{(x+6)^2} = \frac{-100}{(x+6)^3}\) | M1, M1, A1 | |
| \(= \frac{-100}{(14+6)^3} = -\frac{1}{80} \text{ ms}^{-2}\) | A1 | (4) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\frac{dx}{dt} = \frac{10}{x+6} \Rightarrow \int x+6\, dx = \int 10\, dt\) | M1, M1 | |
| \(\left[\frac{x^2}{2} + 6x\right]_2^{14} = \left[10t\right]_1^T\) | M1 A1 | |
| \(\frac{196}{2} + 6\times14 - 2 - 12 = 10T - 10\) | M1 | |
| \(178 = T\), \(T = 17.8 \text{ s}\) | A1 | (6) Total: 10 |
## Question 3:
### Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $a = v\frac{dv}{dx} = \frac{10}{x+6} \times \frac{-10}{(x+6)^2} = \frac{-100}{(x+6)^3}$ | M1, M1, A1 | |
| $= \frac{-100}{(14+6)^3} = -\frac{1}{80} \text{ ms}^{-2}$ | A1 | (4) |
### Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{dx}{dt} = \frac{10}{x+6} \Rightarrow \int x+6\, dx = \int 10\, dt$ | M1, M1 | |
| $\left[\frac{x^2}{2} + 6x\right]_2^{14} = \left[10t\right]_1^T$ | M1 A1 | |
| $\frac{196}{2} + 6\times14 - 2 - 12 = 10T - 10$ | M1 | |
| $178 = T$, $T = 17.8 \text{ s}$ | A1 | (6) Total: 10 |
---3. A particle $P$ is moving in a straight line. At time $t$ seconds, $P$ is at a distance $x$ metres from a fixed point $O$ on the line and is moving away from $O$ with speed $\frac { 10 } { x + 6 } \mathrm {~m} \mathrm {~s} ^ { - 1 }$.
\begin{enumerate}[label=(\alph*)]
\item Find the acceleration of $P$ when $x = 14$
Given that $x = 2$ when $t = 1$,
\item find the value of $t$ when $x = 14$
\end{enumerate}
\hfill \mbox{\textit{Edexcel M3 2012 Q3 [10]}}