| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable Force |
| Type | Given acceleration function find velocity |
| Difficulty | Standard +0.8 This M3 variable force question requires using v dv/dx = a to form and solve a differential equation, then applying boundary conditions across multiple parts. It involves more sophisticated calculus than typical mechanics questions and requires careful algebraic manipulation, but follows a standard M3 pattern for this topic. |
| Spec | 3.02d Constant acceleration: SUVAT formulae6.06a Variable force: dv/dt or v*dv/dx methods |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \(a = v\frac{dv}{dx} = \frac{500-kx}{150}\) | M1 | |
| \(\therefore \int 150v \, dv = \int 500-kx \, dx\) | M1 | |
| \(75v^2 = 500x - \frac{1}{2}kx^2 + c\) | A1 | |
| \(x=0, v=0 \therefore c=0\) | M1 | |
| \(x=40, v=16 \therefore 19200 = 20000 - 800k\) giving \(k=1\) | M1 A1 | |
| (b) \(v_{\max}\) when \(a = 0 \therefore \frac{500-x}{150} = 0\) so \(x=500\) | M1 | |
| \(75v_{\max}^2 = 250000 - 125000\) giving \(v_{\max} = 41 \text{ ms}^{-1} \text{ (2sf)}\) | M1 A1 | |
| (c) when \(v=0, 500x - \frac{1}{2}x^2 = 0\) | M1 | |
| \(\therefore \frac{1}{2}x(1000-x) = 0\) so \(x=0\) (at O) or \(1000 \therefore OA = 1000 \text{ m}\) | M1 A1 | (12) |
**(a)** $a = v\frac{dv}{dx} = \frac{500-kx}{150}$ | M1 |
$\therefore \int 150v \, dv = \int 500-kx \, dx$ | M1 |
$75v^2 = 500x - \frac{1}{2}kx^2 + c$ | A1 |
$x=0, v=0 \therefore c=0$ | M1 |
$x=40, v=16 \therefore 19200 = 20000 - 800k$ giving $k=1$ | M1 A1 |
**(b)** $v_{\max}$ when $a = 0 \therefore \frac{500-x}{150} = 0$ so $x=500$ | M1 |
$75v_{\max}^2 = 250000 - 125000$ giving $v_{\max} = 41 \text{ ms}^{-1} \text{ (2sf)}$ | M1 A1 |
**(c)** when $v=0, 500x - \frac{1}{2}x^2 = 0$ | M1 |
$\therefore \frac{1}{2}x(1000-x) = 0$ so $x=0$ (at O) or $1000 \therefore OA = 1000 \text{ m}$ | M1 A1 | **(12)**
---3. A car starts from rest at the point $O$ and moves along a straight line. The car accelerates to a maximum velocity, $V \mathrm {~ms} ^ { - 1 }$, before decelerating and coming to rest again at the point $A$.
The acceleration of the car during this journey, $a \mathrm {~ms} ^ { - 2 }$, is modelled by the formula
$$a = \frac { 500 - k x } { 150 }$$
where $x$ is the distance in metres of the car from $O$.\\
Using this model and given that the car is travelling at $16 \mathrm {~ms} ^ { - 1 }$ when it is 40 m from $O$,
\begin{enumerate}[label=(\alph*)]
\item find $k$,
\item show that $V = 41$, correct to 2 significant figures,
\item find the distance $O A$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M3 Q3 [12]}}