| Exam Board | WJEC |
|---|---|
| Module | Further Unit 6 (Further Unit 6) |
| Year | 2022 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable Force |
| Type | Given velocity function find force |
| Difficulty | Standard +0.8 This is a Further Maths mechanics question requiring the chain rule for acceleration (v dv/dx), solving a rational equation, and integrating a separable differential equation. While the techniques are standard for FM, the multi-step nature and algebraic manipulation required place it moderately above average difficulty. |
| Spec | 1.07r Chain rule: dy/dx = dy/du * du/dx and connected rates6.06a Variable force: dv/dt or v*dv/dx methods |
\begin{enumerate}
\item A particle is moving along the $x$-axis. At time $t$ seconds the particle is $x$ metres from the origin, $O$, and its velocity $v \mathrm {~ms} ^ { - 1 }$ is given by
\end{enumerate}
$$v = \frac { 24 } { 4 x + 9 }$$
(a) Find, in terms of $x$, an expression for the acceleration of the particle at time $t \mathrm {~s}$.\\
(b) At $t = T$ the acceleration of the particle is $- \frac { 4 } { 3 } \mathrm {~ms} ^ { - 2 }$.\\
(i) Determine the value of $x$ when $t = T$.\\
(ii) Given that $x = - 2$ when $t = 0$, find an expression for $t$ in terms of $x$ and hence find the value of $T$.\\
\hfill \mbox{\textit{WJEC Further Unit 6 2022 Q1 [12]}}