| Exam Board | CAIE |
|---|---|
| Module | Further Paper 3 (Further Paper 3) |
| Year | 2024 |
| Session | November |
| Marks | 4 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable acceleration (1D) |
| Type | Related rates with explicitly given non-geometric algebraic relationships |
| Difficulty | Challenging +1.2 This is a Further Maths mechanics question requiring separation of variables and integration of a differential equation. While it involves manipulating v/x = (3-t)/(1+t) and using the initial condition, the algebraic steps are straightforward: rewrite as dx/dt = x(3-t)/(1+t), separate variables, integrate both sides, and apply the boundary condition. The integration itself is routine (partial fractions or substitution). More challenging than a standard C3/C4 question due to the Further Maths context and differential equation setup, but not requiring deep insight. |
| Spec | 1.07r Chain rule: dy/dx = dy/du * du/dx and connected rates6.06a Variable force: dv/dt or v*dv/dx methods |
| Answer | Marks |
|---|---|
| 5(a) | dx 4 |
| Answer | Marks | Guidance |
|---|---|---|
| x t+1 | M1 | Separate variables, obtain RHS in integrable form. |
| ln x =4ln t+1−t+ A | A1 | |
| t=0, x=5: A=ln5 | M1 | |
| x=5(t+1)4 e−t | A1 |
| Answer | Marks |
|---|---|
| 5(b) | v=(3−t)5(t+1)3 e−t |
| Answer | Marks |
|---|---|
| dt | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| Acceleration = | AEF | |
| F=2acceleration, so at F =10e−t(t+1)2(5−t)(1−t) | M1 | |
| At t=3, magnitude of force is 640e−3 N | A1 | 31.9 N |
| Answer | Marks | Guidance |
|---|---|---|
| Question | Answer | Marks |
Question 5:
--- 5(a) ---
5(a) | dx 4
= −1dt
x t+1 | M1 | Separate variables, obtain RHS in integrable form.
ln x =4ln t+1−t+ A | A1
t=0, x=5: A=ln5 | M1
x=5(t+1)4 e−t | A1
4
--- 5(b) ---
5(b) | v=(3−t)5(t+1)3 e−t
dv ( )
Acceleration = =5e−t −(t+1)3+(3−t)3(t+1)2 −(3−t)(t+1)3
dt | M1
5e−t(t+1)2(5−t)(1−t)
Acceleration = | AEF
F=2acceleration, so at F =10e−t(t+1)2(5−t)(1−t) | M1
At t=3, magnitude of force is 640e−3 N | A1 | 31.9 N
3
Question | Answer | Marks | GuidanceA particle $P$ of mass $2 \text{ kg}$ moving on a horizontal straight line has displacement $x \text{ m}$ from a fixed point $O$ on the line and velocity $v \text{ m s}^{-1}$ at time $t \text{ s}$. The only horizontal force acting on $P$ is a variable force $F \text{ N}$ which can be expressed as a function of $t$. It is given that
$$\frac{v}{x} = \frac{3-t}{1+t}$$
and when $t = 0$, $x = 5$.
\begin{enumerate}[label=(\alph*)]
\item Find an expression for $x$ in terms of $t$. [4]
\end{enumerate}
\hfill \mbox{\textit{CAIE Further Paper 3 2024 Q5 [4]}}