| Exam Board | AQA |
|---|---|
| Module | Paper 2 (Paper 2) |
| Year | 2024 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Differential equations |
| Type | First-order integration |
| Difficulty | Standard +0.3 This is a straightforward mechanics differentiation problem requiring students to find velocity and acceleration by differentiating twice, then substitute given values to solve for constants. Part (a) involves routine differentiation of an exponential function and algebraic manipulation, while part (b) uses an initial condition. The techniques are standard A-level mechanics with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.07i Differentiate x^n: for rational n and sums1.07j Differentiate exponentials: e^(kx) and a^(kx)3.02a Kinematics language: position, displacement, velocity, acceleration |
| Answer | Marks |
|---|---|
| 18(a) | dr |
| Answer | Marks | Guidance |
|---|---|---|
| PI by −0.04qe −0.2t | 3.4 | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| PI by −0.04qe −0.2t | 1.1b | A1 |
| Obtains −0.04qe −0.2t ACF | 1.1b | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| into their expression for a | 1.1a | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| AG | 2.1 | R1 |
| Subtotal | 5 | |
| Q | Marking instructions | AO |
| Answer | Marks |
|---|---|
| 18(b) | Substitutes t = 0, r = 5 and their |
| Answer | Marks | Guidance |
|---|---|---|
| q into r = | 3.4 | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| Obtains AWRT 87 | 1.1b | A1 |
| Subtotal | 2 | |
| Question 18 Total | 7 | |
| Q | Marking instructions | AO |
Question 18:
--- 18(a) ---
18(a) | dr
Uses v= to obtain an
dt
expression for v with one term
correct
PI by −0.04qe −0.2t | 3.4 | M1 | d2r
Using a=
dt2
v=2+0.2qe−0.2t
a =−0.04qe −0.2t
Using a =−1.8when t =3
−1.8=−0.04qe −0.6
q=81.995≈82
2+0.2qe−0.2t
Obtains ACF
PI by −0.04qe −0.2t | 1.1b | A1
Obtains −0.04qe −0.2t ACF | 1.1b | A1
Substitutes t = 3 and a = – 1.8
into their expression for a | 1.1a | M1
Completes reasoned argument
and concludes that q≈82
Accept q rounds to 82
Must show either a correct
expression for q or q = 81.9…
CSO
AG | 2.1 | R1
Subtotal | 5
Q | Marking instructions | AO | Marks | Typical solution
--- 18(b) ---
18(b) | Substitutes t = 0, r = 5 and their
p+2t−qe−0.2t
q into r = | 3.4 | M1 | 5= p−82e −0.2×0
p=5+82
p=87
Obtains AWRT 87 | 1.1b | A1
Subtotal | 2
Question 18 Total | 7
Q | Marking instructions | AO | Marks | Typical solutionA particle is moving in a straight line through the origin $O$
The displacement of the particle, $r$ metres, from $O$, at time $t$ seconds is given by
$$r = p + 2t - qe^{-0.2t}$$
where $p$ and $q$ are constants.
When $t = 3$, the acceleration of the particle is $-1.8$ m s$^{-2}$
\begin{enumerate}[label=(\alph*)]
\item Show that $q \approx 82$
[5 marks]
\item The particle has an initial displacement of 5 metres.
Find the value of $p$
Give your answer to two significant figures.
[2 marks]
\end{enumerate}
\hfill \mbox{\textit{AQA Paper 2 2024 Q18 [7]}}