| Exam Board | AQA |
|---|---|
| Module | AS Paper 1 (AS Paper 1) |
| Year | 2023 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable acceleration (vectors) |
| Type | Find acceleration from velocity |
| Difficulty | Moderate -0.8 This is a straightforward mechanics question requiring only basic calculus operations: differentiate the velocity function to find acceleration, and integrate it to find displacement. Both are standard AS-level techniques with no problem-solving insight needed—students simply apply learned procedures to a polynomial function. |
| Spec | 3.02f Non-uniform acceleration: using differentiation and integration |
| Answer | Marks |
|---|---|
| 16(a) | Differentiates to find expression |
| Answer | Marks | Guidance |
|---|---|---|
| PI by -0.08 or 0.08 | 3.4 | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| PI by -0.08 | 1.1b | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Correct units must be stated. | 3.2a | A1F |
| Subtotal | 3 | |
| Q | Marking instructions | AO |
| Answer | Marks |
|---|---|
| 16(b) | Integrates v with at least one |
| Answer | Marks | Guidance |
|---|---|---|
| PI by 1.96 | 3.1b | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| PI by 1.96 | 1.1b | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| PI by 1.96 | 1.1a | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| Condone omission of units | 1.1b | A1 |
| Subtotal | 4 | |
| Question 16 Total | 7 | |
| Q | Marking instructions | AO |
Question 16:
--- 16(a) ---
16(a) | Differentiates to find expression
for acceleration with at least one
term correct.
PI by -0.08 or 0.08 | 3.4 | M1 | a = 0.16 – 0.12t
a = –0.08 ms–2
Obtains a fully correct
expression for acceleration.
PI by -0.08 | 1.1b | A1
Finds their acceleration of the
boat when t = 2
FT their expression for a
Must have differentiated at least
one term.
Correct units must be stated. | 3.2a | A1F
Subtotal | 3
Q | Marking instructions | AO | Marks | Typical solution
--- 16(b) ---
16(b) | Integrates v with at least one
term correct.
PI by 1.96 | 3.1b | M1 | s = v d t
= 0 .9 + 0 .1 6 t − 0 .0 6 t 2 d t
s = 0 . 9 t + 0 . 0 8 t 2 − 0 . 0 2 t 3 + c
s = 0 when t = 0 so c = 0
s =0.9(2)+0.08(4)−0.02(8)
Displacement = 1.96 m
Obtains a fully correct integral.
Condone omission of constant
PI by 1.96 | 1.1b | A1
Substitutes t = 0 and t = 2 into
their expression for s
Must have integrated at least
one term.
PI by 1.96 | 1.1a | M1
Obtains displacement = 1.96 m
Condone omission of units | 1.1b | A1
Subtotal | 4
Question 16 Total | 7
Q | Marking instructions | AO | Marks | Typical solutionA toy remote control speed boat is launched from one edge of a small pond and moves in a straight line across the pond's surface.
The boat's velocity, $v \text{ m s}^{-1}$, is modelled in terms of time, $t$ seconds after the boat is launched, by the expression
$$v = 0.9 + 0.16t - 0.06t^2$$
\begin{enumerate}[label=(\alph*)]
\item Find the acceleration of the boat when $t = 2$
[3 marks]
\item Find the displacement of the boat, from the point where it was launched, when $t = 2$
[4 marks]
\end{enumerate}
\hfill \mbox{\textit{AQA AS Paper 1 2023 Q16 [7]}}