| Exam Board | AQA |
|---|---|
| Module | AS Paper 1 (AS Paper 1) |
| Year | 2018 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable acceleration (1D) |
| Type | Displacement from velocity by integration |
| Difficulty | Moderate -0.8 Part (a) is straightforward integration of a polynomial with an initial condition - a routine AS-level mechanics exercise. Part (b) is a basic SUVAT calculation (v = u + at with v = 0) that requires simple substitution. Both parts are standard textbook exercises requiring only direct application of learned techniques with no problem-solving insight needed. |
| Spec | 3.02d Constant acceleration: SUVAT formulae3.02f Non-uniform acceleration: using differentiation and integration |
| Answer | Marks | Guidance |
|---|---|---|
| 16(a) | Shows integral of v with respect to t | |
| Condone omission of dt | AO1.1a | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| equivalent form. | AO1.1b | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| displacement of 3. PI. | AO2.2a | M1 |
| Finds fully correct expression for r. | AO1.1b | A1 |
| Answer | Marks |
|---|---|
| 16(b) | Models problem as only force |
| Answer | Marks | Guidance |
|---|---|---|
| v = 0 at highest point. (PI) | AO3.3 | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| negative or u = -3.43 and g positive | AO1.1a | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| to give t = 2.35 s or 2.4s | AO1.1b | A1 |
| Total | 7 | |
| TOTAL | 80 |
Question 16:
--- 16(a) ---
16(a) | Shows integral of v with respect to t
Condone omission of dt | AO1.1a | M1 | ∫0.06(2+𝑡−𝑡2) 𝑑𝑡
= 0.12t + 0.03t 2 – 0.02t3 + c
Using t = 0, c = 3
r = 0.12t + 0.03t2 – 0.02t3 + 3
Integrates with at least two correct
terms (condone omission of
constant at this stage) Any
equivalent form. | AO1.1b | A1
Finds constant using t = 0 and
displacement of 3. PI. | AO2.2a | M1
Finds fully correct expression for r. | AO1.1b | A1
--- 16(b) ---
16(b) | Models problem as only force
acting is that due to gravity so that
v = 0 at highest point. (PI) | AO3.3 | M1 | a = −9.8 m s-2
v = 0 m s-1
0 = 3.43 – 9.8t
max
∴ t = 0.35 s
max
Uses appropriate suvat formula
v = u + at with u = 3.43 and g
negative or u = -3.43 and g positive | AO1.1a | M1
Finds t = 0.35 s (CAO)
max
Condone addition of the 2 seconds
to give t = 2.35 s or 2.4s | AO1.1b | A1
Total | 7
TOTAL | 80A remote-controlled toy car is moving over a horizontal surface. It moves in a straight line through a point $A$.
The toy is initially at the point with displacement $3$ metres from $A$. Its velocity, $v\,\mathrm{m}\,\mathrm{s}^{-1}$, at time $t$ seconds is defined by
$$v = 0.06(2 + t - t^2)$$
\begin{enumerate}[label=(\alph*)]
\item Find an expression for the displacement, $r$ metres, of the toy from $A$ at time $t$ seconds.
[4 marks]
\item In this question use $g = 9.8\,\mathrm{m}\,\mathrm{s}^{-2}$
At time $t = 2$ seconds, the toy launches a ball which travels directly upwards with initial speed $3.43\,\mathrm{m}\,\mathrm{s}^{-1}$
Find the time taken for the ball to reach its highest point.
[3 marks]
\end{enumerate}
\hfill \mbox{\textit{AQA AS Paper 1 2018 Q16 [7]}}