| Exam Board | AQA |
|---|---|
| Module | AS Paper 1 (AS Paper 1) |
| Year | 2021 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable acceleration (1D) |
| Type | Velocity from acceleration by integration |
| Difficulty | Moderate -0.3 This is a straightforward AS-level mechanics question requiring integration of acceleration to find velocity, then substituting boundary conditions. Part (a) is routine integration with constant of integration determined by a given condition. Part (b) is simple algebra. The question requires standard techniques with clear guidance and no problem-solving insight, making it slightly easier than average for AS-level. |
| Spec | 3.02f Non-uniform acceleration: using differentiation and integration |
| Answer | Marks |
|---|---|
| 14(a) | Integrates given expression to |
| Answer | Marks | Guidance |
|---|---|---|
| 𝑣𝑣 | 3.4 | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| omission of + c 𝑣𝑣 | 1.1b | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| integration) | 1.1a | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| t = 0 | 2.1 | R1 |
| Subtotal | 4 | |
| Q | Marking instructions | AO |
| Answer | Marks |
|---|---|
| 14(b) | Forms equation using k, 6.4, |
| Answer | Marks | Guidance |
|---|---|---|
| 𝑘𝑘− | 3.1b | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| Obtains 8 | 1.1b | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Subtotal | 2 | 𝑘𝑘 = |
| Question Total | 6 | |
| Q | Marking instructions | AO |
Question 14:
--- 14(a) ---
14(a) | Integrates given expression to
find with at least one term
correct
𝑣𝑣 | 3.4 | M1 | 𝑣𝑣 = �𝑎𝑎𝑑𝑑𝑡𝑡
3 0.1
2
W = he 𝑡𝑡 n + t = 2 𝑡𝑡 , v + = 𝑐𝑐k
6 0.4
𝑘𝑘 =c k+ + 𝑐𝑐
= 6.4
v 3 0.1 − 6.4
2
S=ince𝑡𝑡 + 𝑡𝑡 w h+en𝑘𝑘 −
the initial velocity is 6.4
𝑣𝑣 = 𝑐𝑐 𝑡𝑡 = 0
𝑘𝑘−
Obtains an expression for with
both terms correct condone
omission of + c 𝑣𝑣 | 1.1b | A1
Substitutes t = 2 and v = k into
their integrated expression
(must include constant of
integration) | 1.1a | M1
Completes rigorous argument
with no slips to obtain v = k – 6.4
when
t = 0 | 2.1 | R1
Subtotal | 4
Q | Marking instructions | AO | Marks | Typical solution
--- 14(b) ---
14(b) | Forms equation using k, 6.4,
and 0.2 or 5
𝑘𝑘− | 3.1b | M1 | 0.2 6.4
𝑘𝑘 = 𝑘𝑘−
8
Obtains 8 | 1.1b | A1
𝑘𝑘 =
Subtotal | 2 | 𝑘𝑘 =
Question Total | 6
Q | Marking instructions | AO | Marks | Typical solutionA particle, P, is moving along a straight line such that its acceleration $a$ m s⁻², at any time, $t$ seconds, may be modelled by
$$a = 3 + 0.2t$$
When $t = 2$, the velocity of P is $k$ m s⁻¹
\begin{enumerate}[label=(\alph*)]
\item Show that the initial velocity of P is given by the expression $(k - 6.4)$ m s⁻¹
[4 marks]
\item The initial velocity of P is one fifth of the velocity when $t = 2$
Find the value of $k$.
[2 marks]
\end{enumerate}
\hfill \mbox{\textit{AQA AS Paper 1 2021 Q14 [6]}}