| Exam Board | AQA |
|---|---|
| Module | AS Paper 1 (AS Paper 1) |
| Year | 2022 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Constant acceleration (SUVAT) |
| Type | Displacement expressions and comparison |
| Difficulty | Moderate -0.3 This is a straightforward kinematics problem requiring application of SUVAT equations. Part (a) involves setting up relative motion equations and algebraic manipulation to reach the given result (4 marks for a 'show that' is generous). Part (b) requires finding t from v=u+at, then substituting into the formula. The problem is slightly easier than average as it's highly structured, uses standard techniques, and the 'show that' format guides students to the answer. |
| Spec | 3.02d Constant acceleration: SUVAT formulae |
| Answer | Marks |
|---|---|
| 16(a) | Uses appropriate constant |
| Answer | Marks | Guidance |
|---|---|---|
| one term cπ‘π‘orrect. | 3.1b | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| 2 | 1.1b | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| π π D o= nπ’π’oπ‘π‘t accπ π e p=t π’π’utπ‘π‘ β+ dππ | 3.1b | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| AG | 2.1 | R1 |
| Subtotal | 4 | |
| Q | Marking instructions | AO |
| Answer | Marks |
|---|---|
| 16(b) | Uses appropriate constant |
| Answer | Marks | Guidance |
|---|---|---|
| value for | 3.1b | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| Accept 10.2 or 10.23 | 1.1b | A1 |
| Subtotal | 2 | |
| Question 16 Total | 6 | |
| Q | Marking instructions | AO |
Question 16:
--- 16(a) ---
16(a) | Uses appropriate constant
acceleration equation to find
expression for displacement of
Jermaine seconds after
acceleration begins, with at least
one term cπ‘π‘orrect. | 3.1b | M1 | Jermaineβs displacement
2
π π = (π’π’β0.2)π‘π‘+0.5Γ2π‘π‘
Reaches Meena when
π π = π’π’π‘π‘+ππ
2
π’π’π‘π‘+ππ = (π’π’β0.2)π‘π‘+π‘π‘
2
π’π’π‘π‘+ππ = π’π’π‘π‘β0.2π‘π‘ +π‘π‘
2
ππ = π‘π‘ β0.2π‘π‘
Obtains fully correct equation for
Jermaineβs displacement PI
Or 2
π π = (π’π’β0.2)π‘π‘+0.5Γ2π‘π‘
2 | 1.1b | A1
π π = (π’π’β0.2)π‘π‘+0.5Γ2π‘π‘ βππ
Forms a correct expression for
Meenaβs displacement.
or
π π D o= nπ’π’oπ‘π‘t accπ π e p=t π’π’utπ‘π‘ β+ dππ | 3.1b | B1
Completes reasoned argument
to obtain given result.
AG | 2.1 | R1
Subtotal | 4
Q | Marking instructions | AO | Marks | Typical solution
--- 16(b) ---
16(b) | Uses appropriate constant
acceleration equation to find
value for | 3.1b | M1 | 7.8= (1.4β0.2)+2π‘π‘
π‘π‘ =3.3 βππ = 10.23
π‘π‘
Uses given equation to find
correct value of d
Accept 10.2 or 10.23 | 1.1b | A1
Subtotal | 2
Question 16 Total | 6
Q | Marking instructions | AO | Marks | Typical solution
Jermaine and his friend Meena are walking in the same direction along a straight path.
Meena is walking at a constant speed of $u$ m s$^{-1}$
Jermaine is walking 0.2 m s$^{-1}$ more slowly than Meena.
When Jermaine is $d$ metres behind Meena he starts to run with a constant acceleration of 2 m s$^{-2}$, for a time of $t$ seconds, until he reaches her.
\begin{enumerate}[label=(\alph*)]
\item Show that
$$d = t^2 - 0.2t$$
[4 marks]
\item When Jermaine's speed is 7.8 m s$^{-1}$, he reaches Meena.
Given that $u = 1.4$ find the value of $d$.
[2 marks]
\end{enumerate}
\hfill \mbox{\textit{AQA AS Paper 1 2022 Q16 [6]}}