AQA AS Paper 1 2020 June — Question 15 7 marks

Exam BoardAQA
ModuleAS Paper 1 (AS Paper 1)
Year2020
SessionJune
Marks7
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicVariable acceleration (1D)
TypeVariable acceleration with initial conditions
DifficultyStandard +0.3 This is a standard kinematics question requiring integration of acceleration to find velocity (part a) and then integration again to find displacement (part b), followed by solving a cubic equation. While it involves multiple steps and algebraic manipulation, the techniques are routine for AS-level mechanics with no novel problem-solving insight required. The 'fully justify' instruction adds minor complexity but this remains slightly easier than average.
Spec1.08d Evaluate definite integrals: between limits3.02f Non-uniform acceleration: using differentiation and integration
A particle, \(P\), is moving in a straight line with acceleration \(a\text{ m s}^{-2}\) at time \(t\) seconds, where $$a = 4 - 3t^2$$
  1. Initially \(P\) is stationary. Find an expression for the velocity of \(P\) in terms of \(t\). [2 marks]
  2. When \(t = 2\), the displacement of \(P\) from a fixed point, O, is 39 metres. Find the time at which \(P\) passes through O, giving your answer to three significant figures. Fully justify your answer. [5 marks]
Question 15:
AnswerMarks
15(a)Integrates to find with at least
one term correct
AnswerMarks Guidance
๐‘Ž๐‘Ž ๐‘ฃ๐‘ฃ3.4 M1
3
Finds fully correct final expression
AnswerMarks Guidance
Condone presence of + c1.1b A1
Subtotal2 ๐‘ฃ๐‘ฃ = 4๐‘ก๐‘กโˆ’๐‘ก๐‘ก
AnswerMarks
15(b)Integrates to find with at least
one term correct
AnswerMarks Guidance
๐‘ฃ๐‘ฃ ๐‘Ž๐‘Ž3.1b M1
2 1 4
๐‘Ž๐‘Ž = 2๐‘ก๐‘ก โˆ’ ๐‘ก๐‘ก +๐‘˜๐‘˜
4
39 = 8โˆ’4+ ๐‘˜๐‘˜ โŸน ๐‘˜๐‘˜ = 35
4 2
0 = ๐‘ก๐‘ก โˆ’8๐‘ก๐‘ก โˆ’140
seconds
๐‘ก๐‘ก = 4.06
Integrates their answer to (a)
correctly including constant of
AnswerMarks Guidance
integration1.1b A1F
Uses given conditions to find
AnswerMarks Guidance
constant3.4 M1
Equates their expression for to
zero and finds a value for
AnswerMarks Guidance
๐‘Ž๐‘Ž1.1a M1
๐‘ก๐‘ก
Obtains correct value of to
required accuracy
AnswerMarks Guidance
๐‘ก๐‘ก1.1b A1
Subtotal5
Question Total7
QMarking Instructions AO 
Question 15:
--- 15(a) ---
15(a) | Integrates to find with at least
one term correct
๐‘Ž๐‘Ž ๐‘ฃ๐‘ฃ | 3.4 | M1 | ๐‘ฃ๐‘ฃ = ๏ฟฝ๐‘Ž๐‘Ž ๐‘‘๐‘‘๐‘ก๐‘ก
3
Finds fully correct final expression
Condone presence of + c | 1.1b | A1
Subtotal | 2 | ๐‘ฃ๐‘ฃ = 4๐‘ก๐‘กโˆ’๐‘ก๐‘ก
--- 15(b) ---
15(b) | Integrates to find with at least
one term correct
๐‘ฃ๐‘ฃ ๐‘Ž๐‘Ž | 3.1b | M1 | ๐‘Ž๐‘Ž = ๏ฟฝ๐‘ฃ๐‘ฃ ๐‘‘๐‘‘๐‘ก๐‘ก
2 1 4
๐‘Ž๐‘Ž = 2๐‘ก๐‘ก โˆ’ ๐‘ก๐‘ก +๐‘˜๐‘˜
4
39 = 8โˆ’4+ ๐‘˜๐‘˜ โŸน ๐‘˜๐‘˜ = 35
4 2
0 = ๐‘ก๐‘ก โˆ’8๐‘ก๐‘ก โˆ’140
seconds
๐‘ก๐‘ก = 4.06
Integrates their answer to (a)
correctly including constant of
integration | 1.1b | A1F
Uses given conditions to find
constant | 3.4 | M1
Equates their expression for to
zero and finds a value for
๐‘Ž๐‘Ž | 1.1a | M1
๐‘ก๐‘ก
Obtains correct value of to
required accuracy
๐‘ก๐‘ก | 1.1b | A1
Subtotal | 5
Question Total | 7
Q | Marking Instructions | AO | Marks | Typical Solution
A particle, $P$, is moving in a straight line with acceleration $a\text{ m s}^{-2}$ at time $t$ seconds, where
$$a = 4 - 3t^2$$

\begin{enumerate}[label=(\alph*)]
\item Initially $P$ is stationary.

Find an expression for the velocity of $P$ in terms of $t$. [2 marks]

\item When $t = 2$, the displacement of $P$ from a fixed point, O, is 39 metres.

Find the time at which $P$ passes through O, giving your answer to three significant figures.

Fully justify your answer. [5 marks]
\end{enumerate}

\hfill \mbox{\textit{AQA AS Paper 1 2020 Q15 [7]}}
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