AQA AS Paper 1 Specimen — Question 16 8 marks

Exam BoardAQA
ModuleAS Paper 1 (AS Paper 1)
SessionSpecimen
Marks8
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Mark schemeDownload PDF ↗
TopicVariable acceleration (1D)
TypeDisplacement from velocity by integration
DifficultyModerate -0.3 This is a straightforward mechanics question requiring standard calculus techniques: differentiate velocity to find acceleration, then apply F=ma for part (a); integrate velocity and solve for displacement=0 in part (b). Both parts follow routine procedures with no conceptual challenges, making it slightly easier than average, though the 8 total marks reflect reasonable working required.
Spec1.07i Differentiate x^n: for rational n and sums1.08f Area between two curves: using integration3.03c Newton's second law: F=ma one dimension
A particle, of mass 400 grams, is initially at rest at the point \(O\). The particle starts to move in a straight line so that its velocity, \(v\) m s⁻¹, at time \(t\) seconds is given by \(v = 6t^2 - 12t^3\) for \(t > 0\)
  1. Find an expression, in terms of \(t\), for the force acting on the particle. [3 marks]
  2. Find the time when the particle next passes through \(O\). [5 marks]
Question 16:
AnswerMarks Guidance
16(a)Differentiates, with at least one
term correctAO1.1a M1
12t36t2
dt
F = ma = 0.4 (12t36t2)
4.8t14.4t2
Selects and applies F = ma to
‘their’ derivative
AnswerMarks Guidance
Condone use of 400 for massAO1.1a M1
Obtains correct expression for
force
FT from ‘their’ F = ma equation,
provided the first M1 has been
awarded
AnswerMarks Guidance
(may be in factorised form)AO1.1b A1F
(b)Integrates v to find r, with at least
one term correctAO3.1b M1
r 2t3 3t4 c
When t = 0, r = 0 so
0202 304+ c
so c = 0
02t3 3t4
t3(23t)
2
t 0 or t 
3
2
Next at O at seconds
3
Obtains correct integral
AnswerMarks Guidance
(condone absence of c)AO1.1b A1
Deduces the value of c using initial
conditions
FT use of ‘their’ integral provided
AnswerMarks Guidance
M1 awardedAO2.2a A1F
Forms and solves an equation for t
AnswerMarks Guidance
(condone numerical slip)AO1.1a M1
Interprets solution, realising that
the non-zero time is required
(Must include units)
FT use of ‘their’ equation for t
provided both M1 marks have been
AnswerMarks Guidance
awardedAO3.2a A1F
Total8
QMarking Instructions AO 
Question 16:
--- 16(a) ---
16(a) | Differentiates, with at least one
term correct | AO1.1a | M1 | dv
12t36t2
dt
F = ma = 0.4 (12t36t2)
4.8t14.4t2
Selects and applies F = ma to
‘their’ derivative
Condone use of 400 for mass | AO1.1a | M1
Obtains correct expression for
force
FT from ‘their’ F = ma equation,
provided the first M1 has been
awarded
(may be in factorised form) | AO1.1b | A1F
(b) | Integrates v to find r, with at least
one term correct | AO3.1b | M1 | r    6t2 12t3 dt
r 2t3 3t4 c
When t = 0, r = 0 so
0202 304+ c
so c = 0
02t3 3t4
t3(23t)
2
t 0 or t 
3
2
Next at O at seconds
3
Obtains correct integral
(condone absence of c) | AO1.1b | A1
Deduces the value of c using initial
conditions
FT use of ‘their’ integral provided
M1 awarded | AO2.2a | A1F
Forms and solves an equation for t
(condone numerical slip) | AO1.1a | M1
Interprets solution, realising that
the non-zero time is required
(Must include units)
FT use of ‘their’ equation for t
provided both M1 marks have been
awarded | AO3.2a | A1F
Total | 8
Q | Marking Instructions | AO | Marks | Typical Solution
A particle, of mass 400 grams, is initially at rest at the point $O$.

The particle starts to move in a straight line so that its velocity, $v$ m s⁻¹, at time $t$ seconds is given by

$v = 6t^2 - 12t^3$ for $t > 0$

\begin{enumerate}[label=(\alph*)]
\item Find an expression, in terms of $t$, for the force acting on the particle. [3 marks]

\item Find the time when the particle next passes through $O$. [5 marks]
\end{enumerate}

\hfill \mbox{\textit{AQA AS Paper 1  Q16 [8]}}
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