AQA Paper 2 2020 June — Question 19 8 marks

Exam BoardAQA
ModulePaper 2 (Paper 2)
Year2020
SessionJune
Marks8
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicDifferential equations
TypeParticle motion - velocity/time (dv/dt = f(v,t))
DifficultyStandard +0.3 This is a standard mechanics differential equation question requiring separation of variables (a = dv/dt = -0.1vΒ²) and integration, followed by applying initial conditions. Part (b) is straightforward substitution. While it involves multiple steps and integration technique, it follows a well-practiced procedure with no novel insight required, making it slightly easier than average.
Spec1.07t Construct differential equations: in context1.08k Separable differential equations: dy/dx = f(x)g(y)6.06a Variable force: dv/dt or v*dv/dx methods
A particle moves so that its acceleration, \(a\text{ ms}^{-2}\), at time \(t\) seconds may be modelled in terms of its velocity, \(v\text{ ms}^{-1}\), as $$a = -0.1v^2$$ The initial velocity of the particle is \(4\text{ ms}^{-1}\)
  1. By first forming a suitable differential equation, show that $$v = \frac{20}{5 + 2t}$$ [6 marks]
  2. Find the acceleration of the particle when \(t = 5.5\) [2 marks]
Question 19:
AnswerMarks
19(a)Forms differential equation
PI by correctly separated variables
AnswerMarks Guidance
or OE3.4 B1
= βˆ’0.1𝑣𝑣
𝑑𝑑𝑑𝑑
1
οΏ½βˆ’ 2 𝑑𝑑𝑣𝑣 =οΏ½0.1 𝑑𝑑𝑑𝑑
𝑣𝑣
1
= 0.1𝑑𝑑+𝑐𝑐
𝑣𝑣
1
= 0.1Γ—(0)+𝑐𝑐
4
1
= 0.1𝑑𝑑+0.25
𝑣𝑣
1
𝑣𝑣 =
0.1𝑑𝑑+0.25
20
1
Separates variables to integrate
βˆ«βˆ’π‘£π‘£ 2 𝑑𝑑𝑣𝑣 =∫0.1 𝑑𝑑𝑑𝑑
AnswerMarks Guidance
Condone sign error1.1a M1
Integrates with one side correct1.1b A1
Fully correct integration
Condone missing constant of
AnswerMarks Guidance
integration1.1b A1
Uses initial conditions to correctly
AnswerMarks Guidance
obtain an equation in v and t only.3.4 M1
Completes rigorous argument to
show given expression
AnswerMarks Guidance
AG2.1 R1
Subtotal6 𝑣𝑣 =
5+2𝑑𝑑
AnswerMarks
19(b)Substitutes in
20
AnswerMarks Guidance
𝑑𝑑 = 5.5 𝑣𝑣 = 5+2𝑑𝑑3.4 M1
𝑣𝑣 = = 1.25
5+1 1
Acceleratiβˆ’on5
π‘Žπ‘Ž = = βˆ’0.15625
32
Obtains correct acceleration
=
βˆ’2
AnswerMarks Guidance
Aβˆ’W0R.1T5 6- 205.1 m56 s1.1b A1
Subtotal2 βˆ’2
= βˆ’0.15625 m s
AnswerMarks
Question Total8 
Question 19:
--- 19(a) ---
19(a) | Forms differential equation
PI by correctly separated variables
or OE | 3.4 | B1 | 𝑑𝑑𝑣𝑣 2
= βˆ’0.1𝑣𝑣
𝑑𝑑𝑑𝑑
1
οΏ½βˆ’ 2 𝑑𝑑𝑣𝑣 =οΏ½0.1 𝑑𝑑𝑑𝑑
𝑣𝑣
1
= 0.1𝑑𝑑+𝑐𝑐
𝑣𝑣
1
= 0.1Γ—(0)+𝑐𝑐
4
1
= 0.1𝑑𝑑+0.25
𝑣𝑣
1
𝑣𝑣 =
0.1𝑑𝑑+0.25
20
1
Separates variables to integrate
βˆ«βˆ’π‘£π‘£ 2 𝑑𝑑𝑣𝑣 =∫0.1 𝑑𝑑𝑑𝑑
Condone sign error | 1.1a | M1
Integrates with one side correct | 1.1b | A1
Fully correct integration
Condone missing constant of
integration | 1.1b | A1
Uses initial conditions to correctly
obtain an equation in v and t only. | 3.4 | M1
Completes rigorous argument to
show given expression
AG | 2.1 | R1
Subtotal | 6 | 𝑣𝑣 =
5+2𝑑𝑑
--- 19(b) ---
19(b) | Substitutes in
20
𝑑𝑑 = 5.5 𝑣𝑣 = 5+2𝑑𝑑 | 3.4 | M1 | 20
𝑣𝑣 = = 1.25
5+1 1
Acceleratiβˆ’on5
π‘Žπ‘Ž = = βˆ’0.15625
32
Obtains correct acceleration
=
βˆ’2
Aβˆ’W0R.1T5 6- 205.1 m56 s | 1.1b | A1
Subtotal | 2 | βˆ’2
= βˆ’0.15625 m s
Question Total | 8
A particle moves so that its acceleration, $a\text{ ms}^{-2}$, at time $t$ seconds may be modelled in terms of its velocity, $v\text{ ms}^{-1}$, as

$$a = -0.1v^2$$

The initial velocity of the particle is $4\text{ ms}^{-1}$

\begin{enumerate}[label=(\alph*)]
\item By first forming a suitable differential equation, show that
$$v = \frac{20}{5 + 2t}$$
[6 marks]

\item Find the acceleration of the particle when $t = 5.5$
[2 marks]
\end{enumerate}

\hfill \mbox{\textit{AQA Paper 2 2020 Q19 [8]}}
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