Question 2 - A-Level Maths

Edexcel M5 2007 June — Question 2 7 marks

Exam Board	Edexcel
Module	M5 (Mechanics 5)
Year	2007
Session	June
Marks	7
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Second order differential equations
Type	Standard non-homogeneous with exponential RHS
Difficulty	Moderate -0.3 This is a standard first-order linear vector differential equation (despite being written in second-order form, it reduces immediately by substituting v = dr/dt). The solution requires routine integration and application of initial conditions—slightly easier than average due to the straightforward structure and lack of non-homogeneous terms, though the vector notation and logarithmic time value add minor complexity.
Spec	1.10a Vectors in 2D: i,j notation and column vectors 1.10b Vectors in 3D: i,j,k notation 4.10d Second order homogeneous: auxiliary equation method 4.10e Second order non-homogeneous: complementary + particular integral

2. At time $t$ seconds, the position vector of a particle $P$ is $\mathbf { r }$ metres, where $\mathbf { r }$ satisfies the differential equation $$\frac { \mathrm { d } ^ { 2 } \mathbf { r } } { \mathrm {~d} t ^ { 2 } } + 3 \frac { \mathrm {~d} \mathbf { r } } { \mathrm {~d} t } = \mathbf { 0 }$$ When $t = 0$, the velocity of $P$ is $( 8 \mathbf { i } - 12 \mathbf { j } ) \mathrm { m } \mathrm { s } ^ { - 1 }$.
Find the velocity of $P$ when $t = \frac { 2 } { 3 } \ln 2$.
(7)

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Question 2:

Answer	Marks	Guidance
Working/Answer	Marks	Guidance
$\frac{d\mathbf{v}}{dt} + 3\mathbf{v} = \mathbf{0}$	B1
IF $= e^{3t} \Rightarrow \frac{d(\mathbf{v}e^{3t})}{dt} = \mathbf{0}$	M1
$\Rightarrow \mathbf{v}e^{3t} = \mathbf{A}$	A1
$t=0, \mathbf{v} = 8\mathbf{i}-12\mathbf{j} \Rightarrow \mathbf{v} = (8\mathbf{i}-12\mathbf{j})e^{-3t}$	M1 A1
$t = \frac{2}{3}\ln 2 \Rightarrow \mathbf{v} = (8\mathbf{i}-12\mathbf{j})e^{-2\ln 2} = (2\mathbf{i}-3\mathbf{j}) \text{ m s}^{-1}$	DM1 A1	(7)

## Question 2:

| Working/Answer | Marks | Guidance |
|---|---|---|
| $\frac{d\mathbf{v}}{dt} + 3\mathbf{v} = \mathbf{0}$ | B1 | |
| IF $= e^{3t} \Rightarrow \frac{d(\mathbf{v}e^{3t})}{dt} = \mathbf{0}$ | M1 | |
| $\Rightarrow \mathbf{v}e^{3t} = \mathbf{A}$ | A1 | |
| $t=0, \mathbf{v} = 8\mathbf{i}-12\mathbf{j} \Rightarrow \mathbf{v} = (8\mathbf{i}-12\mathbf{j})e^{-3t}$ | M1 A1 | |
| $t = \frac{2}{3}\ln 2 \Rightarrow \mathbf{v} = (8\mathbf{i}-12\mathbf{j})e^{-2\ln 2} = (2\mathbf{i}-3\mathbf{j}) \text{ m s}^{-1}$ | DM1 A1 | **(7)** |

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2. At time $t$ seconds, the position vector of a particle $P$ is $\mathbf { r }$ metres, where $\mathbf { r }$ satisfies the differential equation

$$\frac { \mathrm { d } ^ { 2 } \mathbf { r } } { \mathrm {~d} t ^ { 2 } } + 3 \frac { \mathrm {~d} \mathbf { r } } { \mathrm {~d} t } = \mathbf { 0 }$$

When $t = 0$, the velocity of $P$ is $( 8 \mathbf { i } - 12 \mathbf { j } ) \mathrm { m } \mathrm { s } ^ { - 1 }$.\\
Find the velocity of $P$ when $t = \frac { 2 } { 3 } \ln 2$.\\
(7)\\

\hfill \mbox{\textit{Edexcel M5 2007 Q2 [7]}}

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Q1 4 Q2 7 Q3 5 Q4 7 Q5 9 Q6 13 Q7 14 Q8 16