Question 1 - A-Level Maths

Edexcel M2 2024 October — Question 1

Exam Board	Edexcel
Module	M2 (Mechanics 2)
Year	2024
Session	October
Paper	Download PDF ↗
Topic	Variable acceleration (1D)
Type	Position vector from velocity integration
Difficulty	Standard +0.3 This is a straightforward M2 mechanics question requiring integration of velocity to find position (with initial conditions), differentiation to find acceleration, and solving a ratio equation for direction. All techniques are standard with no novel problem-solving required, making it slightly easier than average for A-level.
Spec	1.10c Magnitude and direction: of vectors 1.10d Vector operations: addition and scalar multiplication 3.02f Non-uniform acceleration: using differentiation and integration 3.02g Two-dimensional variable acceleration

In this question you must show all stages of your working. Solutions relying entirely on calculator technology are not acceptable.

At time $t$ seconds, $t \geqslant 0$, a particle $P$ is moving with velocity $\mathbf { v } \mathrm { ms } ^ { - 1 }$, where $$\mathbf { v } = 3 ( t + 2 ) ^ { 2 } \mathbf { i } + 5 t ( t + 2 ) \mathbf { j }$$ Position vectors are given relative to the fixed point $O$ At time $t = 0 , P$ is at the point with position vector $( - 30 \mathbf { i } - 45 \mathbf { j } ) \mathrm { m }$.

Find the position vector of $P$ when $t = 3$
Find the magnitude of the acceleration of $P$ when $t = 3$ At time $T$ seconds, $P$ is moving in the direction of the vector $2 \mathbf { i } + \mathbf { j }$
Find the value of $T$

Show mark scheme Show mark scheme source

Question 1:

Answer	Marks	Guidance
1(a)	Use r =  v d t	M1

Correct integration

Eg ( t 3 + 6 t 2 + 1 2 t ) i +  5 t 3 + 5 t 2  j + ( C )

3  5 

Or ( t + 2 ) 3 i + t 3 + 5 t 2 j + ( K )

Answer	Marks
3	A1

Complete method using t =0,r=−30i−45j(m) and substitute t = 3

Indefinite integration

Use of t = 0 , r = − 3 0 i − 4 5 j (m) to find constant of integration and

substitute t =3.

Definite integration

Use of r = ( − 3 0 i − 4 5 j ) +  3 v d t

Answer	Marks
0	DM1
r = 8 7 i + 4 5 j (m)	A1

(4)

1(a)

Answer	Marks
M1	Integrate to obtain r. Powers increase by 1 in both components.

Condone working with separated components. Condone missing

brackets with i-j notation. M0 for suvat.

Answer	Marks
A1	Correct integration. Condone missing constant of integration.
DM1	• Dependent on the preceding M1.

• Must have a constant of integration before using t = 3 (unless

using definite integration)

• Must use (−30i−45j)

• Must substitute t = 3

Condone working with separated components.

DM0 if substitution of t = 3 occurs before finding + C.

( t3 +6t2 +12t−30 ) i+   5 t3 +5t2 −45  j C =(−30i−45j)

3 

( ( t+2 )3 −38 ) i+   5 t3 +5t2 −45  j K =(−38i−45j)

3 

Answer	Marks
A1	87

Correct answer, accept column vector  

45

87i

A0 for poor notation in final answer eg  

45j

Answer	Marks
ISW if they continue and find	r

Answer	Marks
1(b)	d v

Use a =

Answer	Marks
d t	M1

Correct differentiation

Answer	Marks
a = ( 6 t + 1 2 ) i + ( 1 0 t + 1 0 ) j	A1
Substitute t =3 and find magnitude	DM1
a = 5 0 ( m s − 2 )	A1

(4)

1(b)

Answer	Marks
M1	Differentiate to obtain a. Powers decrease by 1 in both components.

Condone working with separated components and missing brackets

with i-j notation. M0 for suvat

Answer	Marks
A1	Correct differentiation
M1	Dependent on the preceding M1

Use of Pythagoras seen or implied 3 0 2 + 4 0 2

Answer	Marks
A1	Correct only

Answer	Marks
1(c)	3 ( T +2 )2 =10T ( T +2 )
( 3T +6=10T ) ( 7T2 +8T −12=0 )	M1

6 T =

Answer	Marks
7	A1

(2)

(10)

Notes

1(c)

Answer	Marks
M1	Correct method using the ratio of i and j components of v. Form an

equation in T only. Accept working in t or T.

Answer	Marks	Guidance
A1	0.86 or better. Condone t instead of T.
Question	Scheme	Marks

Question 1:
--- 1(a) ---
1(a) | Use r =  v d t | M1
Correct integration
Eg ( t 3 + 6 t 2 + 1 2 t ) i +  5 t 3 + 5 t 2  j + ( C )
3
 5 
Or ( t + 2 ) 3 i + t 3 + 5 t 2 j + ( K )
3 | A1
Complete method using t =0,r=−30i−45j(m) and substitute t = 3
Indefinite integration
Use of t = 0 , r = − 3 0 i − 4 5 j (m) to find constant of integration and
substitute t =3.
Definite integration
Use of r = ( − 3 0 i − 4 5 j ) +  3 v d t
0 | DM1
r = 8 7 i + 4 5 j (m) | A1
(4)
1(a)
M1 | Integrate to obtain r. Powers increase by 1 in both components.
Condone working with separated components. Condone missing
brackets with i-j notation. M0 for suvat.
A1 | Correct integration. Condone missing constant of integration.
DM1 | • Dependent on the preceding M1.
• Must have a constant of integration before using t = 3 (unless
using definite integration)
• Must use (−30i−45j)
• Must substitute t = 3
Condone working with separated components.
DM0 if substitution of t = 3 occurs before finding + C.
( t3 +6t2 +12t−30 ) i+   5 t3 +5t2 −45  j C =(−30i−45j)
3 
( ( t+2 )3 −38 ) i+   5 t3 +5t2 −45  j K =(−38i−45j)
3 
A1 | 87
Correct answer, accept column vector  
45
87i
A0 for poor notation in final answer eg  
45j
ISW if they continue and find |r|
--- 1(b) ---
1(b) | d v
Use a =
d t | M1
Correct differentiation
a = ( 6 t + 1 2 ) i + ( 1 0 t + 1 0 ) j | A1
Substitute t =3 and find magnitude | DM1
a = 5 0 ( m s − 2 ) | A1
(4)
1(b)
M1 | Differentiate to obtain a. Powers decrease by 1 in both components.
Condone working with separated components and missing brackets
with i-j notation. M0 for suvat
A1 | Correct differentiation
M1 | Dependent on the preceding M1
Use of Pythagoras seen or implied 3 0 2 + 4 0 2
A1 | Correct only
--- 1(c) ---
1(c) | 3 ( T +2 )2 =10T ( T +2 )
( 3T +6=10T ) ( 7T2 +8T −12=0 ) | M1
6
T =
7 | A1
(2)
(10)
Notes
1(c)
M1 | Correct method using the ratio of i and j components of v. Form an
equation in T only. Accept working in t or T.
A1 | 0.86 or better. Condone t instead of T.
Question | Scheme | Marks

Show LaTeX source

\begin{enumerate}
  \item In this question you must show all stages of your working. Solutions relying entirely on calculator technology are not acceptable.
\end{enumerate}

At time $t$ seconds, $t \geqslant 0$, a particle $P$ is moving with velocity $\mathbf { v } \mathrm { ms } ^ { - 1 }$, where

$$\mathbf { v } = 3 ( t + 2 ) ^ { 2 } \mathbf { i } + 5 t ( t + 2 ) \mathbf { j }$$

Position vectors are given relative to the fixed point $O$\\
At time $t = 0 , P$ is at the point with position vector $( - 30 \mathbf { i } - 45 \mathbf { j } ) \mathrm { m }$.\\
(a) Find the position vector of $P$ when $t = 3$\\
(b) Find the magnitude of the acceleration of $P$ when $t = 3$

At time $T$ seconds, $P$ is moving in the direction of the vector $2 \mathbf { i } + \mathbf { j }$\\
(c) Find the value of $T$

\hfill \mbox{\textit{Edexcel M2 2024 Q1}}

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