What is M7.3: Constant acceleration formulae?

For constant acceleration, use v=u+at and r=r_0+ut+1/2at^2, with scalar forms available for one-dimensional motion. List the known quantities from s,u,v,a,t, choose an equation containing the required unknown, and keep the sign convention consistent throughout.

How many marks are M7.3 questions worth?

M7.3 practice drills on this page are worth 2–6 marks each; marks in a real exam vary by question.

Which past exam papers has M7.3 appeared in?

We have not yet indexed a verified past-paper appearance for M7.3 — check the Edexcel A-level Maths 9MA0 past-papers hub directly.

M7.3

Understand, use and derive the formulae for constant acceleration for motion in a straight line; extend to 2 dimensions using vectors.

Draft — not yet indexed

Constant acceleration formulae

Worked answers and methods for M7.3 on Edexcel A-level Maths 9MA0.

Explanation

For constant acceleration, use $\mathbf{v}=\mathbf{u}+\mathbf{a}t$ and $\mathbf{r}=\mathbf{r}_0+\mathbf{u}t+\dfrac12\mathbf{a}t^2$ , with scalar forms available for one-dimensional motion. List the known quantities from $s,u,v,a,t$ , choose an equation containing the required unknown, and keep the sign convention consistent throughout.
Eliminating $t$ between $v=u+at$ and $s=\dfrac12(u+v)t$ gives $v^2=u^2+2as$ for motion with constant acceleration.
A common error is to use constant-acceleration formulae when acceleration varies, or to use vector magnitudes before completing the component equations. $v=u+at=4+1.5(6)=13\,\text{m s}^{-1}$ .
Also $s=ut+\tfrac12at^2=4(6)+\tfrac12(1.5)(6^2)=24+27=51\,\text{m}$ .
Confirm that the time is within the stated motion interval.

Worked example

A car moves in a straight line with initial speed $4\,\text{m s}^{-1}$ and constant acceleration $1.5\,\text{m s}^{-2}$ . Find its speed and the distance it travels in the next $6\,\text{s}$ .

1. $v=u+at=4+1.5(6)=13\,\text{m s}^{-1}$ .
2.Also $s=ut+\tfrac12at^2=4(6)+\tfrac12(1.5)(6^2)=24+27=51\,\text{m}$ .

Answer: Speed $=13\,\text{m s}^{-1}$ ; Distance $=51\,\text{m}$

Common mistakes

Don't mix a displacement sign convention with an acceleration value treated as an unsigned magnitude.
Don't select a constant-acceleration formula containing an extra unknown and introduce avoidable algebra.

Exam tip

List the known kinematic quantities with signs, then choose the equation containing only the required unknown.

Worked practice

Tier 1 · Easy

A particle accelerates uniformly from

5\,\text{m s}^{-1}

11\,\text{m s}^{-1}

3\,\text{s}

. Find its acceleration.

(2)

(Total for Question 1 is 2 marks)

Mark scheme

Mark scheme for question 1
Question	Scheme	Marks
1	$2\,\text{m s}^{-2}$	2
Notes
From $v=u+at$ , $11=5+3a$ , so $a=(11-5)/3=2\,\text{m s}^{-2}$ .

(2 marks)

Tier 2 · Standard

A particle has initial velocity

(3\mathbf{i}+4\mathbf{j})\,\text{m s}^{-1}

and constant acceleration

(2\mathbf{i}-\mathbf{j})\,\text{m s}^{-2}

. Find its velocity and displacement after

5\,\text{s}

(4)

(Total for Question 2 is 4 marks)

Mark scheme

Mark scheme for question 2
Question	Scheme	Marks
2	Velocity $(13\mathbf{i}-\mathbf{j})\,\text{m s}^{-1}$ Displacement $(40\mathbf{i}+7.5\mathbf{j})\,\text{m}$	4
Notes
Use the constant-acceleration formulae componentwise. The velocity is $\mathbf{v}=\mathbf{u}+t\mathbf{a}=(3,4)+5(2,-1)=(13,-1)$ . The displacement is $\mathbf{s}=t\mathbf{u}+\tfrac12t^2\mathbf{a}=5(3,4)+12.5(2,-1)=(40,7.5)$ .

(4 marks)

Tier 3 · Hard

A particle starts at the origin with velocity

(3\mathbf{i}+8\mathbf{j})\,\text{m s}^{-1}

and constant acceleration

(2\mathbf{i}-2\mathbf{j})\,\text{m s}^{-2}

. Find the time when its velocity is parallel to

\mathbf{i}

, its displacement then, and its speed then.

(6)

(Total for Question 3 is 6 marks)

Mark scheme

Mark scheme for question 3
Question	Scheme	Marks
3	$t=4\,\text{s}$ Displacement $=(28\mathbf{i}+16\mathbf{j})\,\text{m}$ Speed $=11\,\text{m s}^{-1}$	6
Notes
$\mathbf{v}=\mathbf{u}+\mathbf{a}t=(3+2t)\mathbf{i}+(8-2t)\mathbf{j}$ . It is parallel to $\mathbf{i}$ when $8-2t=0$ , giving $t=4$ . Then $\mathbf{r}=\mathbf{u}t+\tfrac12\mathbf{a}t^2=(12\mathbf{i}+32\mathbf{j})+(16\mathbf{i}-16\mathbf{j})=28\mathbf{i}+16\mathbf{j}$ . The velocity is $11\mathbf{i}\,\text{m s}^{-1}$ , so the speed is $11\,\text{m s}^{-1}$ .

(6 marks)

Tier 1 · Easy

A train travels with initial speed

9\,\text{m s}^{-1}

and constant acceleration

1.2\,\text{m s}^{-2}

. Find the distance it travels in

5\,\text{s}

(2)

(Total for Question 4 is 2 marks)

Mark scheme

Mark scheme for question 4
Question	Scheme	Marks
4	$60\,\text{m}$	2
Notes
$s=ut+\tfrac12at^2=9(5)+\tfrac12(1.2)(5^2)=45+15=60\,\text{m}$ .

(2 marks)

Tier 2 · Standard

The vectors

\mathbf{i}

and

\mathbf{j}

are fixed perpendicular unit vectors. A particle moves with constant acceleration

(2\mathbf{i}+3\mathbf{j})\,\text{m s}^{-2}

. Its displacement during the first

4\,\text{s}

(24\mathbf{i}+4\mathbf{j})\,\text{m}

. Find its initial velocity and its velocity at

t=4

(4)

(Total for Question 5 is 4 marks)

Mark scheme

Mark scheme for question 5
Question	Scheme	Marks
5	Initial velocity $=(2\mathbf{i}-5\mathbf{j})\,\text{m s}^{-1}$ Velocity at $t=4$ is $(10\mathbf{i}+7\mathbf{j})\,\text{m s}^{-1}$	4
Notes
Using $\mathbf{s}=\mathbf{u}t+\tfrac12\mathbf{a}t^2$ , $(24,4)=4\mathbf{u}+8(2,3)=4\mathbf{u}+(16,24)$ . Hence $4\mathbf{u}=(8,-20)$ and $\mathbf{u}=(2,-5)$ . Then $\mathbf{v}=\mathbf{u}+4\mathbf{a}=(2,-5)+(8,12)=(10,7)$ .

(4 marks)

Tier 3 · Hard

The vectors

\mathbf{i}

and

\mathbf{j}

are fixed perpendicular unit vectors. A particle starts from the origin with velocity

(4\mathbf{i}+\mathbf{j})\,\text{m s}^{-1}

. For the first

3\,\text{s}

its acceleration is

(\mathbf{i}-2\mathbf{j})\,\text{m s}^{-2}

. Its acceleration then changes instantaneously to

(-2\mathbf{i}+\mathbf{j})\,\text{m s}^{-2}

for a further

2\,\text{s}

. Find its velocity and position vector at the end of the

5\,\text{s}

(6)

(Total for Question 6 is 6 marks)

Mark scheme

Mark scheme for question 6
Question	Scheme	Marks
6	Velocity $=(3\mathbf{i}-3\mathbf{j})\,\text{m s}^{-1}$ Position $=(\dfrac{53}{2}\mathbf{i}-14\mathbf{j})\,\text{m}$	6
Notes
After $3\,\text{s}$ , $\mathbf{v}_1=(4,1)+3(1,-2)=(7,-5)$ and $\mathbf{r}_1=3(4,1)+\tfrac12(3^2)(1,-2)=(33/2,-6)$ . During the next $2\,\text{s}$ , $\mathbf{v}_2=(7,-5)+2(-2,1)=(3,-3)$ . The second displacement is $2(7,-5)+\tfrac12(2^2)(-2,1)=(10,-8)$ , so $\mathbf{r}_2=(33/2,-6)+(10,-8)=(53/2,-14)$ .

(6 marks)

Tier 2 · Standard

For motion with constant acceleration, eliminate

t

from

v=u+at

and

s=\tfrac12(u+v)t

to derive

v^2=u^2+2as

. Hence find the distance required for a particle moving at

13\,\text{m s}^{-1}

to come to rest with constant acceleration

-2.6\,\text{m s}^{-2}

(5)

(Total for Question 7 is 5 marks)

Mark scheme

Mark scheme for question 7
Question	Scheme	Marks
7	$v^2=u^2+2as$ Stopping distance $=32.5\,\text{m}$	5
Notes
From $v=u+at$ , $t=(v-u)/a$ . Substitution in $s=\tfrac12(u+v)t$ gives $s=(u+v)(v-u)/(2a)=(v^2-u^2)/(2a)$ , so $v^2=u^2+2as$ . With $v=0$ , $u=13$ and $a=-2.6$ , $0=13^2+2(-2.6)s$ , hence $s=169/5.2=32.5\,\text{m}$ .

(5 marks)

Tier 3 · Hard

The vectors

\mathbf{i}

and

\mathbf{j}

are fixed perpendicular unit vectors. A particle starts at the origin with constant acceleration. Its position vectors after

2\,\text{s}

and

5\,\text{s}

are

(10\mathbf{i}+6\mathbf{j})\,\text{m}

and

(40\mathbf{i}+45\mathbf{j})\,\text{m}

respectively. Find its initial velocity, its acceleration and its velocity after

5\,\text{s}

(6)

(Total for Question 8 is 6 marks)

Mark scheme

Mark scheme for question 8
Question	Scheme	Marks
8	Initial velocity $=(3\mathbf{i}-\mathbf{j})\,\text{m s}^{-1}$ Acceleration $=(2\mathbf{i}+4\mathbf{j})\,\text{m s}^{-2}$ Velocity after $5\,\text{s}$ is $(13\mathbf{i}+19\mathbf{j})\,\text{m s}^{-1}$	6
Notes
Write the initial velocity as $\mathbf{u}$ and acceleration as $\mathbf{a}$ . From $\mathbf{r}=\mathbf{u}t+\tfrac12\mathbf{a}t^2$ , $2\mathbf{u}+2\mathbf{a}=10\mathbf{i}+6\mathbf{j}$ and $5\mathbf{u}+\tfrac{25}{2}\mathbf{a}=40\mathbf{i}+45\mathbf{j}$ . Thus $\mathbf{u}+\mathbf{a}=5\mathbf{i}+3\mathbf{j}$ . Substituting $\mathbf{u}=5\mathbf{i}+3\mathbf{j}-\mathbf{a}$ into the second equation gives $\tfrac{15}{2}\mathbf{a}=15\mathbf{i}+30\mathbf{j}$ , so $\mathbf{a}=2\mathbf{i}+4\mathbf{j}$ and $\mathbf{u}=3\mathbf{i}-\mathbf{j}$ . Hence $\mathbf{v}=\mathbf{u}+5\mathbf{a}=13\mathbf{i}+19\mathbf{j}$ .

(6 marks)

Tier 3 · Hard

Particle

A

passes a point

O

at time

t=0

and continues in a straight line with constant speed

10\,\text{m s}^{-1}

. At

t=2\,\text{s}

, particle

B

starts from rest at

O

and moves along the same line with constant acceleration

4\,\text{m s}^{-2}

. Find the exact time after

A

passes

O

when

B

first catches

A

, the distance from

O

where this occurs, and the speed of

B

relative to

A

at that instant. Give all numerical answers exactly.

(6)

(Total for Question 9 is 6 marks)

Mark scheme

Mark scheme for question 9
Question	Scheme	Marks
9	Catch time $=\dfrac{9+\sqrt{65}}{2}\,\text{s}$ Distance from $O=5(9+\sqrt{65})\,\text{m}$ Relative speed $=2\sqrt{65}\,\text{m s}^{-1}$	6
Notes
At global time $t\geq2$ , $A$ is $10t$ metres from $O$ , while $B$ has moved $\tfrac12(4)(t-2)^2=2(t-2)^2$ metres. Equating gives $2(t-2)^2=10t$ , or $t^2-9t+4=0$ . The root exceeding $2$ is $(9+\sqrt{65})/2$ . The meeting distance is $10t=5(9+\sqrt{65})\,\text{m}$ . Particle $B$ then has speed $4(t-2)=2(5+\sqrt{65})=10+2\sqrt{65}$ , so its speed relative to $A$ is $2\sqrt{65}\,\text{m s}^{-1}$ .

(6 marks)

Q10

Tier 3 · Hard

10.

A train moves with constant acceleration. At

t=0

the front of the train passes signal

A

. At

t=4\,\text{s}

the front passes signal

B

, which is

60\,\text{m}

beyond

A

, with speed

18\,\text{m s}^{-1}

. The rear of the train passes signal

A

t=2\,\text{s}

. Find the train's speed when its front passes

A

, its acceleration, the exact length of the train and its speed when its rear passes

A

. Give all numerical answers exactly.

(6)

(Total for Question 10 is 6 marks)

Mark scheme

Mark scheme for question 10
Question	Scheme	Marks
10	Initial speed $=12\,\text{m s}^{-1}$ Acceleration $=1.5\,\text{m s}^{-2}$ Train length $=27\,\text{m}$ Speed when the rear passes $A=15\,\text{m s}^{-1}$	6
Notes
Let the speed at $t=0$ be $u$ and the acceleration be $a$ . Then $18=u+4a$ and $60=4u+8a$ . Substituting $u=18-4a$ into the second equation gives $60=72-8a$ , so $a=1.5$ and $u=12$ . Between $t=0$ and $t=2$ , the front travels $12(2)+\tfrac12(1.5)(2^2)=27\,\text{m}$ ; this is the train's length because the rear then reaches $A$ . Its speed at that time is $12+1.5(2)=15\,\text{m s}^{-1}$ .

(6 marks)

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