What is M7.3: Constant acceleration formulae?

List s, u, v, a, t with a stated positive direction, choose the formula containing the three known quantities and the one required, and substitute with signs. The formulae derive from a straight-line velocity–time graph: the gradient is a, giving v=u+at, and the area is the displacement, giving s=1/2(u+v)t; eliminating t or v gives the others.

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.

Definition

Specification M7.3: understand, use and derive the formulae for constant acceleration for motion in a straight line, $v=u+at$ , $s=ut+\frac12at^2$ , $s=\frac12(u+v)t$ , $v^2=u^2+2as$ and $s=vt-\frac12at^2$ ; extend to two dimensions using vectors.

Explanation

List $s$ , $u$ , $v$ , $a$ , $t$ with a stated positive direction, choose the formula containing the three known quantities and the one required, and substitute with signs. The formulae derive from a straight-line velocity–time graph: the gradient is $a$ , giving $v=u+at$ , and the area is the displacement, giving $s=\frac12(u+v)t$ ; eliminating $t$ or $v$ gives the others.
On a speed–time graph the area under each section is the distance, and a trapezium or triangle is usually quicker than the formulae.
In two dimensions the same formulae hold for vectors, $\mathbf v=\mathbf u+\mathbf at$ and $\mathbf r=\mathbf r_0+\mathbf ut+\frac12\mathbf at^2$ , worked component by component; speed is the magnitude $\sqrt{v_x^2+v_y^2}$ and a direction needs a bearing or an angle.
The formulae apply only while the acceleration is constant; a velocity given as a function of $t$ needs calculus.
Give units with every answer.

Worked example

A particle

P

moves with constant acceleration

(\mathbf i+2\mathbf j)

m s

^{-2}

. At time

t=0

the velocity of

P

(2\mathbf i-5\mathbf j)

m s

^{-1}

. Find the speed of

P

when

t=4

(4)

(Total for Question 1 is 4 marks)

Working

1.Use $\mathbf v=\mathbf u+\mathbf at$ : $\mathbf v=(2\mathbf i-5\mathbf j)+4(\mathbf i+2\mathbf j)$ .
2.Collect components: $\mathbf v=(6\mathbf i+3\mathbf j)$ m s $^{-1}$ .
3.Speed is the magnitude: $\sqrt{6^2+3^2}=\sqrt{45}=3\sqrt5=6.71$ m s $^{-1}$ .

Answer: $3\sqrt5$ m s $^{-1}$ ( $6.7$ m s $^{-1}$ )

Mark scheme

Mark scheme for question 1
Question	Scheme	Marks
1.	Uses $\mathbf v=\mathbf u+\mathbf at$ with the given vectors and $t=4$	M1
	$\mathbf v=6\mathbf i+3\mathbf j$	A1
	Applies Pythagoras, including the square root, to their velocity	M1
	$\sqrt{45}=3\sqrt5$ or $6.7$ or better	A1
Notes
A1: Units required

(4 marks)

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.

Mark-scheme wording to use

“v_B = (−16i − 3j) + 5(2.4i + j) M1 v_B = (−4i + 2j) A1 √((−4)² + 2²) M1 √20 = 2√5 , 4.5 or better ( m s⁻¹) A1 … M1 Use of Pythagoras on their v_B to give a magnitude (need the root)” — Pearson 9MA0, June 2023 Paper 32, Q4(a)
“M1 Any complete suvat method to find a … A1 Any equivalent number with correct units. … 2.5, 5/2 , 10/4 m s⁻² units needed.” — Pearson 9MA0, June 2024 Paper 32, Q2(a)
“s = (1/2) × 3.2 × 5² OR s = ((0 + 16)/2) × 5 OR s = (16 × 5) − (1/2) × 3.2 × 5² OR 16² = 2 × 3.2 × s OR from a v-t graph, s = (1/2) × 5 × 16 M1 s = 40 (m) A1 … M1 Complete method to find an equation in s only, possibly using their ‘16’ … A1 cao. Must be positive.” — Pearson 9MA0, June 2023 Paper 32, Q1(b)

What the examiner wants

Units, and the right tool. The June 2024 Paper 32 report says some lost the second mark on the acceleration for incorrect or missing units, that the area approach to the speed–time graph was more successful than the formulae, and that a few used a constant-acceleration approach where the acceleration was not constant and scored nothing.

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)

Verified exam appearances

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

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