What is M7.4: Calculus in kinematics?

When velocity or acceleration is given as a function of t, the constant-acceleration formulae do not apply:

How many marks are M7.4 questions worth?

M7.4 practice drills on this page are worth 2–8 marks each; marks in a real exam vary by question.

Which past exam papers has M7.4 appeared in?

We have not yet indexed a verified past-paper appearance for M7.4 — check the Edexcel A-level Maths 9MA0 past-papers hub directly.

M7.4

Use calculus in kinematics for motion in a straight line: v = dr/dt, a = dv/dt = d²r/dt², r = ∫v dt, v = ∫a dt; extend to 2 dimensions using vectors.

Draft — not yet indexed

Calculus in kinematics

Worked answers and methods for M7.4 on Edexcel A-level Maths 9MA0.

Definition

Specification M7.4: use calculus in kinematics for motion in a straight line: $v=\dfrac{dr}{dt}$ , $a=\dfrac{dv}{dt}=\dfrac{d^2r}{dt^2}$ , $r=\int v\,dt$ , $v=\int a\,dt$ ; extend to two dimensions using vectors.

Explanation

When velocity or acceleration is given as a function of $t$ , the constant-acceleration formulae do not apply:
differentiate displacement to get velocity and velocity to get acceleration, or integrate the other way, adding a constant of integration fixed by the initial conditions. For vectors do this component by component, keeping the $\mathbf i$ and $\mathbf j$ separate and writing the answer in $\mathbf i$ , $\mathbf j$ form.
Speed is the magnitude of the velocity, $\sqrt{v_x^2+v_y^2}$ , so a speed question is not finished at the velocity; a direction of motion is the angle of the velocity vector.
The particle is momentarily at rest when $v=0$ , moving parallel to a given vector when the components are in that ratio, and has maximum speed where $a=0$ .
Distance travelled is not displacement when the direction changes, so find the times where $v=0$ and add the separate sections. Reject negative times.

Worked example

A particle

P

moves along a straight line. At time

t

seconds,

t\geq0

, the velocity of

P

v

m s

^{-1}

, where

v=3t^2-12t+9

. When

t=0

P

is at the point

A

, where

OA=2

m. Find the total distance travelled by

P

in the interval

0\leq t\leq4

(5)

(Total for Question 1 is 5 marks)

Working

1.Integrate to find displacement from $O$ : $r=t^3-6t^2+9t+c$ , and $r=2$ at $t=0$ gives $c=2$ .
2. $P$ changes direction where $v=0$ : $3(t-1)(t-3)=0$ , so at $t=1$ and $t=3$ .
3.Positions: $r(0)=2$ , $r(1)=6$ , $r(3)=2$ , $r(4)=6$ .
4.Distance $=|6-2|+|2-6|+|6-2|=4+4+4=12$ m.

Answer: $12$ m

Mark scheme

Mark scheme for question 1
Question	Scheme	Marks
1.	Integrates $v$ with respect to $t$ , $t^n\to t^{n+1}$ in at least two terms	M1
	$r=t^3-6t^2+9t+2$	A1
	Sets $v=0$ to find when the direction changes, $t=1$ and $t=3$	M1
	Evaluates $r$ at $t=0,1,3,4$ and adds the moduli of the changes	dM1
	$12$ m	A1
Notes
A1: Uses $r=2$ at $t=0$ for the constant
A1: Displacement $4$ m scores A0

(5 marks)

Common mistakes

Don't differentiate position once and label the result acceleration.
Don't integrate velocity to find displacement but equate net displacement with total distance despite direction changes.

Mark-scheme wording to use

“M1 Both powers of t decreasing by 1 … A1 Correct unsimplified derivative or two correct components … M1 Put t = 4 in their v (must be using an attempted derivative of r) and then use Pythagoras with the root, allow a missing − sign … A1 Accept 3.1 or better” — Pearson 9MA0, June 2024 Paper 32, Q4(b)
“Differentiate v wrt t to give a vector. M1 (2t − 3)i + 4tj A1 … M1 At least one power decreasing by 1 in each component in their v … A1 cao (must be a vector)” — Pearson 9MA0, June 2023 Paper 32, Q3(c)
“t² − 3t + 7 = 2t² − 3 OR (t² − 3t + 7)/(2t² − 3) = 1/1 = 1 M1 t = 2 only A1 … M1 Equating i and j components of v or a ratio of 1:1 to obtain a quadratic in t only.” — Pearson 9MA0, June 2023 Paper 32, Q3(b)

What the examiner wants

Finishing the question and using calculus where the acceleration varies. The June 2024 Paper 32 report says most differentiated correctly and used Pythagoras for the speed, but a minority found the velocity and stopped, losing two marks, and a few tried a constant-acceleration approach without realising the acceleration was not constant.

Exam tip

For total distance, find every zero of velocity in the interval and add the magnitudes of the separate displacements.

Worked practice

Tier 1 · Easy

A particle has position

r=t^3-4t^2+5t

metres at time

t

seconds. Find its velocity and acceleration at

t=2

(3)

(Total for Question 1 is 3 marks)

Mark scheme

Mark scheme for question 1
Question	Scheme	Marks
1	Velocity $=1\,\text{m s}^{-1}$ Acceleration $=4\,\text{m s}^{-2}$	3
Notes
$v=\dfrac{\mathrm{d}r}{\mathrm{d}t}=3t^2-8t+5$ , so $v(2)=12-16+5=1\,\text{m s}^{-1}$ . Then $a=\dfrac{\mathrm{d}v}{\mathrm{d}t}=6t-8$ , giving $a(2)=4\,\text{m s}^{-2}$ .

(3 marks)

Tier 2 · Standard

A particle has acceleration

a=6t-4\,\text{m s}^{-2}

. Initially it is at position

-2\,\text{m}

and moves at

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

. Find its velocity and position as functions of

t

, and find both values when

t=2

(5)

(Total for Question 2 is 5 marks)

Mark scheme

Mark scheme for question 2
Question	Scheme	Marks
2	$v=3t^2-4t+3$ $r=t^3-2t^2+3t-2$ $v(2)=7\,\text{m s}^{-1}$ and $r(2)=4\,\text{m}$	5
Notes
Integrate acceleration: $v=3t^2-4t+C$ . Since $v(0)=3$ , $C=3$ . Integrate again: $r=t^3-2t^2+3t+D$ . Since $r(0)=-2$ , $D=-2$ . Substitution of $t=2$ gives $v(2)=12-8+3=7$ and $r(2)=8-8+6-2=4$ .

(5 marks)

Tier 3 · Hard

A particle has acceleration

\mathbf{a}=(6t\mathbf{i}-4\mathbf{j})\,\text{m s}^{-2}

. Initially its velocity is

(2\mathbf{i}+7\mathbf{j})\,\text{m s}^{-1}

and its position vector is

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

. Find its velocity and position vectors at time

t

. Hence find its position and speed when its velocity is parallel to

\mathbf{i}

(8)

(Total for Question 3 is 8 marks)

Mark scheme

Mark scheme for question 3
Question	Scheme	Marks
3	$\mathbf{v}=(3t^2+2)\mathbf{i}+(7-4t)\mathbf{j}$ $\mathbf{r}=(t^3+2t-1)\mathbf{i}+(3+7t-2t^2)\mathbf{j}$ At $t=\dfrac74$ , $\mathbf{r}=\dfrac{503}{64}\mathbf{i}+\dfrac{73}{8}\mathbf{j}$ metres Speed $=\dfrac{179}{16}\,\text{m s}^{-1}$	8
Notes
Integrating $\mathbf{a}$ and using $\mathbf{v}(0)=2\mathbf{i}+7\mathbf{j}$ gives $\mathbf{v}=(3t^2+2)\mathbf{i}+(7-4t)\mathbf{j}$ . Integrating again and using $\mathbf{r}(0)=-\mathbf{i}+3\mathbf{j}$ gives $\mathbf{r}=(t^3+2t-1)\mathbf{i}+(3+7t-2t^2)\mathbf{j}$ . Parallel to $\mathbf{i}$ requires $7-4t=0$ , so $t=7/4$ . Substitution gives $\mathbf{r}=503\mathbf{i}/64+73\mathbf{j}/8$ . The remaining velocity component is $3(7/4)^2+2=179/16$ , so the speed is $179/16\,\text{m s}^{-1}$ .

(8 marks)

Tier 1 · Easy

A particle has velocity

v=4t-3\,\text{m s}^{-1}

. Find its displacement from

t=1

t=3

(2)

(Total for Question 4 is 2 marks)

Mark scheme

Mark scheme for question 4
Question	Scheme	Marks
4	$10\,\text{m}$	2
Notes
The displacement is $\int_1^3(4t-3)\,\mathrm{d}t=[2t^2-3t]_1^3=(18-9)-(2-3)=10\,\text{m}$ .

(2 marks)

Tier 2 · Standard

A particle moves on a straight line with acceleration

a=4-6t\,\text{m s}^{-2}

. Given that

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

when

t=1

and that its position is

2\,\text{m}

when

t=0

, find expressions for its velocity and position. Hence find its displacement from

t=0

t=3

(5)

(Total for Question 5 is 5 marks)

Mark scheme

Mark scheme for question 5
Question	Scheme	Marks
5	$v=4t-3t^2+4$ $r=2t^2-t^3+4t+2$ Displacement $=3\,\text{m}$	5
Notes
Integrating gives $v=4t-3t^2+C$ . Since $v(1)=5$ , $1+C=5$ and $C=4$ . Integrating again gives $r=2t^2-t^3+4t+D$ . Since $r(0)=2$ , $D=2$ . Thus $r(3)=18-27+12+2=5$ , so the displacement is $r(3)-r(0)=5-2=3\,\text{m}$ .

(5 marks)

Tier 3 · Hard

The vectors

\mathbf{i}

and

\mathbf{j}

are fixed perpendicular unit vectors. During the interval

0<t<1

, the position of a moving particle is

\mathbf{r}=(t^2+2t)\mathbf{i}+(t^3-3t)\mathbf{j}

metres. Find all the times in this interval when its acceleration is perpendicular to its velocity. Find the speed at each of these times, giving exact answers.

(6)

(Total for Question 6 is 6 marks)

Mark scheme

Mark scheme for question 6
Question	Scheme	Marks
6	$t=\dfrac13$ and $t=\dfrac23$ At $t=\dfrac13$ , speed $=\dfrac{8\sqrt2}{3}\,\text{m s}^{-1}$ At $t=\dfrac23$ , speed $=\dfrac{5\sqrt5}{3}\,\text{m s}^{-1}$	6
Notes
$\mathbf{v}=(2t+2)\mathbf{i}+(3t^2-3)\mathbf{j}$ and $\mathbf{a}=2\mathbf{i}+6t\mathbf{j}$ . Perpendicular vectors have zero scalar product, so $2(2t+2)+6t(3t^2-3)=18t^3-14t+4=2(3t-1)(3t-2)(t+1)=0$ . In $0<t<1$ , $t=1/3$ or $2/3$ . The velocities are $(8/3,-8/3)$ and $(10/3,-5/3)$ , whose magnitudes are $8\sqrt2/3$ and $5\sqrt5/3\,\text{m s}^{-1}$ respectively.

(6 marks)

Tier 2 · Standard

A particle moves on a straight line with acceleration

a=6t-8\,\text{m s}^{-2}

. Its velocity is

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

when

t=2

. Find its velocity as a function of

t

and find the exact minimum value of its velocity for

t\geq0

(4)

(Total for Question 7 is 4 marks)

Mark scheme

Mark scheme for question 7
Question	Scheme	Marks
7	$v=3t^2-8t+11\,\text{m s}^{-1}$ Minimum velocity $=\dfrac{17}{3}\,\text{m s}^{-1}$ , attained when $t=\dfrac43\,\text{s}$	4
Notes
Integrating the acceleration gives $v=3t^2-8t+C$ . Since $v=7$ at $t=2$ , $7=12-16+C$ , so $C=11$ . The velocity is stationary when $dv/dt=a=0$ , giving $6t-8=0$ and $t=4/3$ . As the quadratic coefficient is positive this is a minimum, with $v=3(16/9)-8(4/3)+11=17/3\,\text{m s}^{-1}$ .

(4 marks)

Tier 3 · Hard

A particle moves on a straight line with acceleration

a=6t-6\,\text{m s}^{-2}

. It starts at position

0\,\text{m}

and returns to this position when

t=3\,\text{s}

. Find its initial velocity. Hence find the time at which it is furthest from its initial position during

0\leq t\leq3

, and find this greatest distance.

(6)

(Total for Question 8 is 6 marks)

Mark scheme

Mark scheme for question 8
Question	Scheme	Marks
8	Initial velocity $=0\,\text{m s}^{-1}$ The particle is furthest from its initial position at $t=2\,\text{s}$ Greatest distance $=4\,\text{m}$	6
Notes
Integrating gives $v=3t^2-6t+C$ and displacement $s=t^3-3t^2+Ct$ , since $s=0$ at $t=0$ . The return condition $s(3)=0$ gives $27-27+3C=0$ , so $C=0$ and the initial velocity is zero. Stationary positions occur when $v=3t(t-2)=0$ , so the candidates in the closed interval are $t=0,2,3$ . Their displacements are $0,-4,0$ metres respectively, so the particle is furthest from its initial position at $t=2$ and the greatest distance is $4\,\text{m}$ .

(6 marks)

Tier 3 · Hard

A particle moves on a straight line with acceleration

a=6t-4\,\text{m s}^{-2}

. It is instantaneously at rest when

t=1\,\text{s}

and its position is

5\,\text{m}

when

t=2\,\text{s}

. Find its velocity and position as functions of

t

. Hence find all times in

0\leq t\leq2

when the particle is at rest and find the exact total distance travelled during this interval. Give all numerical answers exactly.

(7)

(Total for Question 9 is 7 marks)

Mark scheme

Mark scheme for question 9
Question	Scheme	Marks
9	$v=3t^2-4t+1\,\text{m s}^{-1}$ $r=t^3-2t^2+t+3\,\text{m}$ Rest times $t=\dfrac13\,\text{s}$ and $t=1\,\text{s}$ Total distance $=\dfrac{62}{27}\,\text{m}$	7
Notes
Integrating gives $v=3t^2-4t+C$ . Since $v(1)=0$ , $C=1$ . Integrating again gives $r=t^3-2t^2+t+D$ ; $r(2)=5$ gives $D=3$ . Now $v=(3t-1)(t-1)$ , so the rest times are $1/3$ and $1$ . The positions at $t=0,1/3,1,2$ are $3,85/27,3,5$ . Hence the distance is $\|85/27-3\|+\|3-85/27\|+\|5-3\|=4/27+4/27+2=62/27\,\text{m}$ .

(7 marks)

Q10

Tier 3 · Hard

10.

The vectors

\mathbf{i}

and

\mathbf{j}

are fixed perpendicular unit vectors. A particle has velocity

\mathbf{v}=(2t+1)\mathbf{i}+4\mathbf{j}\,\text{m s}^{-1}

for

t\geq0

and starts at the origin. Find its position vector at time

t

and eliminate

t

to obtain a Cartesian equation for its path, using coordinates

(x,y)

. Find the first point where the path crosses the line

x=6

, and find the exact speed there. Give all numerical answers exactly.

(6)

(Total for Question 10 is 6 marks)

Mark scheme

Mark scheme for question 10
Question	Scheme	Marks
10	$\mathbf{r}=(t^2+t)\mathbf{i}+4t\mathbf{j}\,\text{m}$ $x=\dfrac{y^2}{16}+\dfrac{y}{4}$ First crossing point $(6,8)$ Speed $=\sqrt{41}\,\text{m s}^{-1}$	6
Notes
Integrating the components and using $\mathbf r(0)=\mathbf0$ gives $x=t^2+t$ and $y=4t$ . Thus $t=y/4$ and $x=y^2/16+y/4$ . On $x=6$ , $t^2+t=6$ , so $(t-2)(t+3)=0$ ; the allowed time is $t=2$ , giving $y=8$ . The velocity then is $5\mathbf{i}+4\mathbf{j}\,\text{m s}^{-1}$ , whose magnitude is $\sqrt{25+16}=\sqrt{41}\,\text{m s}^{-1}$ .

(6 marks)

Verified exam appearances

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