What is M7.1: Language of kinematics?

Position locates a particle relative to an origin, displacement is the signed change in position, and distance travelled is the total path length and is always non-negative.

How many marks are M7.1 questions worth?

M7.1 practice drills on this page are worth 2–6 marks each; marks in a real exam vary by question.

Which past exam papers has M7.1 appeared in?

We have not yet indexed a verified past-paper appearance for M7.1 — check the Edexcel A-level Maths 9MA0 past-papers hub directly.

M7.1

Understand and use the language of kinematics: position; displacement; distance travelled; velocity; speed; acceleration.

Draft — not yet indexed

Language of kinematics

Worked answers and methods for M7.1 on Edexcel A-level Maths 9MA0.

Explanation

Position locates a particle relative to an origin, displacement is the signed change in position, and distance travelled is the total path length and is always non-negative.
Velocity is the rate of change of displacement, speed is the magnitude of velocity, and acceleration is the rate of change of velocity; choose a positive direction before assigning signs.
For motion from $x_1$ to $x_2$ , displacement is $x_2-x_1$ ; split reversals into separate path lengths for distance, and use change in velocity divided by time for average acceleration.
A common error is to treat distance and displacement, or speed and velocity, as interchangeable; distance and speed cannot be negative.

Worked example

Taking east as positive, a cyclist's velocity changes uniformly from $+8\,\text{m s}^{-1}$ to $-4\,\text{m s}^{-1}$ in $6\,\text{s}$ . Find the acceleration, the time when the cyclist is instantaneously at rest, and the speed at the end.

1.The acceleration is the change in velocity divided by time: $a=(-4-8)/6=-2\,\text{m s}^{-2}$ .
2.Using $v=u+at$ , rest occurs when $0=8-2t$ , so $t=4\,\text{s}$ .
3.The final velocity is $-4\,\text{m s}^{-1}$ , whose magnitude gives speed $4\,\text{m s}^{-1}$ .

Answer: Acceleration $=-2\,\text{m s}^{-2}$ ; The cyclist is at rest after $4\,\text{s}$ ; Final speed $=4\,\text{m s}^{-1}$

Common mistakes

Don't calculate displacement by adding journey lengths without assigning directions.
Don't report a negative velocity as a negative speed and ignore the chosen positive direction.

Exam tip

Keep signs for displacement, velocity and acceleration, but report distance and speed as non-negative magnitudes.

Worked practice

Tier 1 · Easy

A particle moves on a straight line from position

-3\,\text{m}

5\,\text{m}

and then to

1\,\text{m}

. Find its displacement and the distance it travels.

(3)

(Total for Question 1 is 3 marks)

Mark scheme

Mark scheme for question 1
Question	Scheme	Marks
1	Displacement $=4\,\text{m}$ Distance $=12\,\text{m}$	3
Notes
The displacement is final position minus initial position: $1-(-3)=4\,\text{m}$ . The two path lengths are $5-(-3)=8\,\text{m}$ and $5-1=4\,\text{m}$ , so the distance is $8+4=12\,\text{m}$ .

(3 marks)

Tier 2 · Standard

At one instant a particle moving on a straight line has velocity

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

and acceleration

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

. State its speed and explain whether its speed is increasing or decreasing at that instant.

(3)

(Total for Question 2 is 3 marks)

Mark scheme

Mark scheme for question 2
Question	Scheme	Marks
2	Speed $7\,\text{m s}^{-1}$ The speed is decreasing because velocity and acceleration have opposite signs (equivalently, the acceleration acts opposite to the motion).	3
Notes
Speed is the magnitude of velocity, so it is $\|-7\|=7\,\text{m s}^{-1}$ . The velocity is negative while the acceleration is positive, so the acceleration acts opposite to the direction of motion. The magnitude of the velocity is therefore decreasing at that instant.

(3 marks)

Tier 3 · Hard

A runner travels

120\,\text{m}

east in

15\,\text{s}

50\,\text{m}

west in

5\,\text{s}

and then

30\,\text{m}

east in

10\,\text{s}

. Find the runner's total distance, displacement, average speed and average velocity.

(6)

(Total for Question 3 is 6 marks)

Mark scheme

Mark scheme for question 3
Question	Scheme	Marks
3	Distance $=200\,\text{m}$ Displacement $=100\,\text{m}$ east Average speed $=\dfrac{20}{3}\,\text{m s}^{-1}$ Average velocity $=\dfrac{10}{3}\,\text{m s}^{-1}$ east	6
Notes
Distance adds all path lengths: $120+50+30=200\,\text{m}$ . Taking east as positive, displacement is $120-50+30=100\,\text{m}$ . The total time is $30\,\text{s}$ , so average speed is $200/30=20/3\,\text{m s}^{-1}$ and average velocity is $100/30=10/3\,\text{m s}^{-1}$ east.

(6 marks)

Tier 1 · Easy

Taking upwards as positive, a lift has velocity

-3.5\,\text{m s}^{-1}

. State its speed and direction of motion.

(2)

(Total for Question 4 is 2 marks)

Mark scheme

Mark scheme for question 4
Question	Scheme	Marks
4	Speed $=3.5\,\text{m s}^{-1}$ The lift is moving downwards	2
Notes
Speed is the magnitude of velocity, so it is $\|-3.5\|=3.5\,\text{m s}^{-1}$ . The negative sign means the lift moves opposite to the positive direction, so it is moving downwards.

(2 marks)

Tier 2 · Standard

Taking right as positive, the velocity of a particle changes uniformly from

-6\,\text{m s}^{-1}

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

4\,\text{s}

. Find its acceleration. State whether its speed is initially increasing or decreasing, and whether the particle changes direction during the interval.

(4)

(Total for Question 5 is 4 marks)

Mark scheme

Mark scheme for question 5
Question	Scheme	Marks
5	Acceleration $=2\,\text{m s}^{-2}$ Its speed is initially decreasing It changes direction	4
Notes
The acceleration is $[2-(-6)]/4=2\,\text{m s}^{-2}$ . Initially the velocity and acceleration have opposite signs, so the speed decreases. Since the velocity changes continuously from negative to positive, it passes through zero and the particle changes direction.

(4 marks)

Tier 3 · Hard

A particle starts at position

10\,\text{m}

on a directed straight line. It moves with constant velocity

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

for

4\,\text{s}

, then with velocity

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

for

6\,\text{s}

, and finally with velocity

-2\,\text{m s}^{-1}

for

3\,\text{s}

. Find its final position, total distance travelled, average speed and average velocity over the complete motion.

(6)

(Total for Question 6 is 6 marks)

Mark scheme

Mark scheme for question 6
Question	Scheme	Marks
6	Final position $=22\,\text{m}$ Distance $=48\,\text{m}$ Average speed $=\dfrac{48}{13}\,\text{m s}^{-1}$ Average velocity $=\dfrac{12}{13}\,\text{m s}^{-1}$ in the positive direction	6
Notes
The signed displacements are $-3(4)=-12$ , $5(6)=30$ and $-2(3)=-6$ metres. Their sum is $12\,\text{m}$ , so the final position is $10+12=22\,\text{m}$ . Distance is $12+30+6=48\,\text{m}$ and total time is $13\,\text{s}$ . Hence average speed is $48/13\,\text{m s}^{-1}$ and average velocity is $12/13\,\text{m s}^{-1}$ .

(6 marks)

Tier 2 · Standard

Taking right as positive, a particle has velocity

-6\,\text{m s}^{-1}

at time

t=0

. During the next

2\,\text{s}

its displacement is

-4\,\text{m}

. State its speed at

t=0

and find its average velocity during the interval. Explain why the information given is not sufficient to find its average speed during the interval.

(4)

(Total for Question 7 is 4 marks)

Mark scheme

Mark scheme for question 7
Question	Scheme	Marks
7	Speed at $t=0$ is $6\,\text{m s}^{-1}$ Average velocity $=-2\,\text{m s}^{-1}$ The distance travelled is not known, so the average speed cannot be determined	4
Notes
Speed is the magnitude of velocity, so the initial speed is $\|-6\|=6\,\text{m s}^{-1}$ . Average velocity is displacement divided by elapsed time, giving $-4/2=-2\,\text{m s}^{-1}$ . The displacement does not reveal whether the particle reversed direction, so it does not determine the total distance and hence cannot determine average speed.

(4 marks)

Tier 3 · Hard

A particle starts at position

-5\,\text{m}

on a directed straight line and initially moves in the positive direction. During

8\,\text{s}

it changes direction exactly once and finishes at position

7\,\text{m}

. Its average speed during the interval is

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

. Find the distance it travels in each direction, the position at which it changes direction and its average velocity over the interval.

(6)

(Total for Question 8 is 6 marks)

Mark scheme

Mark scheme for question 8
Question	Scheme	Marks
8	Distance in the positive direction $=20\,\text{m}$ Distance in the negative direction $=8\,\text{m}$ The particle changes direction at position $15\,\text{m}$ Average velocity $=1.5\,\text{m s}^{-1}$ in the positive direction	6
Notes
The total distance is $3.5(8)=28\,\text{m}$ and the displacement is $7-(-5)=12\,\text{m}$ . If $p$ and $n$ are the distances in the positive and negative directions, then $p+n=28$ and $p-n=12$ . Hence $p=20$ and $n=8$ . The single change of direction occurs after the positive motion, at position $-5+20=15\,\text{m}$ . The average velocity is $12/8=1.5\,\text{m s}^{-1}$ in the positive direction.

(6 marks)

Tier 3 · Hard

The vectors

\mathbf{i}

and

\mathbf{j}

are fixed perpendicular unit vectors. A survey robot starts at position

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

and finishes at

(16\mathbf{i}+25\mathbf{j})\,\text{m}

. Its odometer records a total distance of

42\,\text{m}

during the

14\,\text{s}

journey. Find its displacement vector and the magnitude of its displacement. Hence find its average speed and its average velocity vector, and explain why the magnitude of the average velocity is less than the average speed. Give all numerical answers exactly.

(6)

(Total for Question 9 is 6 marks)

Mark scheme

Mark scheme for question 9
Question	Scheme	Marks
9	Displacement $=(18\mathbf{i}+24\mathbf{j})\,\text{m}$ Displacement magnitude $=30\,\text{m}$ Average speed $=3\,\text{m s}^{-1}$ Average velocity $=(\dfrac97\mathbf{i}+\dfrac{12}{7}\mathbf{j})\,\text{m s}^{-1}$ Its magnitude is $15/7\,\text{m s}^{-1}$ , less than the average speed because distance exceeds displacement magnitude	6
Notes
Final position minus initial position gives $(16-(-2))\mathbf{i}+(25-1)\mathbf{j}=18\mathbf{i}+24\mathbf{j}$ metres, with magnitude $\sqrt{18^2+24^2}=30\,\text{m}$ . Average speed is distance divided by time, $42/14=3\,\text{m s}^{-1}$ . Average velocity is displacement divided by time, giving $(9/7)\mathbf{i}+(12/7)\mathbf{j}\,\text{m s}^{-1}$ . Its magnitude is $30/14=15/7\,\text{m s}^{-1}$ ; the winding path length $42\,\text{m}$ exceeds the $30\,\text{m}$ endpoint separation.

(6 marks)

Q10

Tier 3 · Hard

10.

The vectors

\mathbf{i}

and

\mathbf{j}

are fixed perpendicular unit vectors. During a

6\,\text{s}

interval, a particle moves from

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

(20\mathbf{i}-9\mathbf{j})\,\text{m}

. Its velocity changes from

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

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

. Find the particle's average velocity and average acceleration during the interval, and its speed at each endpoint. Explain why these data are insufficient to determine the total distance travelled. Give all numerical answers exactly.

(6)

(Total for Question 10 is 6 marks)

Mark scheme

Mark scheme for question 10
Question	Scheme	Marks
10	Average velocity $=(4\mathbf{i}-2\mathbf{j})\,\text{m s}^{-1}$ Average acceleration $=(\dfrac43\mathbf{i}-2\mathbf{j})\,\text{m s}^{-2}$ Initial speed $=5\,\text{m s}^{-1}$ Final speed $=\sqrt{89}\,\text{m s}^{-1}$ The path between the endpoints is not specified, so its length is not determined	6
Notes
The displacement is $(20-(-4))\mathbf{i}+(-9-3)\mathbf{j}=24\mathbf{i}-12\mathbf{j}$ metres, so average velocity is $(24\mathbf{i}-12\mathbf{j})/6=4\mathbf{i}-2\mathbf{j}\,\text{m s}^{-1}$ . The velocity change is $8\mathbf{i}-12\mathbf{j}$ , giving average acceleration $(4/3)\mathbf{i}-2\mathbf{j}\,\text{m s}^{-2}$ . The endpoint speeds are $\sqrt{(-3)^2+4^2}=5$ and $\sqrt{5^2+(-8)^2}=\sqrt{89}\,\text{m s}^{-1}$ . Endpoint positions and velocities do not specify the intervening path, so they cannot fix its total length.

(6 marks)

Verified exam appearances

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Other points in M7 Kinematics

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