What is M7.2: Kinematics graphs?

The gradient of a displacement-time graph is velocity; the gradient of a velocity-time graph is acceleration, with signs determined by the chosen positive direction.

How many marks are M7.2 questions worth?

M7.2 practice drills on this page are worth 2–7 marks each; marks in a real exam vary by question.

Which past exam papers has M7.2 appeared in?

We have not yet indexed a verified past-paper appearance for M7.2 — check the Edexcel A-level Maths 9MA0 past-papers hub directly.

M7.2

Understand, use and interpret graphs in kinematics for motion in a straight line: displacement against time and interpretation of gradient; velocity against time and interpretation of gradient and area under the graph.

Draft — not yet indexed

Kinematics graphs

Worked answers and methods for M7.2 on Edexcel A-level Maths 9MA0.

Explanation

The gradient of a displacement-time graph is velocity; the gradient of a velocity-time graph is acceleration, with signs determined by the chosen positive direction.
Find displacement from a velocity-time graph using signed area, splitting the graph into rectangles, triangles or trapezia as needed.
For a straight segment from velocity $u$ to $v$ over time $t$ , the area is $\dfrac12(u+v)t$ , which gives the displacement during that interval.
A common error is to add all areas as positive: area below the time axis is negative for displacement, though its magnitude contributes positively to distance travelled.

On a velocity-time graph, gradient gives acceleration and signed area gives displacement.

Worked example

A particle's velocity increases uniformly from $2\,\text{m s}^{-1}$ to $8\,\text{m s}^{-1}$ during the first $3\,\text{s}$ , then remains at $8\,\text{m s}^{-1}$ for $4\,\text{s}$ . Find its acceleration during the first stage and its displacement over all $7\,\text{s}$ .

1.The first gradient is $(8-2)/3=2\,\text{m s}^{-2}$ .
2.The first-stage area is $\tfrac12(2+8)(3)=15\,\text{m}$ and the constant-velocity area is $8(4)=32\,\text{m}$ , giving displacement $15+32=47\,\text{m}$ .

Answer: Acceleration $=2\,\text{m s}^{-2}$ ; Displacement $=47\,\text{m}$

Common mistakes

Don't treat the area under a displacement-time graph as displacement instead of using its gradient for velocity.
Don't read displacement from the final velocity instead of the signed area under the velocity-time graph.

Exam tip

On kinematics graphs, state explicitly whether a gradient or signed area gives the requested quantity.

Worked practice

Tier 1 · Easy

A straight segment of a displacement-time graph joins

(2,5)

(8,23)

, where time is in seconds and displacement in metres. Find the velocity represented by the segment.

(2)

(Total for Question 1 is 2 marks)

Mark scheme

Mark scheme for question 1
Question	Scheme	Marks
1	$3\,\text{m s}^{-1}$	2
Notes
Velocity is the gradient of the displacement-time graph: $v=(23-5)/(8-2)=18/6=3\,\text{m s}^{-1}$ .

(2 marks)

Tier 2 · Standard

A displacement-time graph consists of straight line segments joining

(0,0)

(4,20)

and then

(4,20)

(10,8)

, with time in seconds and displacement in metres. Find the velocity on each segment and the total distance travelled.

(4)

(Total for Question 2 is 4 marks)

Mark scheme

Mark scheme for question 2
Question	Scheme	Marks
2	Velocities $5\,\text{m s}^{-1}$ and $-2\,\text{m s}^{-1}$ Total distance $32\,\text{m}$	4
Notes
Velocity is the gradient of a displacement-time graph. The first gradient is $(20-0)/(4-0)=5\,\text{m s}^{-1}$ ; the second is $(8-20)/(10-4)=-12/6=-2\,\text{m s}^{-1}$ . The particle travels $20\,\text{m}$ out and then $12\,\text{m}$ back, so total distance is $32\,\text{m}$ .

(4 marks)

Tier 3 · Hard

A particle has velocity

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

t=0

. Its velocity increases linearly to

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

t=3

, remains constant until

t=5

, then decreases linearly to

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

t=9

. Find the acceleration during the final stage, the displacement and the total distance travelled from

t=0

t=9

(7)

(Total for Question 3 is 7 marks)

Mark scheme

Mark scheme for question 3
Question	Scheme	Marks
3	Final acceleration $=-3\,\text{m s}^{-2}$ Displacement $=57\,\text{m}$ Distance $=\dfrac{175}{3}\,\text{m}$	7
Notes
The final gradient is $(-2-10)/(9-5)=-3\,\text{m s}^{-2}$ . The signed areas are $\tfrac12(4+10)(3)=21$ , $10(2)=20$ and $\tfrac12(10-2)(4)=16$ , so displacement is $57\,\text{m}$ . In the final stage velocity reaches zero after $10/3\,\text{s}$ ; its positive and negative area magnitudes are $50/3$ and $2/3$ . Hence distance is $21+20+50/3+2/3=175/3\,\text{m}$ .

(7 marks)

Tier 1 · Easy

A particle starts from rest and its velocity increases linearly to

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

after

4\,\text{s}

. Find the displacement during these

4\,\text{s}

(2)

(Total for Question 4 is 2 marks)

Mark scheme

Mark scheme for question 4
Question	Scheme	Marks
4	$14\,\text{m}$	2
Notes
Displacement is the area under the velocity-time graph. The area is a triangle, so it is $\tfrac12(4)(7)=14\,\text{m}$ .

(2 marks)

Tier 2 · Standard

A particle has velocity

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

t=0

. Its velocity increases linearly to

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

t=4

, then decreases linearly to

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

t=10

. The displacement from

t=0

t=10

34\,\text{m}

. Find

V

and the acceleration during each stage.

(5)

(Total for Question 5 is 5 marks)

Mark scheme

Mark scheme for question 5
Question	Scheme	Marks
5	$V=5\,\text{m s}^{-1}$ First acceleration $=0.5\,\text{m s}^{-2}$ Second acceleration $=-\dfrac23\,\text{m s}^{-2}$	5
Notes
Using trapezium areas, $\tfrac12(3+V)(4)+\tfrac12(V+1)(6)=34$ . Thus $2(3+V)+3(V+1)=34$ , so $5V+9=34$ and $V=5$ . The gradients give accelerations $(5-3)/4=0.5\,\text{m s}^{-2}$ and $(1-5)/6=-2/3\,\text{m s}^{-2}$ .

(5 marks)

Tier 3 · Hard

A particle's velocity decreases uniformly from

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

t=0

to a negative value at

t=T

. During this interval its displacement is

20\,\text{m}

and its total distance travelled is

\dfrac{100}{3}\,\text{m}

. Find

T

, the final velocity and the time at which the particle changes direction.

(6)

(Total for Question 6 is 6 marks)

Mark scheme

Mark scheme for question 6
Question	Scheme	Marks
6	$T=10\,\text{s}$ Final velocity $=-4\,\text{m s}^{-1}$ The particle changes direction at $t=\dfrac{20}{3}\,\text{s}$	6
Notes
Let the positive and negative area magnitudes be $A$ and $B$ . Then $A-B=20$ and $A+B=100/3$ , giving $A=80/3$ and $B=20/3$ . The graph consists of similar triangles, so $B/A=(\|v_T\|/8)^2=1/4$ , hence $v_T=-4\,\text{m s}^{-1}$ . The whole trapezium has signed area $\tfrac12(8-4)T=2T=20$ , so $T=10$ . The velocity reaches zero after the fraction $8/(8+4)=2/3$ of the interval, at $t=20/3\,\text{s}$ .

(6 marks)

Tier 2 · Standard

t=4\,\text{s}

, the tangent to a particle's displacement-time graph passes through the points

(2,11)

and

(7,-4)

, where displacement is measured in metres. Find the particle's velocity at

t=4\,\text{s}

. At

t=9\,\text{s}

the graph has a horizontal tangent. State what this shows about the particle's velocity at that instant.

(4)

(Total for Question 7 is 4 marks)

Mark scheme

Mark scheme for question 7
Question	Scheme	Marks
7	Velocity at $t=4$ is $-3\,\text{m s}^{-1}$ Velocity at $t=9$ is zero, so the particle is instantaneously at rest	4
Notes
Velocity is the gradient of the displacement-time graph. The gradient of the tangent at $t=4$ is $(-4-11)/(7-2)=-15/5=-3\,\text{m s}^{-1}$ . A horizontal tangent has gradient zero, so at $t=9$ the velocity is zero and the particle is instantaneously at rest.

(4 marks)

Tier 3 · Hard

A velocity-time graph begins with a straight line segment from

(0,2)

(4,0)

, where time is in seconds and velocity in metres per second. For

4<t<10

, the velocity is modelled by the semicircular arc lying below the time axis, with endpoints

(4,0)

and

(10,0)

and minimum velocity

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

; the model does not specify the acceleration at

t=4

t=10

. Determine the constant acceleration represented by the straight segment, the exact displacement from

t=0

t=10

, and the exact total distance travelled during this interval.

(6)

(Total for Question 8 is 6 marks)

Mark scheme

Mark scheme for question 8
Question	Scheme	Marks
8	Acceleration $=-0.5\,\text{m s}^{-2}$ Displacement $=4-\dfrac{9\pi}{2}\,\text{m}$ Distance $=4+\dfrac{9\pi}{2}\,\text{m}$	6
Notes
The gradient of the straight segment is $(0-2)/(4-0)=-0.5\,\text{m s}^{-2}$ . Its area above the axis is $\tfrac12(4)(2)=4\,\text{m}$ . The semicircle has radius $3$ in the graph coordinates, so the magnitude of its area is $\tfrac12\pi(3^2)=9\pi/2\,\text{m}$ . This area is below the axis, giving displacement $4-9\pi/2\,\text{m}$ . Distance uses both area magnitudes, giving $4+9\pi/2\,\text{m}$ .

(6 marks)

Tier 3 · Hard

A displacement-time graph consists of straight line segments joining

(0,2)

(3,k)

and then

(3,k)

(8,-4)

, where time is in seconds and displacement in metres. The particle travels a total distance of

30\,\text{m}

and

k>2

. Find

k

, the velocity represented by each segment, the average speed and the average velocity over the complete motion. Give all numerical answers exactly.

(6)

(Total for Question 9 is 6 marks)

Mark scheme

Mark scheme for question 9
Question	Scheme	Marks
9	$k=14$ Velocities $4\,\text{m s}^{-1}$ and $-\dfrac{18}{5}\,\text{m s}^{-1}$ Average speed $=\dfrac{15}{4}\,\text{m s}^{-1}$ Average velocity $=-\dfrac34\,\text{m s}^{-1}$	6
Notes
Because $k>2$ and the final displacement is $-4$ , the two distances are $k-2$ and $k+4$ . Thus $(k-2)+(k+4)=30$ , so $2k+2=30$ and $k=14$ . The segment gradients are $(14-2)/3=4\,\text{m s}^{-1}$ and $(-4-14)/(8-3)=-18/5\,\text{m s}^{-1}$ . Average speed is $30/8=15/4\,\text{m s}^{-1}$ , while average velocity is the net displacement $-4-2=-6$ divided by $8$ , giving $-3/4\,\text{m s}^{-1}$ .

(6 marks)

Q10

Tier 3 · Hard

10.

Particles

A

and

B

move on the same directed straight line. At

t=0

A

is at the origin and moves with constant velocity

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

, while

B

6\,\text{m}

ahead. The velocity-time graph for

B

is the straight line through

(0,2)

with gradient

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

. Using areas under the velocity-time graphs, find all positive times at which the particles are at the same position. For each meeting, find the common position and state which particle is moving faster immediately after the meeting. Give all numerical answers exactly.

(7)

(Total for Question 10 is 7 marks)

Mark scheme

Mark scheme for question 10
Question	Scheme	Marks
10	Meeting times $t=2\,\text{s}$ and $t=6\,\text{s}$ Common positions $12\,\text{m}$ and $36\,\text{m}$ Immediately after the first meeting $A$ is faster; immediately after the second meeting $B$ is faster	7
Notes
The area under $A$ 's graph gives $x_A=6t$ . For $B$ , the area under its line from velocity $2$ to $2+t$ is $\tfrac12[2+(2+t)]t=2t+t^2/2$ , so $x_B=6+2t+t^2/2$ . Equating positions gives $6t=6+2t+t^2/2$ , hence $t^2-8t+12=0$ and $t=2$ or $6$ . The common positions are $6t$ , namely $12$ and $36\,\text{m}$ . Particle $B$ has velocity $2+t$ : at $t=2$ this is $4<6$ , while at $t=6$ it is $8>6$ .

(7 marks)

Verified exam appearances

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Other points in M7 Kinematics

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