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Kinematics

Kinematics

Speed-time graphs

Graphs Rules Paper 1 + 2 7 min read

On this page
  1. In this note
  2. Formula bank
  3. The two rules
  4. Finding an unknown time or speed
  5. Ways they test it
  6. Mistakes to avoid
  7. Summary
  8. Practice
  9. Can I do this?
  10. Quiz

In this note

Syllabus 2.9: speed-time graphs
  • 2.9Read speed-time graphs: the gradient is the acceleration; downhill is deceleration.
  • 2.9Find the distance travelled as the area under the graph.
  • 2.9Find an unknown time or speed from a given total distance.

Before you start: Distance-time graphs and areas of triangles and trapezia.

Formula bank

AccelerationLearn it\(a=\dfrac{\text{change in speed}}{\text{time}}\) = gradientunits m/s²; a negative gradient is a deceleration.Use it when a sloping section.
DistanceLearn itdistance = area under the speed-time graphsplit into triangles, rectangles, trapezia.Use it when 'how far'.
TrapeziumLearn it\(\tfrac12(a+b)h\)a and b: the parallel sides; h: the distance between them.Use it when a whole speed-time journey.

The two rules

On a speed-time graph the gradient is the acceleration (how fast the speed changes), and the area under the graph is the distance (speed × time adds up strip by strip).

Speed-time journey51015202530510152025t (s)v (m/s)(8, 20)(20, 20)
A car: 0 to 20 m/s in 8 s, steady for 12 s, stops in 5 s.
LineMeaning
sloping upaccelerating
flatconstant speed
sloping downdecelerating
on the t-axisstationary

General methodHow to answer a speed-time question

  1. Acceleration: gradient = change in speed ÷ time for that section (m/s²). Downhill: say 'deceleration' with a positive number.
  2. Distance: split the area under the graph into triangles, rectangles and trapezia; add them.
  3. Average speed: total distance ÷ total time.
  4. Unknown time or speed: write the area in terms of the letter, set it equal to the distance, solve.

Example For the car above, find the acceleration, the deceleration and the total distance.

Gradients, then the area (two triangles and a rectangle).

  1. acceleration \(20\div8=2.5\) m/s²
  2. deceleration \(20\div5=4\) m/s²
  3. distance \(\tfrac12(8)(20)+12(20)+\tfrac12(5)(20)=80+240+50\)

Answer 2.5 m/s², 4 m/s², 370 m

Distance-time: gradient = speed. Speed-time: gradient = acceleration and AREA = distance. Check which graph it is.

Try it yourself

1
0 to 18 m/s in 6 s: acceleration (m/s²)?
Show answer

Answer 3

2
Distance travelled in those 6 s (m)?
Show answer

Answer 54

Finding an unknown time or speed

Example A train speeds up from 0 to V m/s in 5 s, keeps V for 20 s, then stops in 5 s, covering 300 m. Find V.

The area is a trapezium with parallel sides 30 and 20.

  1. \(\tfrac12(30+20)V=300\)
  2. \(25V=300\)

Answer V = 12 m/s

Try it yourself

1
The car (0 → 20 m/s in 8 s, 20 m/s for T s, 20 → 0 in 5 s) travels 450 m. Find T.
Hint
80 + 20T + 50 = 450.
Show answer

Answer 16

Ways they test it

Type 1 Acceleration

"Find the acceleration during the first 8 seconds." (1-2 marks)

Spot it a sloping section.

  1. Change in speed ÷ time.

Type 2 Total distance

"Find the total distance travelled." (2-3 marks)

Spot it 'distance' with a speed-time graph.

  1. Area: triangles + rectangles, or one trapezium.

Type 3 Unknown time or speed

"The total distance is 450 m. Find T." (3 marks)

Spot it a letter on the graph and a total distance.

  1. Area in terms of the letter = distance.
  2. Solve.

For example \(80+20T+50=450\)

  1. \(20T=320\)

Answer T = 16

Type 4 Two phases

"A runner reaches 12 m/s in 4 s, then runs steadily. When has she run 144 m?" (3 marks)

Spot it a distance part-way through.

  1. Distance in phase 1 from its area.
  2. Remaining distance ÷ steady speed.

For example \(\tfrac12(4)(12)=24\) m

  1. 120 m ÷ 12 = 10 s

Answer 14 s

Mistakes to avoid

  • ✗ Gradient of speed-time = speed✓ accelerationDifferent graph.
  • ✗ Distance = top speed × total time✓ the area under the graphThe speed changes.
  • ✗ Deceleration −4 m/s²✓ deceleration 4 m/s²The word already means slowing down.

Summary

  • Speed-time: gradient = acceleration (m/s²).
  • Area under the graph = distance (m).
  • Average speed = total area ÷ total time.
  • Unknowns: area in terms of the letter = given distance.

If you see …

  • "how far" on a speed-time graphwork out the area.
  • a letter for a time or speedwrite the area with the letter and solve.

Practice

Practice: easy, medium, hard

Easy

1
0 to 20 m/s in 8 s: acceleration (m/s²)?
Show answer

Answer 2.5

2
A horizontal line on a speed-time graph means … ? Type a phrase.
Show answer

Answer constant speed

3
20 m/s to 0 in 5 s: deceleration (m/s²)?
Show answer

Answer 4

Medium

1
A cyclist speeds up uniformly from 10 m/s to 30 m/s in 6 s. Acceleration (m/s², 2 d.p.)?
Guide me

What is it asking? m/s²

What do we know? change 20 m/s

Plan ÷ 6

Show working
  1. \(20\div6=3.33\)

Answer 3.33

2
Distance travelled by that cyclist in the 6 s (m)?
Guide me

What is it asking? a distance

What do we know? trapezium

Plan \(\tfrac12(10+30)\times6\)

Show working
  1. \(20\times6=120\)

Answer 120

3
Total distance for the car (0→20 m/s in 8 s, 20 m/s for 12 s, 20→0 in 5 s), in m?
Guide me

What is it asking? a distance

What do we know? split the area

Plan 80 + 240 + 50

Show working
  1. 80 + 240 + 50 = 370

Answer 370

Hard

1
A train: 0 to V m/s in 5 s, V for 20 s, then stops in 5 s, covering 300 m. Find V.
Guide me

What is it asking? V

What do we know? trapezium: parallel sides 30 and 20

Plan \(\tfrac12(30+20)V=300\)

Show working
  1. \(25V=300\)
  2. \(V=12\)

Answer 12

PartAnswer / workingMarksGuidance
\(\tfrac12(30+20)V\)M1
\(25V=300\)M1
12A1
Total: 3 marks
2
A runner accelerates from rest to 12 m/s in 4 s, then runs at 12 m/s. After how many seconds has she run 144 m?
Guide me

What is it asking? a time

What do we know? first 4 s: 24 m

Plan 120 m more at 12 m/s

Show working
  1. \(\tfrac12\times4\times12=24\)
  2. \(120\div12=10\)
  3. 4 + 10 = 14

Answer 14

PartAnswer / workingMarksGuidance
24 mM1
10 sM1
14 sA1
Total: 3 marks
3
A bus accelerates from rest to 15 m/s in 10 s, travels at 15 m/s for 40 s, then stops in 6 s. Total distance (m)?
Guide me

What is it asking? a distance

What do we know? trapezium: parallel sides 56 and 40

Plan \(\tfrac12(56+40)\times15\)

Show working
  1. \(48\times15=720\)

Answer 720

PartAnswer / workingMarksGuidance
\(\tfrac12(56+40)\times15\)M2
720A1
Total: 3 marks

Can I do this?

Anything unticked: revise that section first. All ticked: take the quiz.

Quiz

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