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AI SL · 13.5

Integration

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13.5.1

Distance as area under a speed-time graph

13.5.1

Distance as area under a speed-time graph

Target: I can explain why the area under a speed-time graph is the distance travelled. (+3 more)

By the end of this section

Key points

Area and distance

  • Speed is the rate of change of distance. Distance accumulates as time passes.
  • The area under a speed-time graph is the distance travelled. 1 hour at 40 mph shades 40 miles.
  • Check the units: seconds × m/s gives metres; hours × km/h gives km.

Straight-line graphs

  • Split the area into rectangles, triangles and trapeziums. Draw a vertical line wherever the graph changes direction.
  • Rectangle: base × height. Triangle: \(\tfrac12\times\) base \(\times\) height. Trapezium: \(\tfrac12(a+b)\times\) width, where a and b are the parallel sides.
  • Add the areas. The sum is the total distance.

Curved graphs

  • Read off values from the curve and join them with straight chords. This makes a triangle (if the graph starts at 0) and trapeziums.
  • The width of each strip is the gap in time between two read-off values. The strips need not be equal.
  • A curve that bends downwards lies above its chords: the estimate is an under-estimate. A curve that bends upwards lies below its chords: the estimate is an over-estimate.

Explore: distance as an area

The graph is made of straight pieces. Drag the orange points up and down to change the speed, and watch the area of each piece.

    Distance =

    Select a point (Tab), then use the up and down arrow keys. Speed is in m/s, time in s, so the area is in m.

    Drag a point to change the speed.

    1. Drag the point at 4 s down to 0. Piece 1 has no area. What shape is piece 2 now? Why does its area have a ½ in it?
    2. Drag the point at 7 s up or down, then make the speed at 7 s equal to the speed at 4 s again. What shape is piece 2, and what is the car doing?
    3. Which piece covers the most distance? Move one point so that a different piece covers the most.
    4. Can you make the total distance exactly 100 m? Is there more than one way?

    Worked examples

    Problems like the ones in class. Pick one, try it, then open the working.
    Distance from a constant speed

    A boat travels at a constant speed of 15 km/h for 4 hours. Find the distance it travels.

    123455101520Time (h)Speed (km/h)
    Working
    1. Rate and total

      1 hour at 15 km/h shades 15 km

      Speed is the rate of change of distance. The area under the speed graph is the distance gained, so area accumulates distance.

    2. The shape

      Rectangle, width 4 h, height 15 km/h

      The graph is a horizontal line, so the area under it is a rectangle.

    3. Area

      \(4\times 15=60\) km

      Units: hours × km/h gives km.

    Video walkthrough coming
    A braking cyclist

    A cyclist brakes steadily from 12 m/s to rest in 8 s. Find the distance she travels while braking.

    123456789102468101214Time (s)Speed (m/s)
    Working
    1. The shape

      Triangle: base 8 s, height 12 m/s

      The line drops straight to the axis, so the area under it is a triangle.

    2. Area

      \(\tfrac12\times 8\times 12=48\) m

      Half base times height.

    Video walkthrough coming
    A tram between two stops

    A tram speeds up steadily from rest to 9 m/s in 4 s, travels at 9 m/s until \(t=10\) s, then slows steadily to rest at \(t=13\) s. Find the total distance.

    123456789101112131424681012Time (s)Speed (m/s)
    Working
    1. Split the area

      Triangle \(0\le t\le 4\), rectangle \(4\le t\le 10\), triangle \(10\le t\le 13\)

      Draw the vertical lines where the graph changes direction.

    2. Each area

      \(\tfrac12(4)(9)=18\), \(6\times 9=54\), \(\tfrac12(3)(9)=13.5\)

      Triangle, rectangle, triangle.

    3. Add

      \(18+54+13.5=85.5\) m

      The total area is the total distance.

    4. Another way

      Trapezium: \(\tfrac12(13+6)\times 9=85.5\)

      The whole shape is a trapezium with parallel sides 13 and 6.

    Video walkthrough coming
    A curved graph: three chords

    The graph shows the speed of a car, in m/s, for the first 16 s. Read-off values: \((0,0)\), \((4,12)\), \((9,18)\) and \((16,24)\). Straight chords join them. Find the area of triangle A (0 to 4 s), trapezium B (4 to 9 s) and trapezium C (9 to 16 s), and use them to estimate the total distance travelled. Is your estimate an over-estimate or an under-estimate?

    24681012141618481216202428Time (s)Speed (m/s)
    Working
    1. Triangle A

      \(\tfrac12\times 4\times 12=24\)

      Base 4 s, height 12 m/s.

    2. Trapezium B

      \(\tfrac12(12+18)\times 5=75\)

      Parallel sides are the speeds at 4 s and 9 s; the width is 5 s.

    3. Trapezium C

      \(\tfrac12(18+24)\times 7=147\)

      Width 9 s to 16 s is 7 s.

    4. Total

      \(24+75+147=246\) m

      Add the strip areas.

    5. Over or under

      Under-estimate

      The curve bends downwards, so each chord lies below it. The shapes hold less area than the curve.

    Video walkthrough coming
    A rising curve: over or under

    A sprinter's speed in m/s is shown for the first 8 s. Read-off values: \((0,2)\), \((4,6)\) and \((8,18)\). Use two trapeziums to estimate the distance, then say whether it is an over-estimate or an under-estimate.

    1234567892468101214161820Time (s)Speed (m/s)
    Working
    1. First strip

      \(\tfrac12(2+6)\times 4=16\)

      Speeds at 0 s and 4 s, width 4 s.

    2. Second strip

      \(\tfrac12(6+18)\times 4=48\)

      Speeds at 4 s and 8 s, width 4 s.

    3. Total

      \(16+48=64\) m

      Estimated distance.

    4. Over or under

      Over-estimate

      The curve bends upwards, so each chord lies above it. The trapeziums hold more area than the curve.

    Video walkthrough coming