7.04c - Areas under graphs

7.04c - Areas under graphs

Can an object travel twenty metres and finish where it started? Learn how areas under velocity-time graphs record motion, and why signed area and total distance can differ. Use rectangles, triangles and trapezia for straight segments, estimate curved areas, and work out what an area means from its units.

Areas under straight-line segments

At a constant speed of 6 m/s6\,\mathrm{m/s} for 55 seconds, an object travels 6×5=306\times5=30 metres. On a speed-time graph this is the area of a rectangle of height six and width five. The graph's area is measured in axis-value products, not square centimetres on the paper.

For a speed increasing uniformly from zero, the region is a triangle. Its area, half base times height, is also average speed times duration: (0+v)/2(0+v)/2 times tt. A straight segment between two nonzero speeds gives a trapezium, with the two vertical speeds as its parallel sides.

Area and motion

For a nonnegative speed-time graph,

distance travelled=area between graph and time axis.\text{distance travelled}=\text{area between graph and time axis}.

For a straight segment over duration hh, starting at speed uu and ending at speed vv,

distance=12(u+v)h.\text{distance}=\frac12(u+v)h.

The trapezium area formula is in the 2027 formula sheet. Its use to interpret a motion graph is not stated there: learn the connection.

Split a piecewise straight graph at each change of slope. Find every shape's width by subtracting times, then add the areas. Do not multiply the maximum speed by the whole duration unless that speed is maintained throughout.

[DIAGRAM: asset_name: Areas for two motion profiles; asset_slug: polygon-motion-areas; caption: Extending only the final slowing-down stage adds triangular area.; description: Two speed-time graphs have vertices zero zero, four eight, ten eight and respectively fourteen zero or eighteen zero. Each area is split into an initial triangle, middle rectangle and final triangle.]
Extending only the final slowing-down stage adds triangular area.

Worked example 1: three straight stages

Question. A speed-time graph joins (0,0)(0,0), (4,8)(4,8), (10,8)(10,8) and (14,0)(14,0), with seconds and metres per second. Find the distance travelled.

Plan. Add the two triangle areas and the central rectangle area.

A1=12×4×8=16A2=(10−4)×8=48A3=12×(14−10)×8=16distance=16+48+16=80 m\begin{aligned} A_1&=\tfrac12\times4\times8=16 \\ A_2&=(10-4)\times8=48 \\ A_3&=\tfrac12\times(14-10)\times8=16 \\ \text{distance}&=16+48+16=80\,\mathrm m \end{aligned}

Answer. 8080 metres.

Check. The whole region is a trapezium with parallel horizontal sides 1414 and 66, separated by 88. Its area is 12(14+6)×8=80\tfrac12(14+6)\times8=80.

Change only the final stopping time from fourteen to eighteen seconds. The first two stages stay unchanged.

Worked example 2: a longer final stage

Question. The new speed-time graph joins (0,0)(0,0), (4,8)(4,8), (10,8)(10,8) and (18,0)(18,0). Find the distance travelled.

Plan. Retain the first two areas and recalculate the final triangle's width.

A3=12×(18−10)×8=32distance=16+48+32=96 m\begin{aligned} A_3&=\tfrac12\times(18-10)\times8=32 \\ \text{distance}&=16+48+32=96\,\mathrm m \end{aligned}

Answer. 9696 metres.

Check. The extra four seconds add 12×4×8=16\tfrac12\times4\times8=16 metres to the triangular area: 80+16=9680+16=96. The average speed is 96/1896/18, less than the maximum of eight.

When all speeds are nonnegative, the area is a distance. The same calculation on a velocity-time graph gives a signed change of position; the distinction matters once the graph goes below the axis.

Watch out

The middle rectangle in the first example has width 10−4=610-4=6, not ten. Coordinates are times at endpoints; area needs elapsed time between them.

Use the area method on a fresh profile.

Estimating curved areas

A curved speed graph cannot usually be split into a few exact triangles. Replace short sections by straight chords and calculate the resulting trapezia. The sum estimates the area under the curve.

Use the actual horizontal widths, which need not be equal. For adjacent readings v1v_1 and v2v_2, the strip estimate is 12(v1+v2)×width\tfrac12(v_1+v_2)\times\text{width}. Shared internal readings are used twice, once for each neighbouring strip; they are not extra measurements.

The graph tells you about the direction of error. If every chord lies above a section of curve, those trapezia overestimate its area. If they lie below, they underestimate it. For a curve changing its bend, inspect each strip; a blanket claim about every trapezium estimate is unsafe.

[DIAGRAM: asset_name: Refining a curved-area estimate; asset_slug: curved-area-estimates; caption: Narrower trapezia follow this upward-bending curve more closely.; description: Two panels show the speed curve v equals two plus t squared from zero to four seconds. The first uses two trapezia of width two, and the second four trapezia of width one. Both chord sets lie above the curve between readings.]
Narrower trapezia follow this upward-bending curve more closely.

Worked example 3: two trapezia

Question. A speed graph has readings (0,2)(0,2), (2,6)(2,6) and (4,18)(4,18). Estimate the distance from zero to four seconds using two trapezia.

Plan. Each strip is two seconds wide. Average its endpoint speeds and multiply by two.

A1=12(2+6)×2=8A2=12(6+18)×2=24estimated distance=8+24=32 m\begin{aligned} A_1&=\tfrac12(2+6)\times2=8 \\ A_2&=\tfrac12(6+18)\times2=24 \\ \text{estimated distance}&=8+24=32\,\mathrm m \end{aligned}

Answer. Approximately 3232 metres.

Check. The shortcut 2[12(2)+6+12(18)]=322[\tfrac12(2)+6+\tfrac12(18)]=32 weights the outer readings once and the shared reading twice. On the displayed upward-bending curve, both chords are above the curve, so this is an overestimate.

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