3.3 - Interpreting real-life and conversion graphs

3.3 - Interpreting real-life and conversion graphs

The same shape can tell different stories when the axis labels change. You will interpret journeys, changing speeds and curved graphs, then draw and use a straight conversion graph. Reading the quantities, units and scales comes before doing any calculation.

Distance-time graphs

Can a horizontal line show someone moving? On one kind of graph it can; on another it shows a stop. Start by checking what the vertical axis measures, rather than guessing from the shape.

You already know

A gradient measures vertical change divided by horizontal change. Speed is distance divided by time; change minutes to seconds if the required unit is metres per second.

On a distance-time graph, time is horizontal and distance is vertical. To read a distance, go vertically from the time to the graph, then horizontally to the distance axis. A straight rising section shows constant speed. A steeper rising section shows greater speed, when the axis scales are unchanged. A horizontal section shows no change in distance, so the object is stopped.

Speed from a distance-time section

speed=distance travelled in the sectionelapsed time in the section\text{speed}=\frac{\text{distance travelled in the section}}{\text{elapsed time in the section}}

For a rising straight section this is its gradient. Not in the Foundation or Higher formulae sheet: learn it.

Read the distance label carefully. Distance from the start can fall during a return journey; the negative gradient then describes movement towards the start, and speed is its magnitude. Total distance travelled cannot fall. A return journey adds to the total distance even while the distance from the start decreases.

[DIAGRAM: asset_slug: m050_journey; caption: Distance from the start during a journey with a two-minute stop; description: Exact straight sections join 0 comma 0, 4 comma 600, 6 comma 600 and 10 comma 1400, with time in minutes and distance in metres; file: m050_journey.png; recommended_method: matplotlib]
Distance from the start during a journey with a two-minute stop

The two moving sections in this graph have equal durations. Their rises differ, so their speeds differ too.

Worked example 1: read a whole journey

Question. A distance-from-start graph joins (0,0)(0,0), (4,600)(4,600), (6,600)(6,600), (10,1400)(10,1400), with time in minutes and distance in metres. Find the two moving speeds, stop duration and overall average speed.

Plan. Use the changes within each section, then divide the whole distance by the whole time.

v1=6004=150 m/minfirst moving sectionstop=6−4=2 minhorizontal sectionv2=1400−60010−6=200 m/minlast section onlyaverage speed=140010=140 m/mininclude the stop\begin{aligned} v_1&=\frac{600}{4}=150\,\mathrm{m/min} &&\text{first moving section}\\ \text{stop}&=6-4=2\,\mathrm{min} &&\text{horizontal section}\\ v_2&=\frac{1400-600}{10-6}=200\,\mathrm{m/min} &&\text{last section only}\\ \text{average speed}&=\frac{1400}{10}=140\,\mathrm{m/min} &&\text{include the stop} \end{aligned}

Why? A section speed uses that section's changes, while the overall average includes every minute, including time stopped.

Answer. Moving speeds 150150 and 200 m/min200\,\mathrm{m/min}; stop 2 min2\,\mathrm{min}; overall average 140 m/min140\,\mathrm{m/min}.

Check. 150(4)+200(4)=1400 m150(4)+200(4)=1400\,\mathrm{m}. Also 600/(4×60)=2.5 m/s600/(4\times60)=2.5\,\mathrm{m/s} for the first section.

A useful surprise: the overall average is lower than both moving speeds. The two-minute stop explains it. Taking the arithmetic mean of the moving speeds would miss that stopped time.

Change only the final distance coordinate to 200200 next. The final section now falls, so it represents a return towards the start.

Worked example 2: a return section

Question. A distance-from-start graph joins (0,0)(0,0), (4,600)(4,600), (6,600)(6,600), (10,200)(10,200), in minutes and metres. Find the final section's speed and the total distance travelled.

Plan. Find how far the person returns, then add that distance to the outward journey.

return distance=600−200=400 mdistance towards the startreturn speed=40010−6=100 m/minuse the positive distancetotal distance=600+400=1000 madd both moving lengths\begin{aligned} \text{return distance}&=600-200=400\,\mathrm{m} &&\text{distance towards the start}\\ \text{return speed}&=\frac{400}{10-6}=100\,\mathrm{m/min} &&\text{use the positive distance}\\ \text{total distance}&=600+400=1000\,\mathrm{m} &&\text{add both moving lengths} \end{aligned}

Why? The final height is the distance from the start, not the total length of the route already travelled.

Answer. Final speed 100 m/min100\,\mathrm{m/min}; total distance 1000 m1000\,\mathrm{m}.

Check. The signed final gradient is (200−600)/4=−100 m/min(200-600)/4=-100\,\mathrm{m/min}. Its magnitude gives the speed; the person finishes 200 m200\,\mathrm{m} from the start.

Both stories use the same graph-reading process. It is the final direction, and the precise vertical-axis label, that change the interpretation.

Watch out

The line v2=1400/10=140 m/minv_2=1400/10=140\,\mathrm{m/min} does not give the last section's speed in example 1. Subtract the section endpoints: (1400−600)/(10−6)=200 m/min(1400-600)/(10-6)=200\,\mathrm{m/min}.

Read each section's horizontal width as an elapsed time, rather than as its ending time.

Speed-time graphs

On a speed-time graph, the height gives speed. A horizontal section above zero therefore shows movement at constant speed: this answers the opening puzzle. A section on the time axis shows a stop. An upward slope means speeding up, and a downward slope means slowing down, even while the object still travels forwards.

The gradient gives acceleration, the change in speed per unit time. With speed in m/s\mathrm{m/s} and time in seconds, its unit is m/s2\mathrm{m/s^2}. A straight slope shows a uniform change in speed.

Acceleration and distance from a speed-time graph

acceleration=change in speedelapsed time\text{acceleration}=\frac{\text{change in speed}}{\text{elapsed time}}

Distance travelled is the area between the speed-time graph and the time axis. These results are not in the Foundation or Higher formulae sheet: learn them. The trapezium area formula is supplied on both sheets.

Why does area give distance? A constant speed makes a rectangle with area speed multiplied by time. For a uniform increase from zero, the average speed is half the final speed, making a triangle. Split more complicated shapes into these pieces. The units check the result: m/s×s=m\mathrm{m/s}\times\mathrm{s}=\mathrm{m}.

[DIAGRAM: asset_slug: m050_speed; caption: Speed rising uniformly, remaining constant and then falling to zero; description: Speed in metres per second against seconds joins 0 comma 0, 4 comma 12, 10 comma 12 and 14 comma 0; file: m050_speed.png; recommended_method: matplotlib]
Speed rising uniformly, remaining constant and then falling to zero

The shaded area contains two triangles and one rectangle. Each width is a time difference.

Worked example 3: acceleration and distance

Question. A speed-time graph joins (0,0)(0,0), (4,12)(4,12), (10,12)(10,12), (14,0)(14,0), with seconds horizontally and m/s\mathrm{m/s} vertically. Find the initial acceleration and total distance.

Plan. Use the initial slope for acceleration and add the three areas for distance.

a=12−04−0=3 m/s2change in speed per secondd1=12(4)(12)=24 mfirst triangled2=(10−4)(12)=72 mrectangled3=12(14−10)(12)=24 mlast triangled=24+72+24=120 madd the areas\begin{aligned} a&=\frac{12-0}{4-0}=3\,\mathrm{m/s^2} &&\text{change in speed per second}\\ d_1&=\frac12(4)(12)=24\,\mathrm{m} &&\text{first triangle}\\ d_2&=(10-4)(12)=72\,\mathrm{m} &&\text{rectangle}\\ d_3&=\frac12(14-10)(12)=24\,\mathrm{m} &&\text{last triangle}\\ d&=24+72+24=120\,\mathrm{m} &&\text{add the areas} \end{aligned}

Why? Height alone gives speed; distance needs both speed and the time for which it acts, represented by area.

Answer. Initial acceleration 3 m/s23\,\mathrm{m/s^2}; total distance 120 m120\,\mathrm{m}.

Check. One trapezium gives 12(14+6)(12)=120 m\frac12(14+6)(12)=120\,\mathrm{m} too.

Keep all times the same and change only the peak speed from 1212 to 8 m/s8\,\mathrm{m/s}. The shape has the same widths, so every area becomes two-thirds as large.

Worked example 4: change the peak speed

Question. A speed-time graph joins (0,0)(0,0), (4,8)(4,8), (10,8)(10,8), (14,0)(14,0), in seconds and m/s\mathrm{m/s}. Find the initial acceleration and total distance.

Plan. Repeat the slope and area calculations with the new height.

a=84=2 m/s2initial gradientd=12(4)(8)+6(8)+12(4)(8)two triangles and a rectangle=16+48+16=80 msum the distances\begin{aligned} a&=\frac84=2\,\mathrm{m/s^2} &&\text{initial gradient}\\ d&=\frac12(4)(8)+6(8)+\frac12(4)(8) &&\text{two triangles and a rectangle}\\ &=16+48+16=80\,\mathrm{m} &&\text{sum the distances} \end{aligned}

Why? Changing the height changes speed, acceleration and distance; it does not change the elapsed times.

Answer. Initial acceleration 2 m/s22\,\mathrm{m/s^2}; distance 80 m80\,\mathrm{m}.

Check. 120×(8/12)=80120\times(8/12)=80, agreeing with the direct area calculation.

The falling part shows braking. Its speed stays non-negative, so its area still adds to distance travelled; it does not subtract distance.

Watch out

Using 10×1210\times12 for the rectangle in example 3 is wrong: 1010 is the ending time. Its width is 10−4=610-4=6, so its area is 6×12=72 m6\times12=72\,\mathrm{m}.

A description in words can supply the same time and speed information as a drawn graph.

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