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The problem of instant speed

Change and motion · level 6, lesson 1 of 6

A question to start

About 2500 years ago, a Greek thinker called Zeno of Elea posed a puzzle that is still quoted today. Think of an arrow in flight. At any single instant, the arrow is in exactly one place, filling a space exactly its own size. In that instant it does not move: there is no time for it to move in. But the flight is made of nothing but instants. So when does the arrow ever move?

Now look at a car's speedometer. It says 50 km/h right now, at this instant. But speed is distance divided by time, and in an instant no time passes and no distance is covered. That is \(0 \div 0\).

So what does the speedometer mean? This whole level is the answer. It is one of the most important ideas in all of mathematics, and it begins with taking the puzzle seriously.

The idea

What we already know: average speed

From level 3 onwards you have used

\[\text{average speed} = \frac{\text{change in distance}}{\text{change in time}}.\]

This needs a stretch of time: from 2 o'clock to 3 o'clock, or from \(t = 1\) to \(t = 3\) seconds. On a distance-time graph, it is the slope of a secant line between two points (level 5, lesson 3).

For steady motion, every stretch gives the same answer, so "the speed" is just that number. The trouble starts when speed changes. A falling stone, a car pulling away from the lights, a sprinter at the start: different stretches give different averages.

020406080100120 km/h

Why "distance over time" breaks at an instant

Try to use the formula at a single instant. The change in time is 0. The change in distance is 0 too: in no time, you cannot go anywhere. So we would need \(0 \div 0\).

Why is \(0 \div 0\) meaningless? Division undoes multiplication: \(12 \div 3 = 4\) because \(4 \times 3 = 12\). So \(0 \div 0\) would be "the number that, times 0, gives 0". But every number does that: \(5 \times 0 = 0\), \(50 \times 0 = 0\), \(-7 \times 0 = 0\). There is no single answer, so \(0 \div 0\) has no value. (And \(6 \div 0\) would be "the number that, times 0, gives 6", and there is none at all.) The formula is not wrong; it just cannot be used with a zero-length stretch of time.

And yet instant speed is real

We cannot simply give up, because instant speed clearly matters:

  • A speed camera fines a driver for going 70 km/h at the moment the photo was taken, not on average over the day.
  • A falling stone hits the ground faster than it started. "How fast was it going when it hit?" is a fair question with a definite answer, and that answer decides how much damage it does.
  • When a ball thrown upwards reaches the top, it is stopped for an instant: its speed at that moment is 0, even though it is moving just before and just after.

How a speedometer gets round it

A real speedometer does not divide 0 by 0. Many cars have a sensor that counts how many times a wheel turns in a short moment, a small fraction of a second, and works out the distance from that. A phone's map app compares your GPS position now with your position a moment ago. Either way, the device measures an average speed over a very short stretch of time.

Over a short stretch, the speed hardly has time to change, so the average is a very good stand-in for the speed at the instant. And a shorter stretch gives an even better one. That suggests a way out.

The way out: watch where the averages are heading

Take the falling stone on Earth, \(d = 4.9t^2\) metres. What is its speed at exactly \(t = 2\) s? Find average speeds over stretches that start at \(t = 2\) and get shorter:

Stretch of time Change in distance (m) Average speed (m/s)
2 to 3 s \(44.1 - 19.6 = 24.5\) 24.5
2 to 2.5 s \(30.625 - 19.6 = 11.025\) 22.05
2 to 2.1 s \(21.609 - 19.6 = 2.009\) 20.09

The averages are falling and seem to be settling down towards some number a little under 20. In the next lesson we push this much further and find that number exactly: 19.6 m/s.

So here is the plan, and it is the heart of calculus:

The speed at an instant is the number that the average speeds get closer and closer to as the stretch of time shrinks towards zero.

We never use a stretch of exactly zero, so we never divide by zero. We only look at where the averages are heading. That is a new kind of thinking: not "what is the answer at 0?" but "what does the answer approach as we get near 0?".

Zeno's arrow is answered the same way. The arrow's speed at an instant is not "distance moved in that instant ÷ time taken in that instant", which is the meaningless \(0 \div 0\). It is the number its average speeds settle on as we look at shorter and shorter stretches of flight around that instant. That number is not zero, so the arrow is moving.

A picture: matching steady motion

Here is another way to think of instant speed. Imagine that at the instant \(t = 2\) s, the stone's gravity were switched off, so it kept going at whatever speed it had, steadily. That steady speed is the instant speed. On the distance-time graph, the steady motion is a straight line that just grazes the curve at \(t = 2\): the tangent line. You will meet it properly in lesson 3.

Worked examples

1. A car pulling away. A dashcam records a car's distance from the traffic lights each second.

Time (s) 0 1 2 3 4 5
Distance (m) 0 5 14 27 44 65

Estimate its speed at \(t = 3\) s.

  1. The second before: from 2 to 3 s, average \(= 27 - 14 = 13\) m/s.
  2. The second after: from 3 to 4 s, average \(= 44 - 27 = 17\) m/s.
  3. The car is speeding up, so at the instant \(t = 3\) it is faster than 13 and slower than 17. A good guess is halfway: about 15 m/s.

2. Better data. A GPS logger records the same car every tenth of a second near \(t = 3\): 25.52 m at 2.9 s, 27 m at 3 s and 28.52 m at 3.1 s. Improve the estimate.

  • Before: \((27 - 25.52) \div 0.1 = 1.48 \div 0.1 = 14.8\) m/s.
  • After: \((28.52 - 27) \div 0.1 = 15.2\) m/s.

The window has narrowed from "13 to 17" to "14.8 to 15.2". The speed at 3 s is very close to 15 m/s. (In fact this car's distance is \(2t^2 + 3t\), and its speed at 3 s is exactly 15 m/s, as you will be able to prove soon.)

3. The stone at 1 second. Find average speeds of the falling stone from 1 s to 2 s, to 1.5 s and to 1.1 s. Where do they seem to be heading?

\(d(1) = 4.9\), \(d(2) = 19.6\), \(d(1.5) = 11.025\), \(d(1.1) = 5.929\).

  • 1 to 2: \(14.7 \div 1 = 14.7\) m/s.
  • 1 to 1.5: \(6.125 \div 0.5 = 12.25\) m/s.
  • 1 to 1.1: \(1.029 \div 0.1 = 10.29\) m/s.

They are heading down towards about 9.8 m/s, half of the 19.6 at \(t = 2\). (At 2 s it has had twice as long to speed up.)

4. The top of a throw. A ball's height is \(h = 20t - 5t^2\) metres. Find its average vertical speed from 1.9 s to 2.1 s.

\(h(1.9) = 38 - 18.05 = 19.95\) and \(h(2.1) = 42 - 22.05 = 19.95\). Average \(= 0 \div 0.2 = 0\) m/s. Over this short stretch it went up a little and came back down by the same amount. At \(t = 2\) the ball is at the top, and its instant vertical speed is 0.

Try it

  1. Why can't you find the speed at an instant by putting "time = 0" into speed = distance ÷ time?
  2. Explain why \(0 \div 0\) has no value, using multiplication.
  3. Using the dashcam table in example 1, estimate the car's speed at \(t = 2\) s and at \(t = 4\) s.
  4. For the stone \(d = 4.9t^2\), find the average speeds from 3 s to 4 s, to 3.5 s and to 3.1 s. What number do they seem to be heading towards?
  5. A runner's distance is 40 m at 8.0 s, 41.1 m at 8.1 s and 42.2 m at 8.2 s. Estimate her speed at 8.1 s.
  6. A phone measures speed by comparing GPS positions 1 s apart. Will it show the speed of a braking car a little too high or a little too low? (Think about which second it uses: the one just gone.)
  7. For the ball \(h = 20t - 5t^2\), find its average speed from 1 s to 1.1 s and from 0.9 s to 1 s. Estimate its speed at exactly 1 s.
  8. Zeno says the arrow cannot move because in each instant it moves no distance. In two or three sentences, explain what is wrong with his argument.
Answers
  1. In zero time you also cover zero distance, so you would need \(0 \div 0\), which has no value. The formula only works for a stretch of time.
  2. \(0 \div 0\) would be the number that gives 0 when multiplied by 0. Every number does that, so there is no one answer.
  3. At 2 s: the second before gives \(14 - 5 = 9\) m/s, the second after \(27 - 14 = 13\) m/s; about 11 m/s. At 4 s: \(44 - 27 = 17\) and \(65 - 44 = 21\); about 19 m/s. (Exactly right for this car: its speed is \(4t + 3\).)
  4. \(d(3) = 44.1\), \(d(4) = 78.4\), \(d(3.5) = 60.025\), \(d(3.1) = 47.089\). Averages: \(34.3\), \(15.925 \div 0.5 = 31.85\), \(2.989 \div 0.1 = 29.89\) m/s. Heading towards about 29.4 m/s.
  5. Before: \(1.1 \div 0.1 = 11\) m/s; after: \(1.1 \div 0.1 = 11\) m/s. About 11 m/s.
  6. Too high. The car was faster at the start of the second just gone, so the average over that second is more than the speed at the end of it, which is "now".
  7. \(h(1) = 15\), \(h(1.1) = 22 - 6.05 = 15.95\), \(h(0.9) = 18 - 4.05 = 13.95\). From 1 to 1.1: \(0.95 \div 0.1 = 9.5\) m/s. From 0.9 to 1: \(1.05 \div 0.1 = 10.5\) m/s. About 10 m/s.
  8. Speed at an instant is not "distance in that instant ÷ time in that instant" (that is \(0 \div 0\)). It is the number the average speeds approach over shorter and shorter stretches around that instant, and for a flying arrow that number is not zero. Being at one place at each instant is not the same as being at the same place at every instant.

Common mistake

Thinking a very short average is the instant speed. The average over a tenth of a second is a good estimate, not the exact answer. The exact instant speed is the number the averages are heading towards, which needs the idea of "getting closer and closer", worked out properly in the next lesson.

Using only the stretch after (or only before). When speed is changing, the average over the next second is too high (if speeding up) and the one over the last second too low. Looking both ways and taking the middle is much better. Lesson 4 shows why.

Where it leads