Motion

The bench follows a car along a straight road for three minutes. It sets off from rest, reaches 20 m/s, or 72 km/h, in 20 s, holds that speed until 120 s, brakes to a halt at 150 s and stays put until the end. Every 20 s the bench notes how far the car is from where it set off, and the ten readings, from 0 to 2,500 m, make up the table to the left of the graph.

Two choices had to be made before the first reading. One is which point of the car to follow, say the tip of the front bumper; since the car does not change length, any other point on it would give readings shifted by a fixed amount. The other is the origin, here the starting point, with distances counted in the direction of travel. An origin 500 m further along would take 500 m off every reading, and what the readings tell us about the motion would stay the same. Once both choices are made, each instant has one position and only one, and the notation \(s = s(t)\) says no more than that; the bench computes this function with a separate formula for each stretch of the journey, and the table records ten of its values.

Plotted as points, with the clock along the horizontal axis and the position up the vertical one, the rows of the table trace a curve that tells the whole journey at a glance. It leaves flat and bends upwards while the car gathers speed, becomes a straight line on the stretch where it covers 400 m every 20 s, bends the other way during braking and ends lying flat at 2,500 m.

The Fall scenario swaps the car for a ball dropped from rest, with no air resistance, and notes the distance fallen every half second. The readings grow faster and faster, from 1.2 m in the first half second to 78.4 m at 4 s, and they obey a simple rule: in twice the time the ball falls four times as far, as the 19.6 m at 2 s shows. The rule is \(s = \tfrac12 g t^2\), with \(g = 9.8\) m/s², and the graph is a branch of a parabola. The secondary-school lesson on free fall shows where this formula comes from; here it serves only as a description.

In both scenarios the table and the graph hold everything there is to say about the motion, yet neither shows directly how fast the body is going at each instant, and reading that off the curve is the business of step 2. The car and the falling ball are the two examples Feynman opens the chapter with, here with numbers of our own.

In Feynman: §8-1 Description of motion ↗

The bench marks a point P on the curve of the fall, at \(t = 1\) s, when the ball has already dropped 4.9 m, and a second point an interval \(\Delta t\) later. With \(\Delta t = 0.1\) s the ball drops a further 1.029 m in that interval, and the ratio \(\Delta s/\Delta t\) comes to 10.29 m/s. This is the average speed over the interval, the one a body would have to keep up, unchanged, to cover the same distance in the same time.

The value depends on the interval. With 1 s the average would be 14.7 m/s, with 0.01 s it comes down to 9.849 m/s and with 0.001 s to 9.8049 m/s. A line of algebra accounts for the sequence: expanding the square in \(\tfrac12 g (t+\Delta t)^2\), the average becomes \(g t + \tfrac12 g\,\Delta t\), and the second term, 4.9 m/s² times \(\Delta t\), is all that separates it from \(g t = 9.8\) m/s. Since that term can be made as small as we please, the averages approach \(g t\) as \(\Delta t\) shrinks, and that value is what we call the speed at the instant \(t\).

On the graph the same calculation has a geometric reading. The average is the slope of the straight line through P and the second point, the secant, and as \(\Delta t\) shrinks the second point slides down the curve towards P and the secant turns. Where that turning ends is the dashed line, the tangent at P, and its slope is the speed at that instant.

In the Car scenario, with P at 120 s, the bench gives 20 m/s, or 72 km/h, which is what the speedometer would show at that moment: it measures the speed at each instant, not an average over an hour on the road. Half a minute later the car has stopped, having covered only another 300 m, and the reading of 72 km/h is still right, because it made no claim about the hour ahead. What it claimed was that over a short enough interval from that moment on the car covers close to 20 m each second, with a discrepancy that shrinks as the interval does.

Shrinking the interval and following where the quotient goes is a procedure with a name, the limit. Newton and Leibniz each arrived at it on their own in the later seventeenth century, and the differential calculus grew out of it. The car also shows something the fall cannot. On the stretch at constant speed the average is the same for any \(\Delta t\), and during braking it falls below the speed at the instant, because the car loses speed during the interval. The secondary-school lesson on average velocity deals with the average over a whole journey; what matters here is what happens as the interval shrinks.

In Feynman: §8-2 Speed ↗

On the bench, a pump fills a spherical balloon at a fixed rate, 120 cm³ every second with the control in its starting position. The volume grows in step with time, but the radius does not keep pace: it reaches 5 cm in 4.4 s and needs another 30.5 s to get from 5 to 10 cm. On the graph of radius against time the curve starts out almost vertical and gradually lies down, and what we are after is its slope at any instant, the rate \(dr/dt\) in centimetres per second.

Geometry relates volume and radius, \(V = \tfrac43\pi r^3\), and since the volume at time \(t\) is the flow rate times \(t\), the relation yields the radius at each instant. The calculation tells us where the edge of the balloon is at every moment, but how fast that edge moves is written nowhere in the formula; to get it we have to compare the radius at two very close instants, as we did with the fall. Posing an inflating balloon as a question about rates is Feynman's idea, which he leaves to the reader in the chapter; the flow rate and the radii here are those of the bench.

The way in is the idea from step 2, looking at a short interval \(\Delta t\). During it the pump adds a volume \(\Delta V\), the flow rate times \(\Delta t\), which settles as a thin shell around the balloon, the orange band on the bench. A shell of thickness \(\Delta r\) has a volume close to the area of the sphere, \(4\pi r^2\), times the thickness, and the approximation improves as the shell thins. With a radius of 10 cm the area is 1,257 cm², and the 12 cm³ that go in during 0.1 s form a layer less than a hundredth of a centimetre thick, 0.0095 cm. Dividing \(\Delta V \approx 4\pi r^2\,\Delta r\) by \(\Delta t\) and letting the interval shrink, the approximation becomes an equality and leaves the rate of the radius, which at 120 cm³/s and a radius of 10 cm is 0.0955 cm/s.

The formula explains why the balloon seems to slow down. The same flow spreads over an area that grows with the square of the radius, and at twice the radius the rate drops to a quarter, from 0.382 cm/s at 5 cm to 0.0955 cm/s at 10 cm. With the animation paused, the Measure button does the calculation numerically: it computes the radius a little later, divides the increase by the interval and sets the result against the formula.

In Feynman: §8-2 Speed ↗

In step 2 the speed of the fall came from expanding a square and watching the term in \(\Delta t\) disappear. With \(s = t^2\) the calculation is the same: \((t+\Delta t)^2 - t^2 = 2t\,\Delta t + \Delta t^2\), the ratio \(\Delta s/\Delta t\) gives \(2t + \Delta t\), and in the limit \(2t\) is what remains. With \(s = t^3\) the expanded cube brings \(3t^2\,\Delta t + 3t\,\Delta t^2 + \Delta t^3\); after dividing by \(\Delta t\) the last two terms still carry \(\Delta t\), vanish in the limit and leave \(3t^2\).

The pattern repeats for any whole-number power. In \((t+\Delta t)^n\) the binomial theorem gives \(t^n\), then \(n\,t^{n-1}\Delta t\), and every term after that has \(\Delta t\) squared or higher, so only the second one survives the division and the limit. It can be shown that the same rule holds for fractional and negative exponents, and it is the source of the rows \(\sqrt t = t^{1/2}\) and \(1/t = t^{-1}\) in the bench's table.

The derivative of \(s\) with respect to \(t\) is written \(ds/dt\), and the d is there to recall that it is the limit of \(\Delta s/\Delta t\). The notation needs handling with some care. \(\Delta s\) is a single number, the difference \(s(t+\Delta t) - s(t)\), and the Δ is not multiplying anything; striking out both Δs in \(\Delta s/\Delta t\) to leave \(s/t\) would give, for the fall at \(t = 1\) s, 4.9 m/s, half the correct 9.8 m/s.

Two rules are enough to combine the rows of the table. The derivative of a sum is the sum of the derivatives, because the \(\Delta s\) of a sum is the sum of the \(\Delta s\) of its parts, and a constant factor comes out of the derivative, because it multiplies every \(\Delta s\); a constant on its own never changes and has zero derivative. With these, the position in the fall, \(\tfrac12 g\,t^2\), has derivative \(\tfrac12 g \cdot 2t = g t\), the result of step 2 without redoing the limit, and \(4t^2 - 3\sqrt t\) has derivative \(8t - \tfrac{3}{2\sqrt t}\).

The bench checks the table with numbers. For the chosen function it computes \(\Delta s/\Delta t\) over an interval of width \(\Delta t\) centred on the point, from \(t - \tfrac12\Delta t\) to \(t + \tfrac12\Delta t\), which approaches the derivative faster than the interval of step 2, and it places the results as dots on the graph of the derivative the table gives. With \(\Delta t = 1\) the dots for \(t^3\) sit 0.25 above the curve; with \(10^{-4}\), all six functions agree with the table to six decimal places. Sine and exponential are not powers of \(t\), and for the moment the bench is the only justification the lesson offers for their rows: the derivative of the sine is the cosine, and the exponential is its own derivative.

In Feynman: §8-3 Speed as a derivative ↗

The graph on this bench is the speedometer of the falling ball, a straight line that starts at zero and reaches 39.2 m/s at 4 s, and the question now runs the opposite way to the earlier steps: knowing only what the speedometer showed, how far did the ball fall? With a reading at the start of each second, 0, 9.8, 19.6 and 29.4 m/s, we may suppose that the ball keeps each reading until the next one. In each second it then covers \(v\,\Delta t\), and the four contributions add up to 58.8 m, well short of the 78.4 m that step 1 gave for the whole fall.

On the graph, each contribution \(v\,\Delta t\) is the area of a rectangle with base \(\Delta t\) and height equal to the reading, and the sum is the area of a staircase. The staircase stays below the line because the ball moves faster at the end of each second than at the start. Reading the speedometer at the end of each interval instead, the staircase rises above the line and the sum goes up to 98 m. As long as the speed only increases, the true distance lies between the two sums, and the gap between them, 39.2 m with 4 rectangles, halves every time the number of rectangles doubles.

With 200 rectangles of 0.02 s the two sums give 78.01 and 78.79 m, and in the limit as \(\Delta t\) goes to zero both approach the area under the line. That limit of sums is called an integral and is written \(\int_0^t v\,dt\), with \(dt\) where the shrinking \(\Delta t\) used to be.

For the fall the area needs no limit. Up to the instant \(t\) the region under the line is a triangle with base \(t\) and height \(g t\), whose area, half the base times the height, is \(\tfrac12 g t^2\), the formula of step 1; at 4 s it gives 78.4 m. In the Middle mode, which reads the speedometer at the midpoint of each interval, the bench hits this value with any number of rectangles, because on a straight line the piece each rectangle loses on one side of the midpoint equals the piece it gains on the other. For the car the area under the graph is a trapezium of 2,500 m, the distance of step 1, and the sums close in on it as the rectangles narrow, not always from the same side.

The area also shows how the integral is tied to the derivative. When the final instant moves on by \(\Delta t\), the area gains a narrow strip of height \(v\) and width \(\Delta t\), and the gain divided by \(\Delta t\) tends to \(v\): the derivative of the accumulated distance is the speed we started from. Integration and differentiation are inverse operations, so a calculation in one direction can be checked in the other, and indeed the derivative of \(\tfrac12 g t^2\) is \(g t\), as in step 4.

In Feynman: §8-4 Distance as an integral ↗

In the Car scenario the bench stacks three graphs of the same journey: the position, the velocity and, at the bottom, the acceleration, which tracks how quickly the velocity changes. In the first 20 s the speedometer climbs 1 m/s every second, an acceleration of 1 m/s², a gain of one metre per second in each second. On the stretch at 72 km/h it is zero, and during braking the velocity drops by 20 m/s in 30 s, which gives −2/3 m/s², or −0.667 m/s², with the sign saying that the velocity is decreasing.

Velocity is the derivative of position, and acceleration is the derivative of velocity, so it is the derivative of the derivative of position, the second derivative, written \(d^2s/dt^2\). The bench computes it from positions alone, at three neighbouring instants: the advance between \(t\) and \(t + \Delta t\) minus the advance between \(t - \Delta t\) and \(t\), divided by \(\Delta t^2\). In the fall, the derivative of \(g t\) is the constant \(g\), which is why 9.8 m/s² goes by the name of the acceleration due to gravity. At 20, 120 and 150 s the car's acceleration jumps from one value to another, because the bench's journey switches from one regime to the next all at once; in a real car the switch would take at least a fraction of a second.

In the Launch scenario, a ball leaves the ground at 5 m/s horizontally and 10 m/s upwards, and its position now takes two numbers, the distance \(x\) along the ground and the height \(y\). Each is a function of time, with its own velocity and its own acceleration, and the two are dealt with separately, as if they were two motions along a line. Without air, nothing pushes or holds back the ball horizontally and \(v_x\) stays at 5 m/s; vertically, \(v_y\) loses 9.8 m/s every second, reaches zero at the top, at 1.02 s and a height of 5.10 m, and the ball lands at 2.04 s, 10.2 m further on.

Since \(x\) grows in step with time, \(t = x/v_x\), and putting this into the formula for \(y\) gives the height as a function of the distance along the ground, a term in \(x\) minus a term in \(x^2\), and the graph is a parabola. The trail, one dot every 0.1 s, says the same thing: the dots are evenly spaced horizontally and bunch up near the top. The arrows show the velocity, always tangent to the trail, turning from 63° above the horizontal to 63° below, while the acceleration points downwards from start to finish. Feynman closes the chapter with a ball thrown horizontally, and here that case is the half of the flight that follows the highest point. The secondary-school lesson on projectiles deals with the angle and the range.

In Feynman: §8-5 Acceleration ↗