Module 04

Motion

6 minute read

Throw a ball straight up and watch its highest moment. Freeze it there, at the very top of the arc. Its speed at that frozen instant is zero. Measure as finely as you like; at the apex, the ball covers no ground.

Now compare it with a ball sitting on a shelf. That one also has speed zero. But the two zeros could not feel more different. The shelf ball's zero is permanent. The apex ball's zero is a formality, gone before it arrives; one instant later the ball is already falling.

Whatever separates those two zeros is invisible to the speedometer, because the speedometer reads the same for both. So there must be another dial, one level deeper. What number tells a ball at rest apart from a ball passing through rest?

The takeaway

Position says where. Velocity says how position is changing. Acceleration says how velocity is changing. A flight is three graphs chained together by the derivative, and the law of nature only ever writes the bottom one.

Three dials, one chain

Module 03 built one machine: feed it a story, get the story's rate. Nothing stops you from feeding the machine its own output. Run it on position and out comes velocity, the rate at which position changes. Run it on velocity and out comes acceleration, the rate at which velocity changes. That second run is the missing dial from the cold open. At the apex, the ball's velocity is zero but its acceleration is not: the velocity is mid-crossing, on its way from up to down, and acceleration is exactly the number that says so. The shelf ball has zero on both dials. The apex ball has zero on one.

Both are vectors, which module 02 made cheap to say: velocity has the direction of travel, and acceleration points wherever velocity is being pushed. They need not agree. Throw the ball up and its velocity points up while its acceleration points down the entire flight, which is precisely why the flight ends.

Figure 4.1 One instant, three readings. At the apex the middle graph touches zero in passing; the bottom graph does not even flinch.

Learn to read the figure the way a musician reads a chord: three staves, one moment. Where the top graph peaks, the middle graph crosses zero. Where the middle graph slopes, the bottom graph sits at that slope's value. The bottom graph is a flat line at , and that flatness is the deepest fact in the picture: through the whole rise and fall, while position and velocity dance, gravity holds one steady note.

Here is a dashboard analogy, and its breaking point in the same paragraph. Velocity is the speedometer; acceleration is what the seat of your chair reports when the car surges, the push in your back. It breaks for free fall: a falling ball accelerates at full strength while feeling no push at all, and a chair-based accelerometer in the ball would read zero. The seat measures the fight against gravity, not gravity itself. File that crack away carefully; an entire theory of gravity gets built inside it, four volumes from now.

And now module 01's postponed promise comes due. The state of the ball, the complete description, is position and velocity together: . Position alone fails, as the cold open showed, because two different futures pass through the same height. Add velocity and the ambiguity dies. Why stop there? Why not carry acceleration in the state too? Because for the ball, acceleration is not free information: it is fixed at by the world, whatever the ball is doing. The state carries what varies; the law dictates the rest.

The math

Definitions first. Velocity is the derivative of position. Acceleration is the derivative of velocity:

v = \frac{d x}{d t}, a = \frac{d v}{d t}

In two dimensions, apply this to each component of the position vector. Nothing new happens.

Now solve the simplest law completely: acceleration held constant at . Integrate once. Velocity accumulates area under a flat line:

v (t) = v_{0} + a t

Integrate again. Position accumulates area under that tilted line: a rectangle from plus a triangle from the growth:

x (t) = x_{0} + v_{0} t + \frac{1}{2} a t^{2}

Check it by running the chain backward. Differentiate : the constant drops, gives , and the power rule turns into . That is . Differentiate again: . The chain closes.

Work the throw. Up is positive, , launch speed . The apex is where velocity crosses zero:

0 = 9.8 - 9.8 t; \Rightarrow; t = 1, x (1) = 9.8 (1) - \frac{1}{2} (9.8) (1)^{2} = 4.9

One second up, 4.9 meters high. The figure was drawn from these numbers.

Play with it

A cart, a track, and the two knobs the universe actually offers: where you start it moving and how hard it is pushed. Run it and watch the three graphs write themselves. Make velocity and acceleration disagree in sign and watch the cart change its mind.

Check yourself

Problem 1. A ball leaves your hand at 20 m/s straight up. Take g = 10 m/s². When does it reach the apex, and how high is that?

Reveal the solution

Apex when v = 0: from v = 20 - 10t, that is t = 2 s.

Height: x = 20(2) - ½(10)(4) = 40 - 20 = 20 m.

Problem 2. Position is x(t) = 3t² + 2t. Find v(t) and a(t).

Reveal the solution

Differentiate once: v = 6t + 2. Differentiate again: a = 6, constant. The chain never needs more than the power rule.

Problem 3. Velocity is v(t) = 8 - 2t from t = 0 to t = 4. Which way does the object move, is it speeding up or slowing down, and how far has it gone by t = 4?

Reveal the solution

v stays positive until t = 4, so it moves forward the whole time while slowing: velocity positive, acceleration -2, signs opposed.

Distance is the area under v: a triangle with base 4 and height 8, so ½ × 4 × 8 = 16 m.

What this did to the demon

Hand the demon a state (x, v) and a rule for acceleration, and it runs the tape forward as far as you like; the constant-acceleration formulas are the demon doing one especially easy tape by hand. But notice what was smuggled in: the rule for acceleration arrived from nowhere, free of charge. Module 05 asks where acceleration actually comes from, and the answer has a name on it.

Next · module 05
Force and Newton's laws