Describing motion

Before anything can be said about why things move, there has to be a language precise enough to say how they move, and that language turns out to be one already built: a single function of time, differentiated twice and integrated back.

This course assumes the calculus in Calculus: derivatives, the chain rule, definite integrals, separation of variables in a simple first order equation, and Taylor expansion about a point. No vector calculus is needed anywhere, and vectors themselves are built from components in the next lesson.

Position, and the trouble with average speed

Fix an origin and a positive direction along a line. The position $x (t)$ of a particle is its signed distance from that origin at time $t$ , in metres, and everything in this lesson is a statement about that one function. Choosing the origin is free: physics never depends on where the zero sits, only on differences, which is why $x$ appears in real equations almost exclusively as $Δ x = x_{2} - x_{1}$ .

The obvious summary of a motion is its average velocity, $Δ x / Δ t$ . Usain Bolt covered 100 m in 9.58 s in Berlin in 2009, so his average velocity was $100 / 9.58 = 10.44$ m s⁻¹. The published 10 metre splits for that run, which sum to exactly 9.58 s, are 1.89, 0.99, 0.90, 0.86, 0.83, 0.82, 0.81, 0.82, 0.83, 0.83. The first ten metres took 1.89 s, an average of only 5.29 m s⁻¹, and the stretch from 60 to 70 m took 0.81 s, an average of 12.35 m s⁻¹. The single number 10.44 describes no instant of the race. It is a fact about the two endpoints and nothing else.

Shortening the interval helps. The 10 m splits are better than the one number, 1 m splits would be better still, and the question that produces calculus is what happens as the interval shrinks towards nothing. On a graph of $x$ against $t$ , the average velocity over an interval is the slope of the chord joining its ends, and the chords approach the tangent.

Velocity as a derivative

Define the instantaneous velocity as the limit of the average:

v (t) = lim_{Δ t \to 0} \frac{x (t + Δ t) - x (t)}{Δ t} = \frac{d x}{d t}

Velocity is signed: $v < 0$ means moving towards decreasing $x$ . Speed is $| v |$ , and losing the distinction is a reliable way to get a sign wrong later. A ball thrown upward has positive velocity going up, negative coming down, and a speed that is zero only at the single instant at the top.

The derivative is what makes a speedometer meaningful. There is no interval short enough to be "the instant", so the reading has to be defined as a limit rather than measured as a ratio, and the limit exists as long as the position function is smooth. It is worth noticing how much physics is packed into the assumption that it is smooth: a particle that jumped from one place to another would have no derivative at the jump, and Newtonian mechanics simply asserts that this does not happen.

Example. A trolley moves along a track with $x (t) = 12 t - 1.5 t^{2}$ , in metres, with $t$ in seconds. Where is it at $t = 2$ s, and how fast is it going?

Differentiate: $v (t) = 12 - 3 t$ . At $t = 2$ , the position is $x = 24 - 6 = 18$ m and the velocity is $v = 12 - 6 = 6$ m s⁻¹. The trolley is still moving forward but has lost half the speed it started with, and $v = 0$ at $t = 4$ s, which is where it turns around.

Now you. A cart has $x (t) = 5 + 8 t - 0.4 t^{3}$ metres. Find its position, velocity and acceleration at $t = 2$ s.

Answer

$v = 8 - 1.2 t^{2}$ and $a = - 2.4 t$ . At $t = 2$ : $x = 5 + 16 - 3.2 = 17.8$ m, $v = 8 - 4.8 = 3.2$ m s⁻¹, and $a = - 4.8$ m s⁻².

Acceleration, and why it is the physical one

Differentiate again and you have the acceleration:

a (t) = \frac{d v}{d t} = \frac{d^{2} x}{d t^{2}}

in metres per second per second, m s⁻². A car going at a steady 30 m s⁻¹ has zero acceleration; one speeding up from rest to 30 m s⁻¹ in 8 s has an average acceleration of 3.75 m s⁻². Acceleration is also signed, and its sign is not "speeding up" or "slowing down" on its own: a car reversing and speeding up has negative velocity and negative acceleration. The rule is that speed increases when $v$ and $a$ share a sign and decreases when they do not.

Nothing so far singles out acceleration as more important than the third derivative or the fourth. What singles it out is a physical fact that this lesson cannot yet prove: forces determine acceleration, and nothing else. That is why the chain stops at two derivatives and why the state of a particle is its position and velocity together. Give both now, plus the forces from now on, and the entire future follows. That claim is the content of the next few lessons.

Two derivatives is also exactly the number a person can feel. Sitting in a cruising aircraft at 250 m s⁻¹ feels like sitting in a chair, because velocity is not detectable from inside; the takeoff run is unmistakable, because acceleration is.

Running the calculus backwards

If acceleration is what physics hands you, the useful direction is the other one. Integrating once recovers velocity and integrating again recovers position:

v (t) = v_{0} + \int_{0}^{t} a (t^{'}) d t^{'} x (t) = x_{0} + \int_{0}^{t} v (t^{'}) d t^{'}

Each integration introduces one constant, fixed by an initial condition, and this is where $v_{0}$ and $x_{0}$ come from. Two integrations, two constants: the future needs the initial position and the initial velocity, and nothing more. A law of motion that gave the third derivative would need three, and a body's history would not be settled by where it is and how fast it is going, which is not the world we live in.

Geometrically, integration says that displacement is the area under the velocity graph, counting area below the axis as negative. A car that drives forward at 20 m s⁻¹ for 30 s and back at 20 m s⁻¹ for 30 s covers 1200 m of road with a displacement of zero, and the two areas cancel exactly. Distance travelled is $\int | v | d t$ , which is a different integral, and confusing the two is the second most common error in kinematics after sign confusion.

The constant acceleration formulas, derived

The special case worth memorising is constant $a$ , because it covers free fall near the ground, braking with a locked wheel, and any short interval over which the force barely changes. With $a$ constant, the first integral is immediate:

v = v_{0} + a t

and integrating that gives

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

These two are the whole set. Everything else is algebra: eliminating $t$ between them gives $v^{2} = v_{0}^{2} + 2 a (x - x_{0})$ , and averaging the velocity gives $x - x_{0} = \frac{1}{2} (v_{0} + v) t$ , which is valid only because a linear velocity has its mean at the midpoint of the interval. There are four formulas in most textbooks and two independent facts.

Free fall is the standard instance. Near the Earth's surface every body released from rest falls with the same downward acceleration $g = 9.81$ m s⁻², independent of its mass, provided air resistance can be ignored. Drop a stone from rest and after 3.0 s it has fallen $\frac{1}{2} (9.81) (9) = 44.1$ m and is moving at $9.81 \times 3 = 29.4$ m s⁻¹. Galileo could not time a fall accurately with the clocks of 1600, so he diluted gravity by rolling balls down inclined planes and found the distance growing as the square of the time, which is the same claim in slow motion. The cleanest demonstration is David Scott's on Apollo 15 in 1971: a hammer and a falcon feather released together on the Moon, where there is no air, hit the surface together, in a fall from 1.6 m taking about 1.41 s at the lunar $g$ of 1.62 m s⁻².

Example. A car brakes from 28 m s⁻¹, roughly 100 km/h, at a steady 7.5 m s⁻². How far does it travel before stopping, and how long does it take?

Take the initial direction as positive, so $a = - 7.5$ m s⁻² and the final $v$ is zero. From $v^{2} = v_{0}^{2} + 2 a Δ x$ with $v = 0$ , $Δ x = - v_{0}^{2} / 2 a = 784 / 15 = 52.3$ m. From $v = v_{0} + a t$ , $t = 28 / 7.5 = 3.73$ s. Note that halving the speed quarters the distance, since $Δ x$ goes as $v_{0}^{2}$ : that quadratic is the whole argument for speed limits.

Now you. The same car brakes from 20 m s⁻¹ at 6.0 m s⁻². Find the stopping distance and the stopping time.

Answer

$Δ x = v_{0}^{2} / (2 | a |) = 400 / 12 = 33.3$ m, and $t = 20 / 6.0 = 3.3$ s.

When time is not wanted

Often the question is about position and speed with no interest in when. The formula $v^{2} = v_{0}^{2} + 2 a Δ x$ answers it for constant $a$ , but the trick behind it survives when $a$ is not constant, and it is worth seeing once. Write acceleration using the chain rule with position as the intermediate variable:

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

so that $a d x = v d v$ . Integrating both sides between two positions gives

\int_{x_{0}}^{x} a d x = \frac{1}{2} v^{2} - \frac{1}{2} v_{0}^{2}

For constant $a$ the left side is $a Δ x$ and the familiar formula drops out. For any $a$ that is known as a function of position, the same integral still works, and time never appears. Multiply through by mass and this identity becomes the work energy theorem, which is where this course goes in a few lessons: the shortcut is not a trick at all but energy conservation in disguise.

Example. A ball is thrown straight up at 22 m s⁻¹. How high does it go, and when does it reach the top?

At the top $v = 0$ . Using $v^{2} = v_{0}^{2} - 2 g h$ with $v = 0$ gives $h = v_{0}^{2} / 2 g = 484 / 19.62 = 24.7$ m. The time is $t = v_{0} / g = 22 / 9.81 = 2.24$ s. The ball then falls back, reaching the thrower's hand at the same speed it left, which the symmetry of the equations guarantees and air resistance spoils by a few per cent.

Now you. A ball is thrown straight up at 16 m s⁻¹. Find its maximum height and the time to reach it.

Answer

$h = 16^{2} / (2 \times 9.81) = 13.0$ m, and $t = 16 / 9.81 = 1.63$ s.

What kinematics cannot do

Everything above is mathematics. Given $a (t)$ , the motion follows; given the motion, the acceleration follows. There is no physics in it, and the proof is that the equations are just as happy with a body whose acceleration is $- 9.81$ m s⁻² as with one whose acceleration is $+ 400$ m s⁻² upward, or one that reverses every second for no reason.

That is the gap. Kinematics is a bookkeeping system with an empty input slot, and the whole of the next lesson but one is about what fills it. The answer, that acceleration is caused by forces and is proportional to their sum divided by the mass, is not derivable from any amount of graph reading. It had to be discovered, and it was discovered late, because the obvious guess, that force determines velocity rather than acceleration, is what everyday experience with friction actually suggests.

First, though, the description has to escape the straight line. Real motion happens in a plane or in space, and the next lesson shows that the escape costs almost nothing: a vector is a bundle of independent copies of what has just been built.