🛤️Motion Along a Line

On the bench at the Tangent Laboratory there is a rail with a single bead on it. The bead can only go two ways: forwards, which everyone calls the positive direction, or backwards, which is the negative one. Everything this craft asks is a reading taken off that rail.

Three functions describe the bead, and each one is the derivative of the one before it.

• ◆s(t) is the position: where the bead is, measured along the rail from a fixed mark, in metres.
• ◆v(t) = s'(t) is the velocity: how fast the position is changing, in metres per second. Its SIGN says which way the bead is going.
• ◆a(t) = v'(t) = s''(t) is the acceleration: how fast the velocity is changing, in metres per second squared.

Two more words are worth pinning down straight away, because half of the mistakes in this topic are really vocabulary mistakes.

Speed is the size of the velocity. A velocity of -6 m/s and a velocity of 6 m/s are different velocities but the same speed, 6 m/s. Speed is never negative; velocity often is.

Position is where, displacement is how far from where it started. If the bead begins at s = 4 and ends at s = 1, its position is 1 and its displacement is -3.

The chain runs the other way too, and that is the half most students meet second. Velocity is an antiderivative of acceleration, and position is an antiderivative of velocity. Going down the chain you differentiate and lose information; coming back up you integrate and have to be given that information again, as an initial condition such as s(0) = 4 or v(0) = -3. Without it you cannot say where the bead is, only how far it has moved.

differentiate: s(t) -> v(t) -> a(t) integrate: a(t) -> v(t) -> s(t) (+ one starting value each time)

The bead is at rest at the instants where v(t) = 0. That is all "at rest" means here: an instant when it is not moving. So finding the times at rest is one job — differentiate s, set v to zero, and solve.

Changing direction is a stronger thing than being at rest. The bead turns round only where v changes sign, from positive to negative or the other way. A velocity that touches zero and comes straight back the way it was is a pause, not a turn.

v(t) = 3(t - 1)(t - 5) zero at t = 1 and t = 5, and the sign flips at each: two turns v(t) = 3(t - 4)² zero at t = 4, positive on both sides: a pause, no turn

That is why a repeated factor is worth spotting. A squared factor is never negative, so it cannot change the sign of anything.

Speeding up and slowing down are about v and a together. Neither one alone settles it, and this is the single most common slip in the whole craft:

• ◆v and a have the same sign: the push agrees with the motion, so the speed rises — the bead is speeding up.
• ◆v and a have opposite signs: the push fights the motion, so the speed falls — the bead is slowing down.
• ◆v = 0: the bead is momentarily at rest.
• ◆a = 0 for every t: the velocity never changes, so the bead is moving at a constant velocity.

Notice what this means: a bead moving backwards (v negative) with a negative acceleration is speeding up, even though every number in sight is negative. It is getting faster in the backwards direction.

To answer "on which times is it speeding up", build one sign table with every cut in it: the zeros of v and the zeros of a. For s(t) = t³ - 15t² + 63t + 9 we get v(t) = 3(t - 3)(t - 7) and a(t) = 6(t - 5), so the cuts are t = 3, t = 5 and t = 7 — three cuts, four stretches. Test one time in each stretch and read off where the two signs agree.

t < 3 v > 0, a < 0 slowing down 3 < t < 5 v < 0, a < 0 speeding up 5 < t < 7 v < 0, a > 0 slowing down t > 7 v > 0, a > 0 speeding up

Over an interval from t = 0 to t = T there are two honest answers to "how far did it go", and they are not the same number.

Displacement is signed and only looks at the two ends.

displacement = s(T) - s(0)

It does not care about the route. If the bead goes forwards 4 m and then back 2 m, its displacement is 2 m.

Total distance is positive and looks at every leg of the journey. Cut the interval at every time the bead turns round, work out the change of position on each leg, throw away the minus signs, and add:

total distance = |s(r) - s(0)| + |s(T) - s(r)| for one turn at t = r

For that same journey the total distance is 4 + 2 = 6 m. The absolute value goes on each leg, before adding — never on the total afterwards. Taking the size of the displacement would give 2, which is not how far the bead travelled.

The two averages follow the same split:

average velocity = displacement / (time taken) average speed = total distance / (time taken)

Average velocity needs no calculus at all: it is two positions and a subtraction. It is not the same as the instantaneous velocity v(a) at either end, which is a derivative read at one instant, and it is only equal to the average speed when the bead never turns round.

Sometimes the velocity is not given as a rule but as a short table of readings. Then an acceleration can only be estimated, by a difference quotient — a change in velocity divided by the time it took:

a(2) is about (v(3) - v(1)) / (3 - 1)

Subtract the earlier reading from the later one, then divide by the gap. Both halves matter: forgetting to divide leaves a change rather than a rate.

Given v(t), integrating gives the family of antiderivatives, all differing by a constant. Only the initial condition picks the right one out.

v(t) = 6t + 4, s(0) = 4 s(t) = 3t² + 4t + C s(0) = C = 4, so s(t) = 3t² + 4t + 4

The constant is the starting position, and leaving it out answers a different question. Without it you have the displacement, not the position. If acceleration is the starting point, the same thing happens twice: integrate once for v with v(0), integrate again for s with s(0).

The classic word problem here is an object thrown straight up. Near the ground, gravity pulls it down at a nearly constant rate, so its height follows

h(t) = -5t² + v0*t + h0 (metres, seconds)

where v0 is the velocity it was thrown with and h0 the height it was thrown from. The -5t² comes from the acceleration of gravity, about -10 m/s², integrated twice. In feet the same model reads h(t) = -16t² + v0*t + h0. Every question about it is one of these four, and each one is a different reading:

• ◆When is it highest? At the top the object is neither rising nor falling, so solve h'(t) = 0. The answer is a TIME.
• ◆How high does it get? Find that time, then substitute it back into h. The answer is a HEIGHT. These first two are different numbers and swapping them is the classic error.
• ◆When does it reach the ground? The ground is h(t) = 0. Solve the quadratic and discard the negative root, which is before the throw.
• ◆How fast is it moving then? Put the landing time into h'. The reading is negative, because the object is falling; the SPEED is its size.

One more reading is worth knowing: the flight is symmetric about the highest point, so an object thrown from a cliff top passes the cliff top again on the way down with the same speed it left with, and the opposite velocity.

Sometimes the velocity arrives as a picture rather than a formula — a graph of v against t made of straight pieces. Three questions can be answered straight off it, and none of them needs a rule for v.

Where does it change direction? Where the graph crosses the time axis, because that is where v changes sign. A high point or a low point on a velocity graph is not a turn: the bead is still going the same way, just faster or slower.

Where is the speed greatest? At the corner whose reading has the biggest SIZE, minus signs ignored. A reading of -6 beats a reading of 4. Between two corners the graph is a straight line, so the extremes are always at corners and only the corners need checking.

How far did it go? Velocity times time is a displacement, so the area between a velocity graph and the time axis is a distance travelled. Each straight piece makes a triangle or a trapezoid with the axis, and a piece of width w whose ends read h1 and h2 covers

((h1 + h2) / 2) * w

Then the same distinction as before decides what to do with the signs:

• ◆displacement: count the pieces above the axis as positive and the pieces below it as negative, and add.
• ◆total distance: take the size of every piece and add, so nothing cancels.

Take the corners (0, 1), (1, 3), (2, 4), (3, 0), (4, -5). The first three pieces sit above the axis and cover (1 + 3)/2 + (3 + 4)/2 + (4 + 0)/2 = 2 + 7/2 + 2 = 15/2. The last piece sits below it and covers (0 + (-5))/2 = -5/2. So the displacement is 15/2 - 5/2 = 5 m, and the total distance is 15/2 + 5/2 = 10 m.