MISSION · Calculus
Position, Velocity and Acceleration
One object on a line, and three functions describing it
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The idea in one sentence
Position, velocity and acceleration are one function and its first two derivatives, so differentiating moves you down the chain and integrating moves you back up.
The chain
\[s(t) \; \xrightarrow{\;\frac{d}{dt}\;} \; v(t) \; \xrightarrow{\;\frac{d}{dt}\;} \; a(t)\] \[a(t) \; \xrightarrow{\;\int\;} \; v(t) \; \xrightarrow{\;\int\;} \; s(t)\]Velocity is the rate at which position changes. Acceleration is the rate at which velocity changes. Nothing here is new — it is the derivative applied to the situation it was invented for.
Going back up, each integration brings a constant that an initial condition pins down. “The particle starts at \(s = 5\)” or “the initial velocity is 3” are not decoration; without them the antiderivative describes a whole family of motions.
Velocity against speed
This distinction carries most of the marks on this topic.
Velocity is signed. Negative velocity means moving in the negative direction — left, or down. Speed is \(|v(t)|\), and is never negative.
So a particle with \(v = -8\) is moving quickly, in the negative direction. Saying it has “negative speed” is wrong, and saying it is “slowing down” is also wrong.
Whether it is speeding up depends on \(v\) and \(a\) together:
- \(v\) and \(a\) with the same sign → speeding up.
- \(v\) and \(a\) with opposite signs → slowing down.
Acceleration alone tells you nothing about it. A ball thrown upward has constant downward acceleration throughout, yet slows on the way up and speeds up on the way down.
Two moments worth naming
At rest means \(v(t) = 0\). The particle has stopped, if only for an instant.
Changes direction means \(v(t)\) changes sign. That is stricter: \(v\) must be zero and then come back with the opposite sign. If \(v\) touches zero and returns the way it came — as \(v(t) = (t-2)^2\) does at \(t = 2\) — the particle pauses without turning round.
The same distinction as a stationary point that is not a maximum or minimum, because it is exactly the same mathematics.
Displacement is not distance
Given \(v(t)\) on \([a, b]\):
\[\text{displacement} = \int_a^b v(t)\,dt \qquad \text{total distance} = \int_a^b |v(t)|\,dt\]Displacement is net change in position — where you ended up relative to where you started. Total distance is how far you actually travelled, counting the backtracking.
Walk 5 m right and 5 m back: displacement 0, distance 10.
The absolute value bars mean you cannot just integrate. Find where \(v = 0\), split the interval there, integrate each piece, and add the absolute values. Skipping the split lets the negative stretch cancel the positive one, and you have computed displacement while answering a distance question.
A worked one
\(v(t) = t^2 - 4t + 3\) on \([0, 4]\), in metres per second.
Factor: \((t-1)(t-3)\), so \(v = 0\) at \(t = 1\) and \(t = 3\). Velocity is positive on \([0,1)\), negative on \((1,3)\), positive again on \((3,4]\) — it turns round twice.
The antiderivative is \(\frac{t^3}{3} - 2t^2 + 3t\), giving \(\frac43\) on \([0,1]\), \(-\frac43\) on \([1,3]\), and \(\frac43\) on \([3,4]\).
\[\text{displacement} = \tfrac43 - \tfrac43 + \tfrac43 = \tfrac43 \text{ m} \qquad \text{distance} = \tfrac43 + \tfrac43 + \tfrac43 = 4 \text{ m}\]One question, two very different answers, separated only by whether you took absolute values.
Common mix-ups
- Integrating \(v\) for total distance without splitting at the zeros.
- Reading \(a > 0\) as “speeding up”. Compare the signs of \(v\) and \(a\).
- Treating “at rest” and “changes direction” as identical.
- Dropping the constant of integration, or forgetting to use the initial condition to find it.
- Confusing average velocity with average speed. The first is displacement over time; the second is distance over time, and they differ whenever the particle turns round.
Check yourself
If \(s(t)\) is position, what is \(s'(t)\)? When is a particle momentarily at rest? And speed is the absolute value of what?
\(s'(t)\) is velocity. The particle is at rest when \(v(t) = 0\). And speed is \(|v(t)|\) — the size of the velocity, with the direction discarded.
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