Velocity and Acceleration

Your phone computes this article's two derivatives, both directions, hundreds of times per second. The IMU (inertial measurement unit) carries a MEMS accelerometer measuring $\mathbf{a}$ directly and a gyroscope for rotation; navigation software then integrates — acceleration to velocity to position — the derivative chain run backward, exactly the area-under-the-curve reading of the graphs above. GPS meanwhile hands over velocity by a completely independent route (the Doppler shift of satellite signals), and a Kalman filter blends the two estimates, weighting each by its trustworthiness: integration drifts over minutes, GPS jitters over seconds, and together they hold a lane. The intrinsic decomposition of eq. (4) is working hardware too. A car's electronic stability control continuously compares the demanded steering acceleration $v^2/\rho$ (from the driver's wheel angle and the measured speed) against what the tyres can plausibly supply; when the budget is about to be blown it brakes individual wheels — adjusting $a_t$ per corner to rescue $a_n$. Worked example 2 is, almost line for line, the calculation an ESC controller runs many times a second. And at the exotic end, accelerometers built from matter-wave interference now measure $\mathbf{a}$ to parts in $10^9$, good enough to navigate submarines for days with no external fix — kinematics as a precision instrument.

Think of position as a moving dot tracing a curve through space as time runs. Velocity is the arrow riding the dot: it points along the curve in the direction of travel, and its length is how fast ground is being covered. Acceleration is the subtler thing — it is how that arrow itself is changing. The arrow can stretch (speeding up), shrink (slowing down), or swing sideways (turning), and each of the three is acceleration. The picture below shows all of it at once: a dot moving along a bend, its velocity tangent to the path, and its acceleration split into the two jobs acceleration can ever do — a piece along the path that changes the speed, and a piece pointing into the bend that changes the direction.

Where do the numbers come from? From graphs, read two ways. On a position–time graph the velocity is the slope; on a velocity–time graph the acceleration is the slope and the displacement is the area underneath. These dual readings — slope and area — are the whole of one-dimensional kinematics, and the bridge between the "average" and "instantaneous" versions of each quantity is one geometric move: a secant line between two points on the curve, pivoting into the tangent as the two points slide together. That move is the definition of the derivative, drawn rather than spoken.

Hold on to one benchmark case before the algebra: the car on the bend. Steady speedometer, yet the velocity arrow swings around the curve, so $\mathrm{d}\mathbf{v}/\mathrm{d}t \neq 0$. The acceleration points inward, toward the centre of the turn, with magnitude $v^2/r$ — this is centripetal acceleration, and it is what the friction of the tyres must supply. If the tyres cannot supply it, the car does not "fly outward"; it simply stops turning and goes straight, which the road unfortunately does not.

tesing Start with the operational definition — the speed camera's division, sharpened. Over a finite interval the ratio $\Delta\mathbf{r}/\Delta t$ is the average velocity, a secant slope; the instantaneous velocity is its limit:

where: $\mathbf{r}(t)$ — position relative to a chosen origin and frame; $\Delta t$ — the sampling interval, taken to zero. The limit does two things at once: the magnitude $|\mathbf{v}|$ becomes the speed (what the speedometer shows), and the direction becomes the tangent to the path (which way the hood points, for a car that doesn't skid). Note what velocity inherits from displacement: over any finite interval, $\Delta\mathbf{r}$ is path-blind — but in the $\Delta t \to 0$ limit every path looks straight, so the instantaneous velocity captures the true local motion, wiggles and all. Apply the same sharpening to the velocity and you get the second derivative of position:

where: $\mathbf{a}(t)$ — the instantaneous acceleration vector. Read it as "how the velocity arrow is changing right now" — in length, in direction, or both. This is the quantity Newton's second law will care about; nature, it turns out, legislates the second derivative and leaves the first as initial data. The finite-difference versions $\bar{\mathbf{v}} = \Delta\mathbf{r}/\Delta t$ and $\bar{\mathbf{a}} = \Delta\mathbf{v}/\Delta t$ remain useful whenever data comes in samples — which is to say, in every experiment and every sensor. To compute, drop the vectors onto axes. In Cartesian coordinates the basis vectors $\hat{\mathbf{x}}, \hat{\mathbf{y}}, \hat{\mathbf{z}}$ are nailed to the frame — they never change — so differentiation passes straight through to the components:

where: the overdots are Newton's notation for time derivatives ($\dot x = \mathrm{d}x/\mathrm{d}t$, $\ddot x = \mathrm{d}^2x/\mathrm{d}t^2$). The physical content is decoupling: each axis runs its own private one-dimensional problem. This is precisely why projectile motion splits into a constant-velocity horizontal story and a free-fall vertical story that share nothing but the clock. Curved paths ask for a different costume. Instead of fixed axes, attach the basis to the motion: a unit tangent $\hat{\mathbf{t}}$ along the velocity and a unit normal $\hat{\mathbf{n}}$ pointing toward the local centre of curvature. In this intrinsic frame the acceleration shows its anatomy:

where: $v = |\mathbf{v}|$ — the speed; $\rho$ — the radius of the circle that best hugs the path at this point (the radius of curvature). The two terms are the hero diagram's two arrows, now with formulas attached: only the $\hat{\mathbf{t}}$ term changes the speedometer, and only the $\hat{\mathbf{n}}$ term steers. Uniform circular motion is the pure-steering limit — $\mathrm{d}v/\mathrm{d}t = 0$, $\rho = r$, leaving the famous $a = v^2/r$ pointed dead at the centre. The other rotating costume is polar coordinates $(r, \theta)$, natural whenever motion winds around a centre — orbits, turntables, pendulums. Here the basis vectors $\hat{\mathbf{r}}$ (outward) and $\hat{\boldsymbol{\theta}}$ (sideways) are attached to the particle's direction from the origin, so they swing as the particle moves, and their motion injects extra terms:

where: $\dot r$ — how fast the particle moves radially; $r\dot\theta$ — how fast it sweeps sideways (angular rate $\dot\theta$ leveraged by the arm length $r$). Differentiating once more — carefully, because now the basis moves too — gives the full polar acceleration:

where: $-r\dot\theta^2$ — the centripetal term, present even when $r$ never changes; $2\dot r\dot\theta$ — the Coriolis term, coupling radial and angular motion, factor of 2 and all. Derivation B builds this from scratch. These two "extra" terms are pure geometry — no force produces them; the coordinate system does — and they are the seed of every fictitious force you will meet in rotating frames. Finally, change the observer. Let frame $S'$ move at constant velocity $\mathbf{u}$ relative to frame $S$. Positions relate by $\mathbf{r}' = \mathbf{r} - \mathbf{u}t$, and differentiating twice:

where: $\mathbf{v}', \mathbf{a}'$ — what $S'$ measures. Velocities disagree by exactly the relative velocity (the everyday rule that a 31 m/s walker on a 30 m/s train is a 1 m/s walker on board); accelerations agree identically, because the constant $\mathbf{u}$ dies under the derivative. Derivation C spells out why this two-line fact is the licence for Newton's laws to hold in every inertial frame at once.

Which equation answers which question

1 · Reading a trajectory function (confidence builder)

A cart on a straight track has position $x(t) = 2t^3 - 3t^2$ (metres, seconds) for $0 \le t \le 2$ s. Find (a) its velocity and acceleration at $t = 2$ s, (b) its average velocity over the full interval, and (c) its average speed over the full interval — and explain why (b) and (c) differ.

Step 1. Differentiate twice: $v(t) = \dot x = 6t^2 - 6t$ and $a(t) = \dot v = 12t - 6$. At $t = 2$: $v = 6(4) - 6(2) = \mathbf{12\ m/s}$ and $a = 12(2) - 6 = \mathbf{18\ m/s^2}$, both positive — moving and accelerating in $+x$.

Step 2. Average velocity is displacement over time. Endpoints: $x(0) = 0$ and $x(2) = 2(8) - 3(4) = 4$ m. So $\bar v = (4 - 0)/2 = \mathbf{2\ m/s}$ — a secant slope, and nothing like the instantaneous 12 m/s at the end.

Step 3. For the average speed we need the distance, and that requires checking for reversals: $v(t) = 6t(t-1) = 0$ at $t = 0$ and $t = 1$ s. Between them $v < 0$ (test $t = 0.5$: $v = -1.5$ m/s) — the cart backs up first.

Step 4. Tally the route: from $x(0) = 0$ down to $x(1) = 2 - 3 = -1$ m (distance 1 m), then forward to $x(2) = 4$ m (distance 5 m). Total distance $= 6$ m, so average speed $= 6/2 = \mathbf{3\ m/s}$.

Step 5. The two averages differ because the cart retraced ground: displacement (4 m) forgave the 1 m detour, distance (6 m) charged for it twice. Note also $a(0) = -6\ \mathrm{m/s^2}$ with $v(0) = 0$: at the very start the cart is momentarily at rest and accelerating — velocity zero, change of velocity decidedly not.

2 · Braking into a bend (exam level)

A car enters a bend of radius $r = 120$ m at $v = 25\ \mathrm{m/s}$ (90 km/h) while braking so that its speed falls at $3.0\ \mathrm{m/s^2}$. Find the magnitude and direction of the total acceleration at that instant, compare it with $g$, and check whether tyres with friction budget $\mu = 0.8$ can deliver it.

Step 1. Two jobs, two terms of eq. (4). Normal (steering) piece: $a_n = v^2/r = 25^2/120 = 625/120 = 5.21\ \mathrm{m/s^2}$, pointing at the bend's centre. Tangential (braking) piece: $a_t = 3.0\ \mathrm{m/s^2}$, pointing backward along the path.

Step 2. The pieces are perpendicular, so Pythagoras assembles the total: $|\mathbf{a}| = \sqrt{5.21^2 + 3.0^2} = \sqrt{27.1 + 9.0} = \sqrt{36.1} = \mathbf{6.0\ m/s^2}$.

Step 3. Direction: the total leans $\arctan(5.21/3.0) = \arctan(1.74) = 60^\circ$ away from the backward tangent, toward the centre — mostly steering, substantially braking.

Step 4. Compare with gravity: $6.0/9.81 = \mathbf{0.61\,g}$. The passengers feel pressed forward-and-outward against the belts at six-tenths of their weight — brisk, but street-legal physics.

Step 5. Feasibility: friction can supply at most $\mu g = 0.8 \times 9.81 = 7.85\ \mathrm{m/s^2}$ of total horizontal acceleration. Demanded: $6.0\ \mathrm{m/s^2} < 7.85\ \mathrm{m/s^2}$ — the manoeuvre is safe, with margin. Had the driver entered at 32 m/s, the steering term alone would be $32^2/120 = 8.5\ \mathrm{m/s^2} > \mu g$: no braking strategy could hold the line, and the car would run wide. This one comparison — demanded $\sqrt{a_t^2 + (v^2/r)^2}$ against available $\mu g$ — is the entire physics of the racing line.

Texts. Morin, Introduction to Classical Mechanics, Ch. 2–3 · Taylor, Classical Mechanics, Ch. 1 · Goldstein, Poole & Safko, Classical Mechanics, 3rd ed., Ch. 1 · Landau & Lifshitz, Mechanics (Vol. 1), §1–3 · Kleppner & Kolenkow, An Introduction to Mechanics, Ch. 1 for the vector-kinematics drills.

Provenance. All definitions and identities are standard textbook kinematics. A verifier should: (1) confirm $\mathbf{v}=\mathrm{d}\mathbf{r}/\mathrm{d}t$, $\mathbf{a}=\mathrm{d}^2\mathbf{r}/\mathrm{d}t^2$ (Morin Ch. 2, Taylor Ch. 1); (2) re-derive the polar acceleration $(\ddot r-r\dot\theta^2)\hat{\mathbf{r}}+(r\ddot\theta+2\dot r\dot\theta)\hat{\boldsymbol{\theta}}$ using $\dot{\hat{\mathbf{r}}}=\dot\theta\hat{\boldsymbol{\theta}}$ (Goldstein Ch. 1, Taylor §1.7); (3) verify centripetal magnitude $v^2/r=\omega^2 r$; (4) check the Galilean result $\mathbf{a}'=\mathbf{a}$ (Landau–Lifshitz §3). The value $g\approx9.81\ \mathrm{m/s^2}$ is the standard surface gravity. Worked-example numbers on this page ($v = 12\ \mathrm{m/s}$, $a = 18\ \mathrm{m/s^2}$, averages 2 and 3 m/s for the cart; $a_n = 5.21$, total $6.0\ \mathrm{m/s^2} = 0.61g$ at $60^\circ$ for the bend) are illustrative arithmetic recomputed from the boxed formulas.

Builds on: Position, Displacement, and Reference Frames · Vectors in Kinematics and Coordinate Systems. Leads to: One-Dimensional Kinematics · Projectile Motion · Circular Motion and Centripetal Acceleration · Newton's Second Law.

• Slope of $x(t)$ = velocity; slope of $v(t)$ = acceleration
• Area under $v(t)$ = displacement; area under $a(t)$ = velocity change
• Flat spot on $x(t)$ = instantaneously at rest (acceleration can still be anything)
• $av > 0$ speeding up, $av < 0$ slowing down — the sign of $a$ alone decides nothing

• Free fall: $g \approx 9.81\ \mathrm{m/s^2}$ — the acceleration at every point of a throw, including the top
• Circular motion: $a = v^2/r$; car at 25 m/s on a 120 m bend needs $5.2\ \mathrm{m/s^2}$ of pure steering
• Friction budget: tyres deliver at most $\mu g \approx 7.8\ \mathrm{m/s^2}$ ($\mu = 0.8$) — total demand $\sqrt{a_t^2 + a_n^2}$ must fit inside

Step 1. Write or read off $\mathbf{r}(t)$; differentiate for $\mathbf{v}$, again for $\mathbf{a}$ — component by component in Cartesian. Step 2. For curved motion, split $\mathbf{a}$ into $\dot v$ along the path and $v^2/\rho$ across it. Step 3. For anything orbiting a centre, go polar and keep all four terms of eq. (6). Step 4. Sanity-check: reversals need $v = 0$ crossings; distance $\ge$ |displacement|; and no inertial-frame change may alter your acceleration.

• Constant speed on a curve IS acceleration ($v^2/r$, inward) — the speedometer sees only half of velocity
• $v = 0$ does not imply $a = 0$: top of the throw, turn-around points
• "Deceleration": check the sign of $av$, not of $a$ alone
• Polar reflexes: forgetting the basis vectors move loses the $-r\dot\theta^2$ and $2\dot r\dot\theta$ terms
• Equal-distance speed averages combine harmonically (40/60 → 48), not arithmetically