Tangential and Radial Acceleration

Write the velocity as a speed times a unit vector along the path: v⃗ = v T̂. The product rule gives a⃗ = (dv/dt) T̂ + v (dT̂/dt). The first term is the speed changing with the direction held fixed. The second is the direction changing with the speed held fixed. Because T̂ has fixed length, dT̂/dt is perpendicular to it, and dT̂/ds = N̂/ρ defines the radius of curvature ρ, so dT̂/dt = (v/ρ) N̂ with N̂ the unit normal pointing toward the centre of the curve. That gives a⃗ = (dv/dt) T̂ + (v²/ρ) N̂. There is no third term: a particle on a path can speed up, turn, or do both.

Speed is the length of the velocity, so differentiate v² = v⃗ · v⃗ to get 2v (dv/dt) = 2 v⃗ · a⃗, which rearranges to dv/dt = a⃗ · v̂. Any part of the acceleration perpendicular to the velocity drops out of that dot product and cannot change the speed, however large it is. That is why uniform circular motion holds its speed: the acceleration is entirely perpendicular to the velocity. Read backwards, an object whose speed is momentarily steady has no tangential part at that instant, so whatever acceleration it does have is purely radial.

ρ is the radius of the circle that best fits the path where the object is now, not the size of the whole trajectory, and radial here means toward that local centre, not away from some fixed origin. On a circular track ρ is the track radius; on a parabola it changes from point to point. It is often easier to get it from the motion than from geometry, since ρ = v²/aᵣ. Take a ball launched at 25 m s⁻¹ at 60° above the horizontal. At the top of the flight the velocity is horizontal at 25 cos 60° = 12.5 m s⁻¹ and gravity is entirely perpendicular to it, so aᵣ = 9.81 m s⁻² and ρ = 12.5²/9.81 = 16 m. The path is locally a 16 m circle even though nothing is going round in circles.

Because the parts are perpendicular, combine them with Pythagoras and take the direction from their ratio. A car takes a bend of radius 50 m at 15 m s⁻¹ while braking at 2.0 m s⁻². The radial part is aᵣ = 15²/50 = 4.5 m s⁻² inward, the tangential part is 2.0 m s⁻² backwards along the road, and the total is √(4.5² + 2.0²) = 4.9 m s⁻². It still points into the bend, but rotated back from the radius by arctan(2.0/4.5) = 24°, leaning against the travel direction because the car is slowing. Accelerating at the same rate would give the same magnitude with the 24° lean forwards.

The acceleration arrow on a curve always points to the concave side, because aᵣ = v²/ρ is never negative. The lean off the radius carries the rest of the story: forward of the radius means speeding up, behind it means slowing, and exactly along the radius means dv/dt = 0, so the speed is momentarily stationary — either steady, as in uniform circular motion, or at a turning point. On a straight run ρ is infinite, aᵣ vanishes, and the arrow lies along the line of travel. Sketching v⃗ and a⃗ at a few points therefore shows where the speed is rising and where it is falling before any number is computed.

Uniform circular motion is this decomposition with aₜ = 0 and ρ fixed, and straight-line kinematics is the same decomposition with aᵣ = 0. Non-uniform circular motion needs both, and Newton's second law splits the same way into ΣFᵣ = mv²/r toward the centre and ΣFₜ = maₜ along the motion. The parts also grow at different rates: a car pulling away from rest around a 50 m circle at a steady aₜ = 2.0 m s⁻² has v² = 2aₜs, so the radial part climbs in step with the distance covered, and with t², while the tangential part holds still. The two match when v = √(raₜ) = √(50 × 2.0) = 10 m s⁻¹, five seconds in. After that, turning takes the larger share of the grip.