Vector Position, Velocity & Acceleration

Pick an origin O and two perpendicular axes, and the particle's location at time t is the arrow r⃗(t) = x(t) î + y(t) ĵ drawn from O to the particle. The trajectory is the set of points that arrow's tip visits, drawn in the x–y plane with the time labels erased. Move O and every r⃗ changes by the same constant vector, so position is partly a choice. Differences of position are not. A buoy drifting from (2.0, 1.0) m to (5.0, 5.0) m has displacement Δr⃗ = (3.0 î + 4.0 ĵ) m, 5.0 m long at 53° above +x, whichever origin you measured from. The distance it actually covered along its curving path is at least 5.0 m and usually more, which is why |Δr⃗| and path length are different quantities.

Dividing a vector by the positive scalar Δt changes its length and units but not its direction, so v⃗_avg = Δr⃗/Δt points along the straight chord between two points on the path, not along the path. Follow r⃗(t) = (2.0 m s⁻¹)t î + (1.0 m s⁻³)t³ ĵ from t = 1.0 s. Over the next 0.50 s, Δr⃗ = (1.00 î + 2.375 ĵ) m, so v⃗_avg = (2.0 î + 4.75 ĵ) m s⁻¹, at 67° above +x. Over 0.10 s it is (2.0 î + 3.31 ĵ) m s⁻¹ at 59°; over 0.010 s, (2.0 î + 3.03 ĵ) m s⁻¹ at 56.6°. The x-part never moves, because x grows steadily; only the y-part, which curves, depends on the interval. The chord is swinging toward a limiting direction.

Let Δt → 0 and the chord shrinks onto the curve, its direction becoming the tangent's: v⃗ = dr⃗/dt. For the path above, dx/dt = 2.0 m s⁻¹ and dy/dt = (3.0 m s⁻³)t², so at t = 1.0 s the velocity is (2.0 î + 3.0 ĵ) m s⁻¹, 3.6 m s⁻¹ at 56.3° above +x, the direction the averages were closing in on. Differentiating component by component is allowed only because î and ĵ are constant: the product rule gives d(x î)/dt = (dx/dt) î + x (dî/dt), and the last term vanishes. Measure along a unit vector that turns with the particle, such as the outward direction r̂ of polar coordinates, and that term survives; it is where the extra pieces of circular motion come from. The tangent property also hands you the path's slope, dy/dx = vᵧ/vₓ, wherever vₓ ≠ 0.

a⃗ = dv⃗/dt, again one component at a time, and it registers a change of direction exactly as it registers a change of size. A car heading east at 12 m s⁻¹ that is heading north at 12 m s⁻¹ four seconds later has not changed speed, yet Δv⃗ = (−12 î + 12 ĵ) m s⁻¹, 17 m s⁻¹ long, so its average acceleration is 4.2 m s⁻² toward the north-west. The part of a⃗ perpendicular to v⃗ points toward the side the velocity is turning to, so on a bending path a⃗ leans into the concave side of the curve. It lies along v⃗ only on a straight path. Whether the particle is also speeding up or slowing down is read from the sign of a⃗ ⋅ v⃗; the later topic on tangential and radial acceleration makes that split exact.

Given a⃗(t), integrate each component on its own, and each integration asks for its own starting value. A drone at the origin with v⃗₀ = (4.0 m s⁻¹) ĵ is given a⃗(t) = (1.5 m s⁻³)t î. Then vₓ = (0.75 m s⁻³)t² while vᵧ stays at 4.0 m s⁻¹, and integrating again gives x = (0.25 m s⁻³)t³ and y = (4.0 m s⁻¹)t. At t = 2.0 s, r⃗ = (2.0 î + 8.0 ĵ) m, v⃗ = (3.0 î + 4.0 ĵ) m s⁻¹ with speed 5.0 m s⁻¹, and a⃗ = (3.0 m s⁻²) î. The three arrows point 76°, 53° and 0° above +x: where a particle is says nothing about which way it is moving, and which way it is moving says nothing about which way it is accelerating. Give the same drone v⃗₀ = (4.0 m s⁻¹) î instead and the identical a⃗(t) keeps it on the x axis for good.

Eliminating t collapses the two component functions into one trajectory equation. For r⃗(t) = (2.0 m s⁻¹)t î + (1.0 m s⁻³)t³ ĵ, substituting t = x/(2.0 m s⁻¹) gives y = x³/(8.0 m²), whose slope at x = 2.0 m is dy/dx = 3x²/(8.0 m²) = 1.5 = vᵧ/vₓ: the same 56° heading found by differentiating in time. What the curve cannot supply is the timing. A second particle at r⃗(t) = (4.0 m s⁻¹)t î + (8.0 m s⁻³)t³ ĵ traces exactly the same curve, but it reaches x = 2.0 m at t = 0.50 s with v⃗ = (4.0 î + 6.0 ĵ) m s⁻¹, the same direction at twice the speed, and aᵧ = 24 m s⁻² instead of 6.0 m s⁻². A trajectory sketch settles directions; only the time functions settle magnitudes.