Position, Velocity & Acceleration Graphs

The value plotted on v–t at any instant is the slope of x–t at that instant, and the value on a–t is the slope of v–t. Read slope with a tangent — not with the height of the curve, and not with a chord. If a tangent laid on an x–t curve at t = 2.0 s passes through (1.0 s, 3.0 m) and (3.0 s, 11.0 m), the velocity at 2.0 s is (11.0 − 3.0) m ÷ (3.0 − 1.0) s = 4.0 m/s. A chord joining two points of the curve gives only the average velocity between them. The sign of the slope is the direction of motion and its steepness is the speed. Height is a different quantity altogether: two carts both at x = 5.0 m at the same instant can have slopes of +3.0 m/s and −1.5 m/s, so a point high on x–t is far from the origin, not fast.

Going the other way, the signed area under v–t between two times is the displacement, and the signed area under a–t is the change in velocity. Area above the t axis counts positive, area below counts negative, and its unit is the product of the axis units: m s⁻¹ × s = m. Take a v–t graph that holds 6.0 m/s for 2.0 s and then falls in a straight line to −2.0 m/s at 6.0 s. The rectangle gives 12 m, the triangle from 2.0 s to the zero crossing at 5.0 s gives ½ × 3.0 × 6.0 = 9.0 m, and the small triangle below the axis gives −½ × 1.0 × 2.0 = −1.0 m. So Δx = +20 m, while the path length is 22 m, because distance counts every area as positive. What area can never supply is the starting value: the same v–t graph fits an object that began at x = 0 or at x = −4.0 m, and only that initial condition decides whether x–t ends at 20 m or at 16 m.

A turning point is where the object reverses, so v passes through zero: x–t has a horizontal tangent at a peak or trough, and v–t crosses the t axis directly beneath it. The acceleration there is usually not zero, because the v–t graph is still sloping as it crosses. For x(t) = 6t² − t³, with x in m and t in s, v = 12t − 3t² is zero at t = 4.0 s, where x peaks at 32 m, yet a = 12 − 6t = −12 m/s² at that instant. One level down the same pattern repeats: v is largest where a crosses zero, at t = 2.0 s with v = 12 m/s, and that instant is an inflection of x–t, where the curve stops bending upward and starts bending downward. Curvature is the visual code for the sign of a: x–t bending upward means a > 0 and bending downward means a < 0, whichever way the object happens to be moving.

Forces can switch on and off abruptly, so an a–t graph may contain steps: a motor cutting out takes a from +1.5 m/s² to 0 in an instant. Velocity cannot jump, because a jump would demand an infinite acceleration; at a step in a, the v–t graph simply changes slope and shows a corner. Position cannot jump either, and wherever v is continuous the x–t graph has no corner at all — only its curvature changes. So a corner in x–t, or a vertical step in v–t, is always an idealisation of something brief. Take a ball that strikes a wall at 5.0 m/s and rebounds at 4.0 m/s, with the approach direction positive: v changes by −9.0 m/s. Drawn as a jump, the a–t graph would need an infinitely tall spike. In reality the contact lasts about 2.0 ms, so the spike has an average height of −9.0 ÷ 0.0020 = −4.5 × 10³ m/s² and an area of exactly −9.0 m/s. The area survives the idealisation; the height does not.

Going down the stack, work feature by feature: mark where the upper graph is flat (zeros below), where it is steepest (extremes below) and where its curvature changes (zeros two levels below). Going up, work interval by interval: start from the known initial value, add each area with its sign, and carry the running total across every boundary. Try an a–t graph reading +2.0 m/s² from 0 to 3.0 s, zero until 5.0 s, then −3.0 m/s² until 7.0 s, for an object starting from rest. The areas give v = 6.0 m/s at 3.0 s, still 6.0 m/s at 5.0 s, and 6.0 − 3.0 × 2.0 = 0 at 7.0 s. Under that v–t graph the areas are 9.0 m, 12.0 m and 6.0 m, so x–t bends upward for 3.0 s, runs straight for 2.0 s, then bends downward and arrives flat at 27.0 m. The last check comes free: the a–t areas, +6.0 and −6.0 m/s, cancel exactly because the object ends at rest.

A motion sensor does not deliver a curve; it delivers positions every Δt, so every slope you read from it is a finite difference. The central difference v ≈ [x(t + Δt) − x(t − Δt)] / (2Δt) is the chord across two samples, and the second difference a ≈ [x(t + Δt) − 2x(t) + x(t − Δt)] / Δt² is the slope of that slope. Dividing by a small Δt magnifies reading noise. With Δt = 0.050 s and independent scatter of 1.0 mm in each position, the velocity estimate scatters by √2 × 0.0010 m ÷ (2 × 0.050 s) ≈ 0.014 m/s, but the acceleration estimate scatters by √6 × 0.0010 m ÷ (0.050 s)² ≈ 0.98 m/s², a tenth of g from one millimetre. Areas work the other way: adding up many samples averages the noise down. That is why an a–t graph computed from a position sensor looks ragged, while x–t built up from an accelerometer looks smooth, until a small offset in a accumulates into a steadily growing drift.