Average and instantaneous acceleration

Average acceleration is the change in velocity divided by the time it took: a_avg = (v₂ − v₁)/(t₂ − t₁). Both velocities keep their signs on the declared axis, so a reversal counts in full. A tennis ball strikes a wall at 30 m s⁻¹ and leaves along the same line at 24 m s⁻¹ after 4.0 ms of contact. Taking the incoming direction as positive, v₁ = +30 m s⁻¹ and v₂ = −24 m s⁻¹, so Δv = −54 m s⁻¹ and a_avg = −54/0.0040 = −1.4 × 10⁴ m s⁻², roughly 1400 g, directed away from the wall. Subtracting speeds, 24 − 30 = −6 m s⁻¹, underestimates the change ninefold: before the ball can move back at all, its 30 m s⁻¹ of incoming velocity has to be removed. The unit carries the idea, since m s⁻² is metres per second of velocity change in each second.

The average needs only two readings. The instantaneous acceleration asks what happens as the second reading is brought toward the first: a = lim Δv/Δt as Δt → 0, the derivative dv/dt, which is the slope of the tangent to the velocity–time graph. A test drone moves vertically with v(t) = 6.0 − 1.5t² (v in m s⁻¹, t in s, up positive). For intervals starting at t = 2.0 s, Δv/Δt = −1.5(4.0 + Δt) m s⁻²: −7.5 for Δt = 1.0 s, −6.15 for 0.10 s and −6.015 for 0.010 s. The chord slopes close in on the tangent slope dv/dt = −3.0t = −6.0 m s⁻². Since v is itself dx/dt, acceleration is the second derivative d²x/dt², so its sign is the concavity of the position graph: an x–t curve bending upward means a is positive, whether x is increasing, decreasing or below zero.

A component of acceleration is positive when the change in velocity points along +x and negative when it points along −x. Speed appears nowhere in that statement. Reverse the axis and every kinematic component flips together: x becomes −x, v becomes −v and a becomes −a. A car heading east at 20 m s⁻¹ and braking at 4.0 m s⁻² has v = +20 m s⁻¹ and a = −4.0 m s⁻² with east positive, but v = −20 m s⁻¹ and a = +4.0 m s⁻² with west positive. The car has not changed; the bookkeeping has. So declare the positive direction before writing any number, mark it on the sketch, and keep it for the whole problem. Braking and deceleration are words about speed falling, not synonyms for a minus sign. The product va is the one combination the flip leaves alone, which is why it is the right tool for the next step.

Speed is |v|, and the clean way to follow it is through v², which rises and falls with it: d(v²)/dt = 2v(dv/dt) = 2va. So the speed increases where va is positive and decreases where va is negative, which gives four cases rather than two. Moving along +x with positive a, or along −x with negative a, the object speeds up; moving along +x with negative a, or along −x with positive a, it slows down. A bus reversing out of a depot, with forward taken as positive, has v = −2.0 m s⁻¹ and a = −1.5 m s⁻². After 2.0 s at that acceleration Δv = aΔt = −3.0 m s⁻¹, so v = −5.0 m s⁻¹: its speed has grown from 2.0 to 5.0 m s⁻¹ under a negative acceleration. Where v passes through zero while a is not zero, va changes sign and the object turns round. The drone of the previous section is at rest at t = 2.0 s with a = −6.0 m s⁻², and a moment later v and a are both negative, so it speeds up on the way down.

Because Δv = ∫ a dt taken from t₁ to t₂, the average acceleration is the time average of a(t): a_avg = (1/Δt) ∫ a dt. Very different histories can share one average. A courier van at 9.0 m s⁻¹ brakes at 2.5 m s⁻² for 2.0 s, rolls through a junction at 4.0 m s⁻¹ for 5.0 s, then pulls away at 1.0 m s⁻² for 5.0 s. It ends at 9.0 m s⁻¹, so a_avg over the 12.0 s is zero, although the acceleration was −2.5 m s⁻² for two of those seconds and +1.0 m s⁻² for five. If v(t) is smooth, the mean value theorem promises at least one instant where a equals a_avg, without saying when. Nor is a_avg the mean of the accelerations at the two ends, unless a changes linearly in time. With a = 3.0t² (a in m s⁻², t in s) over the first 2.0 s, the end values average 6.0 m s⁻², but Δv = ∫ 3.0t² dt = 8.0 m s⁻¹, so a_avg = 4.0 m s⁻².

Sensors report velocity or position at discrete times, so the limit has to be approximated. The forward difference [v(t + h) − v(t)]/h is a chord over [t, t + h]; for smooth motion it tracks the acceleration at the midpoint t + h/2 far better than at t. Centring the pair puts the estimate where it is wanted: a(t) ≈ [v(t + h) − v(t − h)]/2h. For the drone at t = 2.0 s with h = 0.50 s, the forward difference gives −6.75 m s⁻² and the central difference exactly −6.0 m s⁻², because a central difference is exact whenever a changes linearly in time. With positions alone, the second difference [x(t + h) − 2x(t) + x(t − h)]/h² does the same job. The catch is noise. A reading error δ enters divided by h, or by h² for positions, so shrinking h is not automatically an improvement: below some interval the estimate stops converging on the tangent and starts reporting the sensor.