Average acceleration vs instantaneous acceleration: Worked examples

Before anything else, let me re-earn the two tools we use everywhere, in plain words:

The scenario matrix

Every acceleration problem is one (or a blend) of these cells. The right-hand column names the example that covers it.

#	Case class	What makes it tricky	Covered by
A	Positive average, 1-D, speeding up	none — the friendly base case	Ex 1
B	Negative average (slowing down, still moving forward)	sign of $a$ opposite to motion	Ex 2
C	Direction reversal (velocity flips sign, speed unchanged)	speed constant yet $Δ v \neq = 0$	Ex 3
D	Instantaneous ≠ average (varying $a$ )	derivative vs secant slope differ	Ex 4
E	Zero / degenerate input ( $Δ t \to 0$ , or $v$ constant)	division by zero, undefined vs zero	Ex 5
F	2-D vector $Δ v$ (turning corner)	must subtract vectors, not speeds	Ex 6
G	Constant speed, curved path (circular)	acceleration points sideways	Ex 7
H	Real-world word problem (units, hidden signs)	translate English → symbols	Ex 8
I	Exam twist (find when $a = 0$ from a position function)	second derivative + solve	Ex 9

The examples

Cell A — friendly positive average

Cell B — negative average, still moving forward

Cell C — direction reversal, speed unchanged

Worked example Ex 3 — a ball bounces straight back

A ball hits a wall at $20 m/s$ east and rebounds at $20 m/s$ west. Contact lasts $0.05 s$ . Take east as $+$ .

Forecast: Speed is $20$ before and after — is the acceleration zero? (Watch the trap.)

Write velocities with sign. $v_{i} = + 20$ , $v_{f} = - 20$ . Why this step? West is the opposite direction, so it carries the opposite sign. Speed ignores sign; velocity does not.
$a_{a v g} = \frac{- 20 - ( + 20 )}{0.05} = \frac{- 40}{0.05}$ . Why this step? The subtraction $v_{f} - v_{i}$ is where the reversal shows up — the two 20's add in magnitude.
$a_{a v g} = - 800 m/s^{2}$ .

Verify: Huge and negative (pointing west, toward the wall's push). Even though speed never changed, $Δ v = - 40 m/s$ is large, so acceleration is large. This is the steel-manned mistake from the parent note made concrete: constant speed $\neq =$ zero acceleration.

Cell D — instantaneous vs average differ

Worked example Ex 4 — a rocket with rising acceleration

A rocket has velocity $v (t) = 2 t^{2} + 3 t$ (m/s). Find (a) the instantaneous acceleration at $t = 3 s$ , and (b) the average acceleration over $[0, 3] s$ .

Forecast: Since the acceleration is increasing with time, will the average over $[0, 3]$ be bigger or smaller than the instantaneous value at the end $t = 3$ ?

Differentiate for instantaneous $a$ . $a (t) = \frac{d v}{d t} = 4 t + 3$ . Why this step? Instantaneous acceleration is the tangent slope = derivative. Term-by-term: $\frac{d}{d t} 2 t^{2} = 4 t$ , $\frac{d}{d t} 3 t = 3$ .
Evaluate at the instant. $a (3) = 4 (3) + 3 = 15 m/s^{2}$ .
For the average, get endpoint velocities. $v (0) = 0$ , $v (3) = 2 (9) + 3 (3) = 27 m/s$ . Why this step? Average uses only endpoints (secant slope), so we need the two $v$ values, not the derivative.
$a_{a v g} = \frac{27 - 0}{3 - 0} = 9 m/s^{2}$ .

Verify: $a_{a v g} = 9 < a (3) = 15$ , matching the forecast — because acceleration climbs, the end is steeper than the whole-interval average. See the tangent (red) vs secant (yellow) below.

Cell E — zero and degenerate inputs

Worked example Ex 5 — the edge cases

Consider two separate situations. (a) A car cruises at a perfectly steady $20 m/s$ for $10 s$ . (b) You try to compute $a_{a v g}$ over $Δ t = 0$ (same instant, twice).

Forecast: Which of these gives zero acceleration, and which gives something undefined? They are not the same.

Case (a): $v_{f} = v_{i} = 20$ , so $a_{a v g} = \frac{20 - 20}{10} = \frac{0}{10} = 0 m/s^{2}$ . Why this step? Numerator is exactly zero, denominator is a real positive time → a genuine, well-defined zero.
Case (b): $a_{a v g} = \frac{v ( t ) - v ( t )}{0} = \frac{0}{0}$ . Why this step? When $Δ t = 0$ the formula divides by zero — this is undefined, not zero. You cannot measure a "change over no time."
The resolution: to talk about an instant, you don't set $Δ t = 0$ ; you take the limit as $Δ t \to 0$ (see the parent's Step 2). The $\frac{0}{0}$ form is exactly why the derivative is a limit, not a plain division.

Verify: (a) gives $0$ (defined). (b) is undefined until rescued by a limit. Two different "nothings" — never confuse them.

Cell F — 2-D vector change (turning a corner)

Worked example Ex 6 — a car rounds a right-angle bend

A car moves at $10 m/s$ east, then $2 s$ later moves at $10 m/s$ north (same speed). Find $a_{a v g}$ : its size and direction.

Forecast: Same speed before and after — but this is a 90° turn. Do you expect zero acceleration or not? Which compass direction should $a_{a v g}$ roughly point?

Write each velocity as components $(east, north)$ . $v_{i} = (10, 0)$ , $v_{f} = (0, 10)$ . Why this step? Vectors must be subtracted component-by-component; you cannot subtract speeds. See Vectors: addition and subtraction.
Subtract. $Δ v = v_{f} - v_{i} = (0 - 10, 10 - 0) = (- 10, 10) m/s$ . Why this step? This $Δ v$ points north-west — toward the inside of the turn, not along either velocity.
Divide by time. $a_{a v g} = \frac{Δ v}{Δ t} = \frac{( - 10 , 10 )}{2} = (- 5, 5) m/s^{2}$ .
Magnitude. $∣ a_{a v g} ∣ = (- 5)^{2} + 5^{2} = 50 \approx 7.07 m/s^{2}$ . Why this step? The size of a vector comes from the Pythagorean theorem on its components.

Verify: Direction $(- 5, 5)$ is $45°$ north of west — pointing into the bend, as forecast. Speed unchanged yet $∣ a_{a v g} ∣ \approx 7.07 \neq = 0$ . Figure below shows the vector triangle.

Cell G — constant speed, curved path (circular)

Worked example Ex 7 — a stone on a string

A stone whirls in a circle of radius $r = 2 m$ at constant speed $v = 4 m/s$ . Find its instantaneous acceleration (magnitude and direction).

Forecast: Speed is constant, so is the acceleration zero? If not, where does it point — along the motion, or somewhere else?

Recall that in Uniform circular motion the speed is constant but the velocity vector keeps turning, so $\frac{d v}{d t} \neq = 0$ . Why this step? Acceleration reacts to the change in the velocity vector, and here direction changes even though magnitude doesn't.
The centripetal (centre-pointing) acceleration has magnitude $a = \frac{v ^{2}}{r}$ . Why this step? Geometry of the turning velocity vector (a small $Δ v$ points toward the centre) gives this size — derived fully in the Uniform circular motion note.
Plug in. $a = \frac{4 ^{2}}{2} = \frac{16}{2} = 8 m/s^{2}$ , directed toward the centre.

Verify: Nonzero acceleration despite constant speed ✓. It points inward (perpendicular to velocity), so it changes direction, not speed. Figure shows the velocity (tangent) vs acceleration (inward).

Cell H — real-world word problem

Worked example Ex 8 — an airplane on the runway

"A jet touches down at $70 m/s$ and must stop within $14 s$ using reverse thrust. Assuming the deceleration is roughly constant, what average acceleration does the runway record, and is it within a comfortable $- 6 m/s^{2}$ limit?"

Forecast: "Stop" means final velocity is what number? Will the answer be negative? Bigger or smaller in size than $6$ ?

Translate English to symbols. $v_{i} = 70 m/s$ , $v_{f} = 0$ (stopped), $Δ t = 14 s$ , forward $= +$ . Why this step? The hidden data is " $v_{f} = 0$ " — "stop" is not a speed word you plug in blindly; it means the velocity becomes zero.
$a_{a v g} = \frac{0 - 70}{14} = \frac{- 70}{14} = - 5 m/s^{2}$ . Why this step? Slowing while moving forward → negative sign, as in Cell B.
Compare to the limit. $∣ - 5∣ = 5 < 6$ .

Verify: $- 5 m/s^{2}$ is within the $- 6 m/s^{2}$ comfort limit ✓. Units check: $\frac{m/s}{s} = m/s^{2}$ ✓. The negative sign is braking, exactly what reverse thrust does.

Cell I — exam twist (find where $a = 0$ )

Worked example Ex 9 — "at what time is the acceleration zero?"

A particle's position is $x (t) = t^{3} - 9 t^{2} + 15 t$ (m). At what time(s) is the instantaneous acceleration zero, and is it positive or negative before that instant?

Forecast: Acceleration is the second derivative of position. A cubic $x$ → what shape is $a (t)$ , and how many zeros should it have?

First derivative → velocity. $v (t) = \frac{d x}{d t} = 3 t^{2} - 18 t + 15$ . Why this step? One derivative of position gives velocity (Derivatives as rates of change).
Second derivative → acceleration. $a (t) = \frac{d v}{d t} = 6 t - 18$ . Why this step? Instantaneous acceleration is $\frac{d ^{2} x}{d t ^{2}}$ — differentiate velocity again. A cubic $x$ → linear $a$ → exactly one zero, matching the forecast.
Set $a = 0$ . $6 t - 18 = 0 \Rightarrow t = 3 s$ .
Sign before $t = 3$ : pick $t = 0$ , $a (0) = 6 (0) - 18 = - 18 < 0$ . Why this step? Testing a point on one side tells you the acceleration is negative before the switch, positive after.

Verify: $a (3) = 6 (3) - 18 = 0$ ✓. For $t < 3$ acceleration is negative (velocity decreasing), for $t > 3$ positive (velocity increasing) — so $t = 3$ is where velocity bottoms out.

Recall

Recall Cover the answers and test yourself

Cell B (Ex 2) — cyclist slows $15 \to 3$ in $4$ s, $a_{a v g}$ ? ::: $- 3 m/s^{2}$ Cell C (Ex 3) — ball $+ 20 \to - 20$ in $0.05$ s, $a_{a v g}$ ? ::: $- 800 m/s^{2}$ Cell D (Ex 4) — $v = 2 t^{2} + 3 t$ ; $a (3)$ and $a_{a v g}$ over $[0, 3]$ ? ::: $15$ and $9 m/s^{2}$ Cell E — why is $a_{a v g}$ over $Δ t = 0$ undefined not zero? ::: it is $0/0$ , division by zero; only a limit rescues it Cell F (Ex 6) — 90° turn at $10$ m/s in $2$ s, $∣ a_{a v g} ∣$ ? ::: $50 \approx 7.07 m/s^{2}$ Cell G (Ex 7) — stone, $v = 4$ , $r = 2$ , centripetal $a$ ? ::: $8 m/s^{2}$ toward centre Cell I (Ex 9) — $x = t^{3} - 9 t^{2} + 15 t$ , when is $a = 0$ ? ::: $t = 3 s$

Connections

Average acceleration vs instantaneous acceleration — the parent this page drills.
Average velocity vs instantaneous velocity — same secant/tangent logic one rung down.
Velocity-time graphs — Ex 4's tangent-vs-secant figure lives here.
Derivatives as rates of change — the differentiation used in Ex 4 and Ex 9.
Equations of motion (constant acceleration) — why "un-averaging" verifies Ex 1 and Ex 8.
Uniform circular motion — the source of Ex 7's $v^{2} / r$ .
Vectors: addition and subtraction — the component subtraction in Ex 6.