Acceleration

A cricket ball rises off the bat, slows, hangs for one heartbeat at the top of its flight — and in that instant its velocity is exactly zero while its acceleration is a full $9.8\ \text{m/s}^2$ pointing down. Most students meet that sentence and refuse to believe it: how can something be accelerating while it isn't even moving? The confusion comes from treating acceleration as "how fast you're going". It isn't. Acceleration is how fast your velocity is changing — and a velocity can be changing furiously at the very moment it happens to pass through zero. Car advertisements have trained you to read acceleration already: "0 to 100 km/h in 4 seconds" is an average acceleration of about $6.9\ \text{m/s}^2$ , dressed up for a showroom. Exams dress it differently — a ball rebounding off a wall, a train braking into a station, a velocity that flips sign mid-question — and then test one thing over and over: do you know the difference between velocity decreasing and speed decreasing? The single most reliable mark-loser in this chapter is writing "negative acceleration" when you mean "slowing down". You already own velocity — magnitude plus direction, sign carrying the direction in one dimension. This module divides its change by time and studies the result: the sign rules, the average-versus-instantaneous split you have seen once before, and the one test that never lies about speeding up versus slowing down.

Key takeaways

Average acceleration = change in velocity ÷ time taken: $\bar{a} = \Delta v/\Delta t$ , a vector built from the signed velocities, in $\text{m/s}^2$ .
$\Delta v = v - v_0$ is a subtraction of signed values. A ball reversing from $+15$ m/s to $-10$ m/s has $\Delta v = -25$ m/s — not $5$ .
Instantaneous acceleration is the limit of $\Delta v/\Delta t$ over a shrinking time window — the value the average settles on as the interval closes onto one instant.
The sign of $a$ alone tells you nothing about slowing down. The test: same signs of $a$ and $v$ → speeding up; opposite signs → slowing down.
Constant acceleration means velocity changes by equal amounts in equal times — $a = 5\ \text{m/s}^2$ means the velocity gains 5 m/s every second, whatever it currently is.
Zero acceleration means constant velocity, not rest; zero velocity permits any acceleration (top of a throw).
Key idea: acceleration measures the change in velocity per second — so it lives entirely in the comparison between now and a moment later, and an object's current speed, even zero, puts no limit on it.

The core insight

Acceleration is the rate of change of velocity — $\bar{a} = \Delta v / \Delta t$ over an interval, its shrinking-window limit at an instant — so it inherits velocity's vector character, and whether an object speeds up or slows down is decided not by the sign of $a$ alone but by whether $a$ and $v$ point the same way.

Where this lives today

Your phone measures acceleration hundreds of times per second. A MEMS accelerometer — a microscopic mass on silicon springs — feels every change in velocity: that is how the screen knows you rotated it, how a fitness app counts steps, and how a dropped laptop parks its hard-drive head mid-fall. A car's airbag computer runs the same physics with grimmer thresholds: a crash is an average acceleration of order $-100g$ for a few hundredths of a second — equation (1) evaluated in real time, the decision to fire made before your body reaches the steering wheel. At the human end, an F1 car braking from 300 km/h pulls about $5g$ of deceleration, while a rising rocket throttles to keep its crew near $3g$ . Engineers quote it all in $g$ 's because $9.8\ \text{m/s}^2$ is the acceleration every human body has calibrated since birth.

The physical picture

Photograph a car once every second and draw its velocity as an arrow at each flash. Two very different films can come out of the same road.

Figure

Two films of the same road — acceleration is the second-by-second change in the velocity arrow

In Film 1 every arrow is 3 m/s longer than the one before: the acceleration is $+3\ \text{m/s}^2$ , pointing with the motion. In Film 2 every arrow is 3 m/s shorter: $-3\ \text{m/s}^2$ , pointing against it. Notice what your eye actually compared — not the arrows, but the difference between consecutive arrows. That is the entire concept. Acceleration never describes where you are or how fast you go; only how the velocity arrow is being edited from moment to moment. One more frame of Film 2 makes the chapter's famous point. At $t = 4$ s the velocity reaches zero — and the editing continues. If the acceleration stays $-3\ \text{m/s}^2$ , the next second produces $v = -3$ m/s: the car (now, say, a ball on a slope) starts moving backwards. At the instant $v = 0$ the object was stationary and accelerating — its velocity was zero and changing. That is exactly the top of a thrown ball's flight, and exactly the situation exams love.

The definition, made exact

Average acceleration compares the velocity at the two ends of an interval and divides by the elapsed time. It is the velocity's version of average velocity itself — one differencing, one division:

\bar{a} = \frac{\Delta v}{\Delta t} = \frac{v - v_0}{t - t_0} \qquad (1)

where: $v_0$ — velocity at the start of the interval (sign included); $v$ — velocity at the end (sign included); $\Delta t$ — the time elapsed, always positive. The unit is $(\text{m/s})/\text{s} = \text{m/s}^2$ : "metres per second, per second" — how many m/s of velocity arrive each second. A showroom's "0 to 100 km/h in 4 s" is equation (1) with the units tidied: $100$ km/h $= 27.8$ m/s, so $\bar{a} = 27.8/4 = 6.9\ \text{m/s}^2$ . The subtraction $v - v_0$ is where marks die, because both symbols are signed. A ball that hits a wall at $+15$ m/s and rebounds at $10$ m/s in the opposite direction has $v = -10$ m/s, so

\Delta v = (-10) - (+15) = -25\ \text{m/s},

not $5$ and not $25$ . The change in velocity can be far larger than either velocity, because reversing direction is a huge change. Speed barely moved (15 down to 10); velocity swung through 25. Every rebound, bounce and turnaround problem is this one subtraction wearing different clothes. Averages blur, as always. A drag car can average $+10\ \text{m/s}^2$ over four seconds while touching $+15$ at launch and $+6$ near the end. To speak of the acceleration right now, shrink the window: average over one second around this instant, then a tenth, then a hundredth, and watch the value settle. The settled value is the instantaneous acceleration,

a = \lim_{\Delta t \to 0} \frac{\Delta v}{\Delta t} \qquad (2)

where: the limit means exactly the settling process just described. Geometrically this is the slope of the tangent to the velocity–time curve — the same secant-to-tangent move you met for instantaneous velocity, applied one level up (the graphs module drills that reading; the calculus module writes it as $a = dv/dt$ ). For this module you need the idea: instantaneous acceleration is what the average acceleration becomes when the interval closes onto one moment. When acceleration is constant, average and instantaneous agree everywhere and equation (1) is all you need — which is why boards lean so hard on constant (uniform) acceleration. Constant acceleration has a clean verbal signature: equal changes of velocity in equal times. At $a = 5\ \text{m/s}^2$ , an object at rest reads 5, 10, 15, 20 m/s at one-second ticks — a perfectly linear climb. It also has an honest scope: it is an idealisation, holding when the driving push is steady (a dropped stone before air resistance matters, a train under constant braking force). The suvat equations of the next module are the algebra of exactly this case.

Speeding up or slowing down: the only test that works

Here is the chapter's central discipline. The sign of $a$ tells you which way the velocity is being pushed — toward the positive or negative direction. Whether the object speeds up depends on whether that push agrees with the current motion:

current velocity	acceleration	verdict	example
$v > 0$	$a > 0$	speeding up	car pulling away, gas pedal down
$v > 0$	$a < 0$	slowing down	car braking while moving forward
$v < 0$	$a < 0$	speeding up	ball falling faster (down taken negative)
$v < 0$	$a > 0$	slowing down	ball thrown downward being braked by a net upward push

Same signs → speed grows. Opposite signs → speed shrinks. That is the whole test, and the third row is the one that breaks intuition: a negative acceleration making an object go faster, because the object was already moving the negative way and the push agrees with it.

Figure

The sign test at a glance — compare the directions of v and a, not the sign of a alone

The thrown ball ties the table together in one motion. Take up as positive. Rising: $v > 0$ , $a = -g$ — opposite signs, slowing. At the top: $v = 0$ , $a$ still $-g$ . Falling: $v < 0$ , $a$ still $-g$ — same signs, speeding up. One unchanging acceleration, both verdicts; the turnaround happens because the acceleration never pauses.

Derivation AInstantaneous acceleration by shrinking the window

We want the acceleration at the single instant $t = 1$ s for an object whose velocity follows $v = 4t^2$ (m/s with $t$ in seconds) — a genuinely non-uniform case, so the average over any finite window disagrees with the instant and the limit has real work to do. The strategy mirrors the tangent construction from the velocity module, one level up: compute $\bar{a}$ over a window starting at $t = 1$ s, shrink the window, watch the numbers converge, then do all windows at once with algebra. Watch how the answer separates into a fixed part and a part proportional to the window size — that structure is why the limit exists.

Step 1. Average acceleration from $t = 1$ to $t = 1 + \Delta t$ uses equation (1) with velocities read from the law: $v(1) = 4$ m/s and $v(1+\Delta t) = 4(1+\Delta t)^2$ .

Step 2. Try windows. $\Delta t = 1$ s: $v(2) = 16$ , so $\bar{a} = (16-4)/1 = 12\ \text{m/s}^2$ . $\Delta t = 0.1$ s: $v(1.1) = 4(1.21) = 4.84$ , so $\bar{a} = 0.84/0.1 = 8.4\ \text{m/s}^2$ . $\Delta t = 0.01$ s: $v(1.01) = 4(1.0201) = 4.0804$ , so $\bar{a} = 0.0804/0.01 = 8.04\ \text{m/s}^2$ . The sequence 12, 8.4, 8.04 is closing in on 8.

Step 3. Do every window at once. Expand the square and divide:

\bar{a} = \frac{4(1+\Delta t)^2 - 4}{\Delta t} = \frac{8\,\Delta t + 4\,\Delta t^2}{\Delta t} = 8 + 4\,\Delta t

Why this is the punchline: the messy quotient collapses to "a constant plus something that dies with the window". No limiting machinery is needed to see where it is heading.

Step 4. Shrink: as $\Delta t \to 0$ , the average $8 + 4\Delta t$ settles on exactly $8\ \text{m/s}^2$ . That is the instantaneous acceleration at $t = 1$ s — and each tenfold shrink of the window cut the excess tenfold (4, 0.4, 0.04), exactly as the formula $8 + 4\Delta t$ predicts. Verified: the tangent-slope rule for $v = 4t^2$ gives $a = 8t$ , hence $8\ \text{m/s}^2$ at $t = 1$ s; the three window values recompute correctly from $16$ , $4.84$ , $4.0804$ .

Go deeper · Derivation B · 1 minWhy the sign test works — same signs speed you up

Derivation BWhy the sign test works — same signs speed you up

We want to prove the table above rather than memorise it: an object speeds up exactly when its velocity and acceleration carry the same sign. The strategy is to track the square of the speed, $v^2$ , because squaring erases direction and leaves pure "how fast" — if $v^2$ grows, the object is speeding up, whatever the signs. We compare $v^2$ now with $v^2$ a tiny time $\Delta t$ later, using only algebra. Watch which term survives when $\Delta t$ is small — that term carries the whole verdict.

Step 1. In a short time $\Delta t$ the velocity updates from $v$ to $v + a\,\Delta t$ (that is what acceleration means: $a\,\Delta t$ of new velocity arrives in time $\Delta t$ ).

Step 2. Compare the squares — the change in $v^2$ is

(v + a\,\Delta t)^2 - v^2 = 2\,v\,a\,\Delta t + a^2\,\Delta t^2

Why square: $v^2 = |v|^2$ is speed squared, blind to sign — the perfect referee for "faster or slower" when directions are muddying the picture.

Step 3. Let the window be small. The term $a^2 \Delta t^2$ shrinks like $\Delta t^2$ — a hundredth of a second makes it ten-thousand times smaller — while $2va\,\Delta t$ shrinks only like $\Delta t$ . For any small window, the sign of the change in $v^2$ is the sign of the product $v\,a$ .

Step 4. Read the verdict. If $v$ and $a$ have the same sign, $va > 0$ : speed squared grows — speeding up. Opposite signs, $va < 0$ : speed squared shrinks — slowing down. If $a = 0$ or $v = 0$ momentarily, $v^2$ is momentarily not changing (the top of a throw is the turnaround between the two verdicts). Verified: Film 2 of the hero figure has $v = +6$ m/s, $a = -3\ \text{m/s}^2$ at $t = 2$ s, so $va = -18 < 0$ — slowing, as drawn; the falling ball has $v = -5$ , $a = -9.8$ , $va = +49 > 0$ — speeding up, as observed.

Limiting cases

case	result	meaning
$a = 0$ throughout	velocity constant	uniform motion — not necessarily rest
$v = 0$ at one instant, $a \neq 0$	turnaround point	top of a throw: stationary and accelerating
$a$ constant	equal $\Delta v$ in equal $\Delta t$	the suvat regime of the next module
$\Delta t \to 0$	$\bar{a} \to a$	average settles on the instantaneous value
$v$ and $a$ same sign	speed grows	"negative acceleration" can mean faster
$v$ reverses during the interval	$\lvert\Delta v\rvert$ can exceed both speeds	rebound: $15 \to 10$ backwards is a change of 25

The second row rewards a pause: at the top of a throw, velocity passes through zero but its rate of change never does. If acceleration were zero there too, the ball would stay put forever — zero velocity plus zero change of velocity is permanent rest. The ball comes down precisely because $a \neq 0$ at the top.

Common trapNegative acceleration means the object is slowing down

This is the most-tested lie in kinematics. The sign of $a$ states a direction (which way the velocity is being pushed), not a verdict about speed. A ball falling with down taken negative has $a = -9.8\ \text{m/s}^2$ and is speeding up every moment. A car moving in the negative direction with $a = +2\ \text{m/s}^2$ is slowing down despite the cheerful plus sign. Before writing "decelerates", run the test: same signs of $v$ and $a$ → speeding up; opposite → slowing down. And when a question says an object "decelerates at 5 m/s²", translate it as: the acceleration has magnitude 5 and points against the current velocity — then choose its sign from your own axis, not from the word.

Worked examples

Board level — braking train, both readings (CBSE/ICSE pattern)

A train moving at 72 km/h is brought uniformly to rest in 10 s. Find (a) its acceleration, (b) the velocity 4 s after the brakes are applied.

Step 1. Convert first, always: $72$ km/h $\times \tfrac{5}{18} = 20$ m/s. Take the direction of motion as positive: $v_0 = +20$ m/s, $v = 0$ , $\Delta t = 10$ s.

Step 2. Equation (1): $\bar{a} = (0 - 20)/10 = \boxed{-2\ \text{m/s}^2}$ . The minus sign is the answer, not decoration: the acceleration points opposite to the motion. In board language, the train has a retardation of $2\ \text{m/s}^2$ .

Step 3. Uniform acceleration means 2 m/s of velocity is removed every second. After 4 s, $v = 20 + (-2)(4) = \boxed{+12\ \text{m/s}}$ — still moving forward, still braking.

Step 4. Sanity: at this steady rate the train dies at $t = 10$ s exactly ( $20 - 2 \times 10 = 0$ ) ✓, matching the given data.

Competitive level — ball rebounding off a wall (NEET pattern)

A ball moving horizontally at 15 m/s strikes a wall and rebounds along the same line at 10 m/s. The contact lasts 0.05 s. Find the average acceleration during contact.

Step 1. Set the axis before touching numbers: let the initial motion (toward the wall) be positive. Then $v_0 = +15$ m/s and — the whole question — $v = -10$ m/s, because the rebound reverses direction.

Step 2. The signed subtraction: $\Delta v = (-10) - (+15) = -25$ m/s. Not 5. The velocity changed by twenty-five metres per second because reversal is a huge change.

Step 3. Equation (1): $\bar{a} = -25/0.05 = \boxed{-500\ \text{m/s}^2}$ , directed away from the wall (the negative direction). That is about $51g$ — collisions compress big velocity changes into tiny times, so enormous accelerations are normal, not a sign of error. Timing note: 45 seconds. The only decision is Step 1; options 100 m/s² (from $\Delta v = 5$ ) and 500 m/s² are both on the grid precisely to price that decision. If your $\lvert\Delta v\rvert$ in any rebound problem is smaller than both speeds, you forgot the sign flip.

Competitive level — reversal inside the interval (JEE-Main pattern)

A particle moves along the x-axis with velocity $+12$ m/s. Under constant acceleration, 5 s later its velocity is $-8$ m/s. Find (a) the acceleration, (b) the change in speed over the same interval, and (c) state when the particle was slowing down.

Step 1. $\Delta v = (-8) - (+12) = -20$ m/s over 5 s, so $a = -20/5 = \boxed{-4\ \text{m/s}^2}$ , constant and negative throughout.

Step 2. Change in speed is a different subtraction: final speed 8, initial speed 15 — no: initial speed 12, final speed 8, so the speed changed by $8 - 12 = \boxed{-4\ \text{m/s}}$ . Velocity changed by 20; speed changed by 4. Two different questions about the same motion.

Step 3. Locate the reversal: velocity hits zero when $12 - 4t = 0$ , i.e. $t = 3$ s. Before $t = 3$ s: $v > 0$ , $a < 0$ — opposite signs, slowing down. After: $v < 0$ , $a < 0$ — same signs, speeding up. The particle slows for the first 3 s only, then accelerates backwards. $\boxed{0 < t < 3\ \text{s}}$ Timing note: 60–75 seconds. Assertion–reason variants ask "the particle decelerates for the whole 5 s" — false, and the sign test dispatches it in one line. Any question whose velocity changes sign mid-interval is really asking whether you own that test.

Exam traps

Deceleration read off the sign of a. The word belongs to the relationship between $a$ and $v$ , not to the sign of $a$ . Falling objects with down-negative have $a < 0$ and speed up. Run the same-sign test every time; never answer "slowing" from the sign of $a$ alone.

The rebound subtraction. $\Delta v = v - v_0$ with signed values. Reversals make $\lvert\Delta v\rvert$ bigger than either speed (15 in, 10 out → 25). The wrong answer $\lvert\Delta v\rvert = 5$ is always among the options.

Change in speed ≠ change in velocity. Speed is the magnitude; its change can be tiny while the velocity's change is huge (the rebound again). Read which one the question wants — papers alternate deliberately.

km/h fed straight into formulas. Convert with $\tfrac{5}{18}$ before dividing by seconds: $72$ km/h $= 20$ m/s. An answer of $7.2\ \text{m/s}^2$ for the braking train is the tell-tale of skipped conversion.

"Retardation" (CBSE/ICSE wording). It means the magnitude of an acceleration that opposes the motion. "Retardation of 5 m/s²" for a body moving in $+x$ means $a = -5\ \text{m/s}^2$ — but if the body moves in $-x$ , it means $a = +5\ \text{m/s}^2$ . Translate the word into a signed $a$ on your own axis, then compute.

Zero confusions, both directions. $a = 0$ does not mean at rest (it means constant velocity); $v = 0$ does not mean $a = 0$ (top of a throw: $v = 0$ , $a = g$ downward). MCQs test both within the same paper.

Value of g. CBSE, JEE and NEET numericals normally intend $g = 10\ \text{m/s}^2$ unless the paper states 9.8; AP, IB, IGCSE and A-Level mark schemes use $9.8\ \text{m/s}^2$ . State the value you take at the top of vertical-motion answers — it is a free line and protects you when options differ by 2%.

Origins · 1 minIsaac Newton and how this was discovered

Isaac Newton

1643–1727 · Woolsthorpe, Cambridge, London

Newton is the reason acceleration — not position, not velocity — is the star quantity of mechanics. His second law, published in the Principia of 1687, declares that force does not set how fast a body moves but how fast its velocity changes: push twice as hard, change the velocity twice as quickly. That one legislative choice explains why nature lets a ball keep rising after it leaves your hand (no force needed to maintain velocity) and why the ball's velocity is edited by the same $9.8$ m/s each second whether it is going up, pausing at the top, or coming down. Every accelerometer, from an airbag trigger to the chip in your phone, is a small monument to his insight: to know what forces are doing, measure the second-by-second change in velocity — nothing else will do.

History

1638

— Galileo's Two New Sciences defines uniformly accelerated motion: equal additions of speed in equal times.

1687

— Newton's Principia makes acceleration the quantity that forces control, via F = ma.

1784

— Atwood's machine slows gravity's acceleration down enough to measure it in a lecture room.

1971

— Apollo 15's hammer and feather fall together on the Moon: one acceleration for all masses, on live TV.

2007

— MEMS accelerometer chips ship in the first iPhone, putting instantaneous acceleration sensing in every pocket.

Going further

Texts. NCERT Physics XI Part 1, Ch. 3 (Motion in a Straight Line, acceleration sections) — the CBSE/NEET canonical treatment · H.C. Verma, Concepts of Physics Vol. 1, Ch. 3 · Halliday, Resnick & Walker, Ch. 2 · Galileo, Two New Sciences, Third Day, for the original definition of uniform acceleration. The full vector treatment with derivatives is the UG deep dive Velocity & Acceleration (aq-mech tier).

Provenance. Standard kinematics per NCERT Ch. 3. Verifier recomputes: $27.8/4 = 6.9$ ; window on $v = 4t^2$ at $t=1$ : $12, 8.4, 8.04 \to 8$ vs $a = 8t$ ; $(v+a\Delta t)^2 - v^2 = 2va\Delta t + a^2\Delta t^2$ , checks $va = -18, +49$ ; train $72 \times \tfrac{5}{18} = 20$ , $-20/10 = -2$ , $20-8 = 12$ ; rebound $-25/0.05 = -500 \approx 51g$ ; reversal $-20/5 = -4$ , zero at $t = 3$ s; quiz 4: $54 \times \tfrac{5}{18} = 15$ , $-15/3 = -5$ .

Prerequisites. Speed & Velocity: Average and Instantaneous — the signed, vector velocity whose changes this module measures. Leads to. Velocity–Time Graphs — acceleration as slope, displacement as area · Kinematic Equations for Uniform Acceleration — the algebra of constant $a$ .

How can the ball at the top of its flight be accelerating if it isn't moving?

Because acceleration measures the change of velocity, not velocity itself. One frame before the top the ball moved up slowly; one frame after, it moves down slowly. Something edited the velocity across that instant — steadily, at 9.8 m/s per second — and that editor never paused. If acceleration really were zero at the top, the velocity would stop changing while equal to zero, and the ball would hover forever. The proof that $a \neq 0$ up there is simply that the ball comes back down.

Is deceleration the same thing as negative acceleration?

No — and the distinction is worth marks several times a year. "Deceleration" means the speed is decreasing, which happens when acceleration points against the current velocity, whatever the signs on your axis. A falling stone with down chosen negative has $a = -9.8\ \text{m/s}^2$ and is emphatically not decelerating. Exams also use "retardation" (CBSE/ICSE) for the magnitude of an opposing acceleration: translate it into a signed $a$ against your own axis before computing, and never diagnose slowing-down from the sign of $a$ alone — use the same-sign test.

Why is the change in velocity 25 m/s when the speeds are only 15 and 10?

Because velocity is signed and the direction flipped. On an axis pointing toward the wall, the ball's velocity went from $+15$ to $-10$ m/s; the subtraction $v - v_0$ crosses zero on the way, so the pieces add: 15 to bring it to rest, then 10 more to send it back. Reversal is the largest kind of velocity change available for given speeds. If you ever compute a rebound's $\lvert\Delta v\rvert$ and get something smaller than one of the speeds, you subtracted magnitudes instead of signed values.

Does constant speed mean zero acceleration?

Only for straight-line motion. In one dimension, constant speed means constant velocity, so $a = 0$ . But a car rounding a bend at a steady 60 km/h has a velocity whose direction changes every instant — and a changing velocity is an acceleration, full stop. That sideways acceleration is what the Motion in a Plane and circular-motion modules quantify. Within this 1D module, though, the equivalence holds: flat speedometer ⟺ zero acceleration.

What does "metres per second squared" actually mean?

Read it unsquared: metres per second, per second — how many m/s of velocity arrive (or leave) each second. An acceleration of $5\ \text{m/s}^2$ means the velocity gains 5 m/s every second: 0, 5, 10, 15… The "squared" is just the algebra of dividing m/s by s; nothing in the physics is being squared. Keeping this reading in mind also protects you in unit conversions: $\text{km/h per second}$ is a legitimate acceleration unit ( $1\ \text{km/h/s} = \tfrac{5}{18}\ \text{m/s}^2$ ), and questions occasionally use it.

Do I need calculus to handle instantaneous acceleration?

Not in this module, and not for boards. "The value the average acceleration settles on as the window shrinks" plus "slope of the tangent on a velocity–time graph" answers every conceptual question, and the shrinking-window algebra in the first derivation is a complete method for polynomial laws. JEE expects you to recognise $a = dv/dt$ as compact notation for that limit; the Calculus in Kinematics module makes the notation a working tool. Until then, uniform acceleration plus equation (1) covers the numericals.

Can average acceleration be zero while the object accelerated the whole time?

Yes. Average acceleration only compares endpoint velocities. A pendulum bob swinging left turning point to right and back returns to (momentarily) zero velocity, so over the full swing $\Delta v = 0$ and $\bar{a} = 0$ — despite non-stop accelerating in between. Averages blur; the instant tells the truth.

Which value of g should I use in numericals?

Follow your board's convention and say so. CBSE, JEE-Main/Advanced and NEET numericals are normally set with $g = 10\ \text{m/s}^2$ (papers state 9.8 when they want 9.8). AP Physics 1, IB, IGCSE and A-Level mark schemes use $9.8\ \text{m/s}^2$ . Writing "taking $g = 10$ " costs three seconds and earns method marks even if a slip follows.