Average acceleration vs instantaneous acceleration: Worked examples

1.1.15 · D3 · Physics › Measurement, Vectors & Kinematics › Average acceleration vs instantaneous acceleration

Kuch bhi shuru karne se pehle, main woh do tools dobara earn karta hoon jo hum har jagah use karte hain, plain words mein:

The scenario matrix

Har acceleration problem in cells mein se ek (ya blend) hoti hai. Right-hand column us example ka naam deta hai jo use cover karta hai.

#	Case class	Tricky kya hai	Covered by
A	Positive average, 1-D, speeding up	kuch nahi — yeh friendly base case hai	Ex 1
B	Negative average (slow ho raha hai, phir bhi forward move kar raha hai)	sign of $a$ opposite to motion	Ex 2
C	Direction reversal (velocity sign flip karti hai, speed unchanged)	speed constant yet $Δ v \neq = 0$	Ex 3
D	Instantaneous ≠ average (varying $a$ )	derivative vs secant slope differ karte hain	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)	speeds nahi, vectors subtract karne honge	Ex 6
G	Constant speed, curved path (circular)	acceleration sideways point karta hai	Ex 7
H	Real-world word problem (units, hidden signs)	English → symbols translate karo	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, phir bhi forward move kar raha hai

Worked example Ex 2 — ek cyclist brake lagata hai

Ek cyclist $v_{i} = 15 m/s$ se $v_{f} = 3 m/s$ tak slow hota hai $Δ t = 4 s$ mein, phir bhi forward roll karta hua. Forward ko $+$ lo.

Forecast: Phir bhi forward move kar raha hai, lekin slow ho raha hai — $a_{a v g}$ positive hoga ya negative? (Hint: velocity mein change kis taraf point karta hai?)

$a_{a v g} = \frac{v _{f} - v _{i}}{Δ t} = \frac{3 - 15}{4}$ . Yeh step kyun? $v_{f} < v_{i}$ hai, isliye numerator negative hai chahe bike abhi bhi forward move kar rahi ho.
$a_{a v g} = \frac{- 12}{4} = - 3 m/s^{2}$ .

Verify: Negative sign ka matlab hai velocity mein change backward point karta hai (ek braking push), jabki velocity khud abhi bhi forward point karti hai. Sanity check: $15 + (- 3) (4) = 3 = v_{f}$ ✓. Lesson: $a$ ka sign velocity change ki direction batata hai, motion ki direction nahi.

Cell C — direction reversal, speed unchanged

Worked example Ex 3 — ek ball seedha wapas bounce karti hai

Ek ball wall se $20 m/s$ east pe takraati hai aur $20 m/s$ west pe rebound karti hai. Contact $0.05 s$ tak rehta hai. East ko $+$ lo.

Forecast: Speed $20$ pehle bhi hai aur baad mein bhi — kya acceleration zero hai? (Trap dekho.)

Velocities ko sign ke saath likho. $v_{i} = + 20$ , $v_{f} = - 20$ . Yeh step kyun? West opposite direction hai, isliye woh opposite sign carry karta hai. Speed sign ignore karta hai; velocity nahi karta.
$a_{a v g} = \frac{- 20 - ( + 20 )}{0.05} = \frac{- 40}{0.05}$ . Yeh step kyun? Subtraction $v_{f} - v_{i}$ mein hi reversal dikhta hai — do 20 magnitude mein add ho jaate hain.
$a_{a v g} = - 800 m/s^{2}$ .

Verify: Bahut bada aur negative (west ki taraf point karta hai, wall ke push ki taraf). Chahe speed kabhi nahi badli, $Δ v = - 40 m/s$ bada hai, isliye acceleration bada hai. Yeh parent note ki steel-manned mistake ko concrete banata hai: constant speed $\neq =$ zero acceleration.

Cell D — instantaneous vs average differ karte hain

Worked example Ex 4 — rising acceleration wala rocket

Ek rocket ki velocity $v (t) = 2 t^{2} + 3 t$ (m/s) hai. Dhundho (a) $t = 3 s$ pe instantaneous acceleration, aur (b) $[0, 3] s$ par average acceleration.

Forecast: Kyunki acceleration time ke saath increase ho raha hai, kya $[0, 3]$ ka average end $t = 3$ ke instantaneous value se bada hoga ya chhota?

Instantaneous $a$ ke liye differentiate karo. $a (t) = \frac{d v}{d t} = 4 t + 3$ . Yeh step kyun? Instantaneous acceleration tangent slope = derivative hai. Term-by-term: $\frac{d}{d t} 2 t^{2} = 4 t$ , $\frac{d}{d t} 3 t = 3$ .
Instant pe evaluate karo. $a (3) = 4 (3) + 3 = 15 m/s^{2}$ .
Average ke liye, endpoint velocities nikalo. $v (0) = 0$ , $v (3) = 2 (9) + 3 (3) = 27 m/s$ . Yeh step kyun? Average sirf endpoints (secant slope) use karta hai, isliye humein do $v$ values chahiye, derivative nahi.
$a_{a v g} = \frac{27 - 0}{3 - 0} = 9 m/s^{2}$ .

Verify: $a_{a v g} = 9 < a (3) = 15$ , forecast se match karta hai — kyunki acceleration climb karta hai, end poore interval ke average se steeper hai. Neeche tangent (red) vs secant (yellow) dekho.

Cell E — zero aur degenerate inputs

Worked example Ex 5 — edge cases

Do alag situations consider karo. (a) Ek car bilkul steady $20 m/s$ pe $10 s$ ke liye cruise karti hai. (b) Tum $Δ t = 0$ par $a_{a v g}$ compute karne ki koshish karte ho (same instant, do baar).

Forecast: In mein se kaunsa zero acceleration deta hai, aur kaunsa kuch undefined deta hai? Yeh same nahi hain.

Case (a): $v_{f} = v_{i} = 20$ , isliye $a_{a v g} = \frac{20 - 20}{10} = \frac{0}{10} = 0 m/s^{2}$ . Yeh step kyun? Numerator exactly zero hai, denominator ek real positive time hai → ek genuine, well-defined zero.
Case (b): $a_{a v g} = \frac{v ( t ) - v ( t )}{0} = \frac{0}{0}$ . Yeh step kyun? Jab $Δ t = 0$ hota hai toh formula zero se divide karta hai — yeh undefined hai, zero nahi. "Koi time nahi" par "change" measure nahi kar sakte.
Resolution: kisi instant ki baat karne ke liye, tum $Δ t = 0$ set nahi karte; tum limit lete ho jab $Δ t \to 0$ (parent ka Step 2 dekho). $\frac{0}{0}$ form exactly yahi reason hai ki derivative ek limit hai, plain division nahi.

Verify: (a) $0$ deta hai (defined). (b) undefined hai jab tak limit rescue nahi karta. Do alag "nothings" — inhe kabhi confuse mat karo.

Cell F — 2-D vector change (corner turn karna)

Worked example Ex 6 — ek car right-angle bend leti hai

Ek car $10 m/s$ east move karti hai, phir $2 s$ baad $10 m/s$ north move karti hai (same speed). $a_{a v g}$ dhundho: uska size aur direction.

Forecast: Speed pehle aur baad mein same hai — lekin yeh $90°$ turn hai. Kya tum zero acceleration expect karte ho ya nahi? $a_{a v g}$ roughly kis compass direction mein hona chahiye?

Har velocity ko components $(east, north)$ mein likho. $v_{i} = (10, 0)$ , $v_{f} = (0, 10)$ . Yeh step kyun? Vectors ko component-by-component subtract karna hota hai; speeds subtract nahi kar sakte. Dekho Vectors: addition and subtraction.
Subtract karo. $Δ v = v_{f} - v_{i} = (0 - 10, 10 - 0) = (- 10, 10) m/s$ . Yeh step kyun? Yeh $Δ v$ north-west point karta hai — turn ke andar ki taraf, kisi bhi velocity ke along nahi.
Time se divide karo. $a_{a v g} = \frac{Δ v}{Δ t} = \frac{( - 10 , 10 )}{2} = (- 5, 5) m/s^{2}$ .
Magnitude. $∣ a_{a v g} ∣ = (- 5)^{2} + 5^{2} = 50 \approx 7.07 m/s^{2}$ . Yeh step kyun? Vector ka size uske components par Pythagorean theorem se aata hai.

Verify: Direction $(- 5, 5)$ west se $45°$ north hai — bend ke andar point karta hai, jaise forecast tha. Speed unchanged phir bhi $∣ a_{a v g} ∣ \approx 7.07 \neq = 0$ . Neeche figure vector triangle dikhata hai.

Cell G — constant speed, curved path (circular)

Worked example Ex 7 — ek string par stone

Ek stone $r = 2 m$ radius ke circle mein constant speed $v = 4 m/s$ pe ghoomta hai. Uska instantaneous acceleration dhundho (magnitude aur direction).

Forecast: Speed constant hai, toh kya acceleration zero hai? Agar nahi, toh yeh kahan point karta hai — motion ke along, ya kahi aur?

Yaad karo ki Uniform circular motion mein speed constant hai lekin velocity vector turning karta rehta hai, isliye $\frac{d v}{d t} \neq = 0$ . Yeh step kyun? Acceleration velocity vector ke change pe react karta hai, aur yahan direction change hoti hai chahe magnitude nahi badlti.
Centripetal (centre-pointing) acceleration ka magnitude $a = \frac{v ^{2}}{r}$ hota hai. Yeh step kyun? Turning velocity vector ki geometry (ek chhota $Δ v$ centre ki taraf point karta hai) yeh size deti hai — Uniform circular motion note mein puri derivation hai.
Plug in karo. $a = \frac{4 ^{2}}{2} = \frac{16}{2} = 8 m/s^{2}$ , centre ki taraf directed.

Verify: Constant speed ke baavajood nonzero acceleration ✓. Yeh inward point karta hai (velocity ke perpendicular), isliye direction change karti hai, speed nahi. Figure velocity (tangent) vs acceleration (inward) dikhata hai.

Cell H — real-world word problem

Worked example Ex 8 — runway par ek airplane

"Ek jet $70 m/s$ pe touch down karta hai aur reverse thrust use karke $14 s$ mein rukna chahiye. Maano deceleration roughly constant hai, toh runway kya average acceleration record karega, aur kya yeh comfortable $- 6 m/s^{2}$ limit ke andar hai?"

Forecast: "Stop" ka matlab final velocity kya number hai? Kya answer negative hoga? Size mein $6$ se bada hoga ya chhota?

English ko symbols mein translate karo. $v_{i} = 70 m/s$ , $v_{f} = 0$ (ruk gaya), $Δ t = 14 s$ , forward $= +$ . Yeh step kyun? Hidden data hai " $v_{f} = 0$ " — "stop" ek speed word nahi hai jo blindly plug in karo; iska matlab hai velocity zero ho jaati hai.
$a_{a v g} = \frac{0 - 70}{14} = \frac{- 70}{14} = - 5 m/s^{2}$ . Yeh step kyun? Forward move karte hue slow hona → negative sign, jaise Cell B mein.
Limit se compare karo. $∣ - 5∣ = 5 < 6$ .

Verify: $- 5 m/s^{2}$ , $- 6 m/s^{2}$ comfort limit ke andar hai ✓. Units check: $\frac{m/s}{s} = m/s^{2}$ ✓. Negative sign braking hai, exactly wahi jo reverse thrust karta hai.

Cell I — exam twist (dhundho kahan $a = 0$ hai)

Worked example Ex 9 — "kis time pe acceleration zero hai?"

Ek particle ki position $x (t) = t^{3} - 9 t^{2} + 15 t$ (m) hai. Kis time(s) pe instantaneous acceleration zero hai, aur us instant se pehle yeh positive hai ya negative?

Forecast: Acceleration position ka second derivative hai. Cubic $x$ → $a (t)$ ki shape kya hogi, aur kitne zeros hone chahiye?

Pehla derivative → velocity. $v (t) = \frac{d x}{d t} = 3 t^{2} - 18 t + 15$ . Yeh step kyun? Position ka ek derivative velocity deta hai (Derivatives as rates of change).
Doosra derivative → acceleration. $a (t) = \frac{d v}{d t} = 6 t - 18$ . Yeh step kyun? Instantaneous acceleration $\frac{d ^{2} x}{d t ^{2}}$ hai — velocity ko dobara differentiate karo. Cubic $x$ → linear $a$ → exactly ek zero, forecast se match.
$a = 0$ set karo. $6 t - 18 = 0 \Rightarrow t = 3 s$ .
$t = 3$ se pehle sign: $t = 0$ pick karo, $a (0) = 6 (0) - 18 = - 18 < 0$ . Yeh step kyun? Ek side par ek point test karne se pata chalta hai ki acceleration switch se pehle negative hai, baad mein positive.

Verify: $a (3) = 6 (3) - 18 = 0$ ✓. $t < 3$ ke liye acceleration negative hai (velocity decrease ho rahi hai), $t > 3$ ke liye positive (velocity increase ho rahi hai) — isliye $t = 3$ woh jagah hai jahan velocity bottom out karti hai.

Recall

Recall Answers cover karo aur khud ko test karo

Cell B (Ex 2) — cyclist $15 \to 3$ slow hota hai $4$ s mein, $a_{a v g}$ ? ::: $- 3 m/s^{2}$ Cell C (Ex 3) — ball $+ 20 \to - 20$ in $0.05$ s, $a_{a v g}$ ? ::: $- 800 m/s^{2}$ Cell D (Ex 4) — $v = 2 t^{2} + 3 t$ ; $a (3)$ aur $a_{a v g}$ over $[0, 3]$ ? ::: $15$ aur $9 m/s^{2}$ Cell E — $Δ t = 0$ par $a_{a v g}$ undefined kyun hai, zero nahi? ::: yeh $0/0$ hai, division by zero; sirf limit use se rescue hota hai Cell F (Ex 6) — $10$ m/s pe $90°$ turn $2$ s mein, $∣ a_{a v g} ∣$ ? ::: $50 \approx 7.07 m/s^{2}$ Cell G (Ex 7) — stone, $v = 4$ , $r = 2$ , centripetal $a$ ? ::: $8 m/s^{2}$ centre ki taraf Cell I (Ex 9) — $x = t^{3} - 9 t^{2} + 15 t$ , $a = 0$ kab? ::: $t = 3 s$

Connections

Average acceleration vs instantaneous acceleration — woh parent jise yeh page drill karta hai.
Average velocity vs instantaneous velocity — same secant/tangent logic ek rung neeche.
Velocity-time graphs — Ex 4 ka tangent-vs-secant figure yahan rehta hai.
Derivatives as rates of change — Ex 4 aur Ex 9 mein use ki gayi differentiation.
Equations of motion (constant acceleration) — kyun "un-averaging" Ex 1 aur Ex 8 ko verify karta hai.
Uniform circular motion — Ex 7 ke $v^{2} / r$ ka source.
Vectors: addition and subtraction — Ex 6 mein component subtraction.