Average acceleration vs instantaneous acceleration: Question bank

1.1.15 · D5 · Physics › Measurement, Vectors & Kinematics › Average acceleration vs instantaneous acceleration

Shuru karne se pehle, is poori page ka vocabulary yahan plain words mein hai:

Definition Woh words jo hum baar baar use karenge

Position $r$ — ek fixed origin se object ki abhi ki jagah tak ka arrow. Uska tip time ke saath path trace karta hai.
Velocity $v$ — kitni tezi aur kis direction mein move kar rahe ho. Ye ek arrow hai: length = speed, arrowhead = direction. Ye batata hai $r$ kitni tezi se change hota hai: $v = d r / d t$ .
Speed — bas us arrow ki length, koi direction nahi. Ek single positive number.
Subscripts $i$ aur $f$ — ye sirf ek time interval ke do endpoints tag karte hain: $i$ = initial (shuruat), $f$ = final (ant). Toh $v_{i}$ shuruat ki velocity hai, $v_{f}$ ant ki velocity, aur $t_{i}$ , $t_{f}$ do clock readings hain.
Symbol $Δ$ ("delta") — matlab "change in": $Δ v = v_{f} - v_{i}$ (final minus initial) aur $Δ t = t_{f} - t_{i}$ . Ye hamesha end value minus start value hota hai, ulta nahi.
Function notation $v (t)$ — padho " $v$ of $t$ ": ye velocity hai us particular time par jo tum daalo. Toh $v (3)$ matlab "velocity at $t = 3 s$ ", aur $a (t)$ woh acceleration hai jo kisi bhi instant $t$ par evaluate ki ja sake.
Average acceleration $a_{a v g} = \frac{Δ v}{Δ t} = \frac{v _{f} - v _{i}}{t _{f} - t _{i}}$ — velocity change ko time gap par evenly baant do. Sirf do endpoint arrows $v_{i}, v_{f}$ par depend karta hai.
Instantaneous acceleration $a = \frac{d v}{d t}$ — ek frozen instant par change rate, gap ko zero tak shrink karke milti hai.
Secant — curve par do points ko join karne wali straight chord.
Tangent — ek straight line jo curve ko ek point par graze kare; uska slope wahan ka derivative hai.
$g$ — Earth ki surface ke paas free fall ki acceleration, magnitude $\approx 9.8 m/s^{2}$ , hamesha downward (zameen ki taraf). Hum poori page mein "up" ko positive lete hain.

Is page ka notation promise: ek arrow ( $a$ , $v$ ) matlab poora vector (magnitude aur direction). Jab hum arrow hata dete hain ( $a$ , $v$ ) toh matlab object ek seedhi line mein chal raha hai aur humne direction ke liye $+$ / $-$ sign chun liya hai, toh ek signed number sab kuch carry karta hai — ek scalar component, bare speed nahi. Arrows dhyaan se dekho: ye "kitna" aur "kitna, kis direction mein" ke beech ka fark hai.

Neeche har trap asal mein yahi hai: "kya tumne direction arrow mein rakha, aur kya tumne average aur instant alag rakhe?"

Do pictures poori page carry karti hain. Har item ke liye inhe dhyaan mein rakho.

Figure — Average acceleration vs instantaneous acceleration

Ek hi v–t curve par do lines dekho: coral chord (secant) $a_{a v g}$ hai; ek point ko graze karne wali lavender line (tangent) instantaneous $a$ hai. Ye tabhi milte hain jab curve seedha ho.

Ek frame mein do aur traps: left mein, circular motion mein ek object — velocity arrow ki length same rehti hai lekin turn karta hai, toh $Δ v$ (mint arrow) nonzero hai chahe speed constant ho. Right mein, $a$ vs $v$ ke chaar sign combinations (straight-line motion) jo decide karte hain "speeding up" ya "slowing down".

True or false — justify karo

Constant speed ka matlab hamesha zero acceleration hota hai

False. Speed sirf arrow ki length hai; agar arrow turn kare (curve), toh

v

change hota hai chahe length same ho, toh

a = d v / d t \neq = 0

. Dekho Uniform circular motion aur figure s02 ka left panel.

Agar kisi body ki velocity ek instant par zero ho, toh uski acceleration bhi zero honi chahiye

False. Throw ki gayi ball ke sabse upar wale point par

v = 0

lekin

a = g

downward (

\approx 9.8 m/s^{2}

, up positive liya hai) — velocity sirf zero se guzar rahi hai, wahan ruki nahi, toh abhi bhi change ho rahi hai.

Average aur instantaneous acceleration tabhi equal hote hain jab acceleration constant ho

True. Agar

a

interval mein constant hai, toh constant ka average wahi constant hai; koi bhi variation hogi toh endpoint-based average har moment ki value miss kar dega.

Ek interval par

a_{a v g}

, shuruat aur ant mein

a

ke arithmetic mean ke barabar hota hai

Saamaanya mein False. Ye mean tabhi equal hoga jab $a$ time mein linearly vary kare.

a (t)

ke curved hone par, sahi

a_{a v g} = Δ v /Δ t

\frac{1}{2} (a_{i} + a_{f})

se alag ho sakta hai.

Agar ek instant par instantaneous acceleration zero ho, toh velocity momentarily change nahi ho rahi

True.

a = d v / d t = 0

matlab velocity graph wahan flat hai —

v (t)

ka turning point ya straight-flat stretch.

Negative acceleration ka matlab hamesha object slow down ho raha hai

False. Matlab slow down tabhi hai jab velocity positive ho. Agar velocity bhi negative hai, toh negative acceleration use speed up karti hai (dono same way point karte hain).

a

ka sign vs

v

ka sign decide karta hai — dekho figure s02, right panel.

Average acceleration zero ho sakta hai chahe object poore time accelerating raha ho

True. Agar woh apni starting velocity par wapas aa jaye (

v_{f} = v_{i}

), toh

Δ v = 0

aur

a_{a v g} = 0

, chahe beech mein speed up aur slow down hua ho.

Instantaneous acceleration velocity–time graph par tangent ki slope hai

True. Ye definition

a = d v / d t

ko geometrically padhna hai — dekho figure s01, Velocity-time graphs aur Derivatives as rates of change.

Average acceleration v–t curve par tangent ki slope hai

False. Ye do endpoint times ko join karne wale secant (chord) ki slope hai; tangent instantaneous value deta hai (figure s01).

Agar v–t graph ek straight line hai, toh kisi bhi sub-interval ka average har jagah instantaneous ke barabar hai

True. Seedha v–t line matlab constant slope = constant acceleration, toh har secant slope har tangent slope ke barabar hai.

Error dhundho

"Speed ek bend mein 30 m/s constant hai, toh

a = 0

." Galti batao

Galti hai speed ko velocity maanna. Direction change hoti hai, toh

v

change hota hai;

a

bend ke centre ki taraf point karta hai aur nonzero hai.

a_{a v g} = \frac{v}{t} = \frac{25}{4}

ek bus ke liye jo 5→25 m/s in 4 s." Galti batao

Usne final velocity use ki jaise bus rest se start hui ho. Sahi hai

Δ v /Δ t = (25 - 5) /4 = 5 m/s^{2}

25/4

nahi.

"t = 3 s par instantaneous acceleration bas

v (3) /3

hai." Galti batao

Ye ek average from rest hai, instant nahi. Instantaneous ke liye derivative/tangent slope

d v / d t

chahiye jo

t = 3

par evaluate ho, na ki current values ka ratio.

"Car east mein 20 m/s, phir west mein 20 m/s 5 s mein; speed same rahi toh

a_{a v g} = 0

." Galti batao

Velocity reverse ho gayi:

Δ v = (- 20) - (+ 20) = - 40

m/s. Diye gaye

5

s mein,

a_{a v g} = - 40/5 = - 8 m/s^{2}

. Direction flip ek real velocity change hai. Dekho Vectors: addition and subtraction.

a_{a v g}

= har second ki accelerations ka average." Galti batao

a_{a v g}

velocities se bana hai (

Δ v /Δ t

), accelerations se nahi.

a

-values ka average tabhi match karta hai jab

a

t

mein linear ho.

v (t) = 3 t^{2}

, toh acceleration

3 t^{2} / t = 3 t

hai." Galti batao

Usne differentiate karne ki jagah velocity ko time se divide kiya (average-from-rest trick). Sahi instantaneous

a = d v / d t = 6 t

hai — galat answer se double. Sirf ratio compute hua, rate nahi.

"Velocity graph ek parabola hai, toh ek interval ka average acceleration us interval ke midpoint par tangent slope ke barabar hai." Galti batao

Quadratic

v (t)

ke liye ye actually sach hai, lekin koi mystical reason nahi —

v = α t^{2} + β t + γ

ke liye slope

v^{'} = 2 α t + β

linearly

t

mein hai, aur ek linear function ka

[t_{i}, t_{f}]

par average midpoint

\frac{t _{i} + t _{f}}{2}

par uski value ke barabar hota hai (midpoint symmetry / Mean Value Theorem centre par exactly land karta hai). Error ye hai ki ise general rule bata diya: cubic ya higher

v (t)

ke liye slope linear nahi rahta aur midpoint kaam nahi karta.

Why questions

Acceleration ek vector kyun hai, sirf number kyun nahi?

Kyunki ye

d v / d t

hai aur

v

direction carry karta hai; sirf direction mein change (constant speed, turning arrow) ek genuine acceleration hai, aur ek plain number ye record nahi kar sakta ki velocity rotate hui — sirf ek vector kar sakta hai.

Limit

Δ t \to 0

"window par average" ko "instant par value" mein kyun convert karta hai?

v–t graph par,

Δ v /Δ t

do points ke beech secant ki slope hai. Doosre point ko pehle ki taraf slide karne par woh secant pivot karta hai aur tangent ban jaata hai — shrinking window par average physically rotate hokar instantaneous slope ban jaata hai.

Hum instantaneous acceleration ko directly average jaisi measure kyun nahi kar sakte?

Kisi bhi real measurement ke liye do velocity readings aur ek clock chahiye, jo hamesha ek gap hai — ek average. Instant chhote se chhote gaps ke mathematical limit hai, toh hum increasingly fine averages measure karke approach karte hain, ek single reading se kabhi nahi.

a_{a v g}

aur

a

constant acceleration mein agree kyun karte hain, warna nahi?

Constant

a

v–t graph ko ek straight line banata hai, jiski slope har jagah same hai — toh endpoint secant aur har tangent ek slope share karte hain.

v (t)

mein koi bhi curvature matlab tangent slope point-to-point vary karta hai, aur single secant slope sirf us variation ke average se match kar sakta hai, har instant se nahi.

Negative acceleration ka matlab automatically "braking" kyun nahi hai?

"Braking" matlab speed decrease, jiske liye

a

v

ke opposite point karna hoga. Negative

a

sirf tabhi motion ke opposite hai jab

v

positive ho; agar

v

bhi negative hai toh dono align hote hain aur object negative direction mein speed up karta hai (figure s02, right).

a = d^{2} r / d t^{2}

position ka "second derivative" kyun hai?

Velocity position arrow ki pehli rate of change hai (

v = d r / d t

), aur acceleration us ki rate of change hai (

d v / d t

). Do rate-of-change operations stack karne ka matlab position ko do baar differentiate karna hai — likha jaata hai

d^{2} r / d t^{2}

Average acceleration beech ki violent moment kyun hide kar sakta hai?

a_{a v g} = Δ v /Δ t

sirf do endpoint velocities padhta hai, toh ye net velocity change ko poore time mein flat spread karke report karta hai. Geometrically, ek acceleration–time graph par sahi velocity change curve ke neeche ka area hai; ek tall spike phir dip utna hi total area enclose kar sakta hai jitna ek gentle constant band, toh ek huge instantaneous

a

ek mild average mein wash out ho sakta hai.

Edge cases

Seedha upar throw kiya hua object: sabse upar wale point par uski acceleration kya hai?

Phir bhi

g

downward (

\approx 9.8 m/s^{2}

, up positive liya hai). Velocity momentarily zero hai lekin sign change kar rahi hai, toh acceleration unbroken hai.

Ek ball bounce karti hai: velocity near-zero contact time mein down se up flip hoti hai. Contact mein

a_{a v g}

ka kya hota hai?

Bahut bada upward

a_{a v g} = Δ v /Δ t

Δ v

bada hai (sign flip) aur

Δ t

tiny hai, toh ratio spike karta hai — impulsive force.

Constant speed par uniform circular motion: ek poore loop mein

a_{a v g}

zero hai?

Haan. Ek poore loop ke baad

v_{f} = v_{i}

, toh

Δ v = 0

aur

a_{a v g} = 0

— chahe instantaneous

a

(centripetal) har instant nonzero tha.

Uniform circular motion: kya aadhe loop mein

a_{a v g}

zero hai?

Nahi. Aadhe loop ke baad velocity ne direction reverse kar li hai, toh

Δ v \neq = 0

;

a_{a v g}

circle ke across point karta hai aur nonzero hai.

Average formula mein

Δ t = 0

: kya hota hai?

Ye undefined hai — zero se division. Instantaneous value "zero plug in" karna nahi hai; ye limit hai jab

Δ t

shrink hota hai, jo finite rehta hai.

Ek particle hamesha perfectly still baitha hai: average aur instantaneous acceleration?

Dono zero.

v

constant hai (zero par), toh

Δ v = 0

aur

d v / d t = 0

har instant par.

Ek straight line mein constant nonzero velocity:

a_{a v g}

aur

a

kya hain?

Dono zero. Velocity magnitude ya direction mein change nahi hoti, toh koi bhi measure kuch detect nahi kar sakta.

Velocity graph mein ek sharp corner (kink) hai — wahan instantaneous acceleration kya hai?

Corner par undefined: left se tangent slope right se alag hai, toh

d v / d t

ka us instant par koi single value nahi (sirf ek physical idealisation).

Recall One-line survival summary

Har velocity arrow mein direction rakho, average (secant, endpoints) ko instant (tangent, limit) se alag rakho, aur "speeding up" ya "slowing down" bolne se pehle $a$ ka sign $v$ ke sign se check karo.

Connections

Average velocity vs instantaneous velocity — same secant-vs-tangent traps ek rung neeche.
Velocity-time graphs — jahan secant aur tangent slopes rehte hain.
Derivatives as rates of change — limit ek instant kyun ban jaata hai.
Equations of motion (constant acceleration) — woh ek case jahan do accelerations milte hain.
Uniform circular motion — flagship "constant speed, nonzero acceleration" trap.
Vectors: addition and subtraction — $Δ v$ kaise compute karein jab directions flip hon.