Average velocity vs instantaneous velocity: Visual walkthrough

1.1.14 · D2 · Physics › Measurement, Vectors & Kinematics › Average velocity vs instantaneous velocity

Step 1 — Position–time graph hota kya hai

KYA. Slopes ki baat karne se pehle, hum woh cheez draw karte hain jiska slope chahiye. Hum horizontal axis par time $t$ rakhte hain aur vertical axis par position $x$ . Is picture par ek dot kehta hai: "is waqt par, object is jagah tha."

KYUN. Hum position ko time ke against plot karte hain kyunki velocity hai "position mein ek second mein kitna change." Position vs time ka graph us idea ko — change-per-second — ek aisi cheez mein badal deta hai jo tum literally steepness ke roop mein dekh sakte ho. Steep = position tezi se badal rahi hai = tezi se chal raha hai. Flat = position muskil se badal rahi hai = slow ya ruka hua hai.

PICTURE. Neeche, warm curve hai $x (t) = 3 t^{2} - 2 t$ . Ise left se right padhein ek story ki tarah: yeh thoda zero se neeche jaata hai shuruaat mein (particle pehle peeche drift karta hai), ek lowest point tak pahunchta hai, phir chadh jaata hai. Do moments mark kiye gaye hain: $t = 1$ (position $x = 1$ ) aur $t = 3$ (position $x = 21$ ).

Figure — Average velocity vs instantaneous velocity

Dekho Position-time graphs ke liye yeh generally kaise padhe jaate hain.

Step 2 — Do instants chuno aur chord draw karo

KYA. Do times chuno, $t_{1} = 1$ aur $t_{2} = 3$ . Do curve points mark karo $P = (1, 1)$ aur $Q = (3, 21)$ . Unke through ek straight line draw karo. Woh straight line secant line kehlaati hai (secant ka matlab sirf "ek line jo curve ke do points se guzre").

KYUN. Average velocity sirf do moments ki parwah karti hai — start aur end ki. Beech mein kya hua, woh kabhi nahi dekhti. Do points ke beech ek straight line perfect visual hai: woh wiggle ko bhool jaata hai aur sirf "kahan se shuru kiya, kahan khatam hua, kitna waqt laga" rakhta hai.

PICTURE. Orange chord $P$ aur $Q$ ko jodta hai. Notice karo yeh curve se seedha guzarta hai; yeh curve se chipakta nahi.

Step 3 — Us chord ka slope hi average velocity hai

KYA. Slope ka matlab hai "rise over run": kitna upar jaate ho ( $Δ x$ , position mein change) divided by kitna across jaate ho ( $Δ t$ , time mein change).

v_{a v g} = rise over run \frac{Δ x}{Δ t} = \frac{x ( t _{2} ) - x ( t _{1} ) upar: metres gained}{across: seconds passed t _{2} - t _{1}}

Term by term:

$Δ x = x (t_{2}) - x (t_{1})$ — $P$ aur $Q$ ke beech vertical gap; position kitni move hui.
$Δ t = t_{2} - t_{1}$ — horizontal gap; usme kitna waqt laga.
unka ratio — metres per second — exactly steepness hai, aur $x$ – $t$ line ki steepness hi velocity hai.

KYUN yahi ratio aur kuch nahi? Kyunki "velocity" ka matlab literally hai position-change per time. Rise ( $Δ x$ ) position change hai; run ( $Δ t$ ) time hai. Unhe divide karo aur units force karti hain ki answer $m/s$ ho. Koi doosra combination rate nahi deta.

PICTURE. Hum rise $Δ x = 20$ ko ek vertical plum bar ki tarah aur run $Δ t = 2$ ko ek horizontal teal bar ki tarah draw karte hain, ek right triangle banate hue jiska hypotenuse chord hai. Chord ka slope = (plum bar) / (teal bar).

v_{a v g} = \frac{21 - 1}{3 - 1} = \frac{20}{2} = 10 m/s

Step 4 — $Q$ ko $P$ ki taraf slide karo: chord tilt hona shuru ho jaata hai

KYA. $P$ ko $t_{1} = 1$ par fixed rakho. Ab doosra point $Q$ paas karo: try karo $t_{2} = 2.5$ , phir $2$ , phir $1.5$ . Har nayi $Q$ ek nayi, chhoti chord deti hai apne slope ke saath.

KYUN. Ek wide interval ki chord motion ke fast aur slow parts ko aapas mein mila deti hai — uska slope ek average hai jo journey ko chhupaata hai. Ek single instant $t_{1} = 1$ par velocity dhundhne ke liye, hume far-away endpoint ko answer pollute karna band karna hoga. Isliye hum $Q$ ko $P$ ki taraf kheeche hain aur dekhte hain ki slope kis par settle hota hai.

PICTURE. Usi $P$ se teen chords draw ki gayi hain, har ek ek paas wale $Q$ par khatam hoti hai. Jaise $Q$ slide karta hai, chords swing karti hain aur unke slopes change hote hain. Unke slopes ( $v_{a v g} = Δ x /Δ t$ se compute kiye) hain:

$t_{2}$	$Δ t$	$x (t_{2})$	$Δ x$	slope $= Δ x /Δ t$
$3$	$2$	$21$	$20$	$10$
$2.5$	$1.5$	$13.75$	$12.75$	$8.5$
$2$	$1$	$8$	$7$	$7$
$1.5$	$0.5$	$3.75$	$2.75$	$5.5$

Slopes ek target value ki taraf march kar rahe hain. Inhe dekho: $10, 8.5, 7, 5.5, \dots$

Step 5 — Limit: chord tangent ban jaati hai

KYA. Hum algebraically wahi karte hain jo aankhein already dekh rahi hain. Doosra time $t + Δ t$ ho, aur $Δ t$ ko $0$ ki taraf shrink hone do. Yahi derivative ki definition hai — secant slope ki limit.

v (t) = Δ t \to 0 lim \frac{x ( t + Δ t ) - x ( t ) shrinking rise}{shrinking run Δ t}

KYUN limit aur seedha " $Δ t = 0$ plug in" nahi? Agar hum $Δ t = 0$ seedha set kar dein toh $\frac{0}{0}$ milta hai — meaningless, kyunki rise aur run dono saath mein gayab ho jaate hain. Limit ek subtler sawal puchti hai: "ratio us waqt kis value ki taraf ja raha hai jab window band ho rahi hai?" Woh target exist karta hai even though endpoint $Δ t = 0$ khud forbidden hai.

$x (t) = 3 t^{2} - 2 t$ ke liye work out karo:

$x (t + Δ t) = 3 (t + Δ t)^{2} - 2 (t + Δ t) = 3 t^{2} + 6 t Δ t + 3 Δ t^{2} - 2 t - 2 Δ t$ .
$x (t) = 3 t^{2} - 2 t$ subtract karo: numerator $6 t Δ t + 3 Δ t^{2} - 2 Δ t$ tak collapse ho jaata hai.
$Δ t$ se divide karo (allowed kyunki limit se pehle $Δ t \neq = 0$ hai): $6 t + 3 Δ t - 2$ .
$Δ t \to 0$ hone do: $3 Δ t$ term mar jaata hai, bacha $v (t) = 6 t - 2$

$6 t - 2$ mein term by term: $6 t$ kehta hai velocity khud time ke saath badhti hai (particle speed karta rehta hai); $- 2$ shuruaat ke backward drift ka leftover hai.

Hamare point $t = 1$ par: $v (1) = 6 (1) - 2 = 4 m/s$ — exactly wahi number jis par shrinking slopes $10, 8.5, 7, 5.5, \dots$ close kar rahe the.

PICTURE. Limiting line ab curve ko do points par nahi kaatti — woh sirf $P$ par use chhuti hai. Woh chhuuta hua line tangent hai. Uska slope $v (1) = 4$ hai.

Step 6 — Degenerate case A: ek straight-line graph (constant velocity)

KYA. Maano position graph pehle se hi ek straight line hai, jaise $x (t) = 2 t$ . Tab har secant, chahe kitni bhi wide ya narrow ho, exactly usi line par lie karti hai.

KYUN matters karta hai. Yeh edge case hai jahan average aur instantaneous velocity har interval ke liye identical hain. Koi wiggle nahi hai jo smear ho, isliye window chhota karne se kuch nahi badalata. Secant slope $=$ tangent slope $= 2 m/s$ everywhere.

PICTURE. Ek clean line; uspar ek wide chord aur ek narrow chord drawn hain aur slope mein indistinguishable hain.

Recall Average aur instantaneous velocity

sab intervals ke liye kab equal hote hain? Exactly tab jab $x$ – $t$ graph ek straight line ho — yaani constant velocity. ::: Secant aur tangent har jagah coincide karte hain.

Step 7 — Degenerate case B: ek flat tangent (ek turning point)

KYA. Hamare curve $x = 3 t^{2} - 2 t$ ke bilkul bottom par graph momentarily neeche jaana band kar deta hai aur upar jaana shuru kar deta hai. Set karo $v (t) = 6 t - 2 = 0 \Rightarrow t = \frac{1}{3} s$ . Us instant par tangent horizontal hai.

KYUN matters karta hai. Horizontal tangent ka slope $0$ hota hai, isliye instantaneous velocity zero hai even though particle sirf ruk raha hai, permanently stopped nahi hai. Yahi parent ki mistakes list ka "turning point" hai — throw ke upar ball. Particle pehle peeche ja raha tha, baad mein aage, aur ek instant ke liye uski velocity genuinely $0$ hai.

PICTURE. Curve ke lowest point par ek flat teal tangent line rest kar rahi hai; wahan velocity $= 0$ hai.

Ek-picture summary

Sab ek saath: fixed point $P$ , secants ki ek family jo tangent par collapse ho rahi hai, average ke liye chhota rise/run triangle, aur turning point mark karta flat tangent. Ise left se right padhein secant → shrink → tangent.

secant slope = v_{a v g} \frac{Δ x}{Δ t} slide Q into P Δ t \to 0 tangent slope = v (t) \frac{d x}{d t}

Recall Feynman: poora walkthrough seedhe shabdon mein

Draw karo tum kahan ho clock ke against — woh curve hai. Do moments chuno aur unhe ek ruler se jodo: us ruler ki tilt hai teri average velocity, rise (metres gained) over run (seconds passed). Lekin woh ruler cheating karta hai — woh sirf start aur end dekhta hai aur beech wala ignore karta hai. Toh pehla pinned dot still rakho aur doosra dot uski taraf slide karo. Ruler swing karta hai aur uski tilt ek number par settle ho jaati hai. Puri tarah slide karo aur ruler curve ko kaatna band kar deta hai aur sirf chhuta hai — woh chhuta hua line tangent hai, aur uski tilt hai teri speed us exact tick par: instantaneous velocity, $\frac{d x}{d t}$ . Do special cases tumhe honest rakhte hain: agar curve pehle se hi ek straight line thi, ruler ko swing karna hi nahi pada — average har jagah instant ke barabar hai. Aur agar chhuta hua line bilkul flat aata hai, teri speed ek instant ke liye zero hai — woh throw ka top hai, jahan tum mudh jaate ho.

Connections

Average velocity vs instantaneous velocity (parent)
Position-time graphs
Slope and tangent in calculus
Position and displacement vectors
Distance vs displacement
Average speed vs instantaneous speed
Acceleration as derivative of velocity
Equations of motion under uniform acceleration