Average velocity vs instantaneous velocity: Worked examples

1.1.14 · D3 · Physics › Measurement, Vectors & Kinematics › Average velocity vs instantaneous velocity

Kuch bhi compute karne se pehle, ek reminder plain words mein. Position $r (t)$ = "cheez $t$ time par kahan hai." One dimension mein hum sirf $x (t)$ likhte hain, ek signed number: positive matlab ek taraf (jaise, right/east), negative matlab doosri taraf. Displacement $Δ x = x (end) - x (start)$ start se finish tak ka net arrow hai. Do operations se hume velocity milti hai:

Speed ek alag cheez hai, aur kyunki Cases C, D, H, I usi par depend karte hain, hum ise abhi pin down karte hain:

The scenario matrix

Har velocity problem in case classes mein se ek hai (ya inka blend). Table mein yeh listed hain; har row neeche ek example se covered hai.

#	Case class	Tricky kya banata hai	Covered by
A	Ek direction mein motion, positive $x$	baseline — intuition build karo	Ex 1
B	Ek direction mein motion, negative velocity	answer ka sign	Ex 2
C	Reversal (turn around karna)	$∥ v_{a v g} ∥ \neq =$ avg speed; $v = 0$ at turn	Ex 3
D	Zero displacement (round trip)	$v_{a v g} = 0$ but avg speed $\neq = 0$	Ex 4
E	Degenerate: interval ek point tak shrink ho	limit ki zaroorat kyun hai	Ex 5
F	Instant $v$ from the definition (no shortcut)	derivative prove karna	Ex 6
G	Midpoint theorem for uniform acceleration	$v_{a v g} = v (t_{mi d})$	Ex 7
H	Real-world word problem (multi-leg trip)	words ko $Δ x, Δ t$ mein translate karna	Ex 8
I	Exam twist: given $v_{a v g}$ , find unknown time	reverse-solving	Ex 9
J	Turning-point case, graph-read	flat tangent, sign flip of $v$	Ex 10

Hum poore mein Slope and tangent in calculus, Position-time graphs, Distance vs displacement aur Equations of motion under uniform acceleration par depend karte hain.

Example 1 — Case A: steady motion, positive direction

Figure — position line, uska secant $[0, 5]$ par (dashed coral), aur note ki yahan secant = tangent, slope $4$ m/s:

Figure — Average velocity vs instantaneous velocity

Example 2 — Case B: negative velocity

Example 3 — Case C: motion that reverses

Worked example Statement

$x (t) = t^{2} - 4 t$ (metres) on $[0, 4]$ . $[0, 4]$ par $v_{a v g}$ , total distance, average speed, aur jis instant particle turn karta hai woh nikalo.

Forecast: parabola neeche dip karti hai phir wapas aati hai. Guess karo: kya $∣ v_{a v g} ∣$ yahan average speed ke barabar hoga?

$v (t) = \frac{d}{d t} (t^{2} - 4 t) = 2 t - 4$ . Pehle kyun? Turning point wahan hai jahan $v = 0$ .
$v = 0 \Rightarrow 2 t - 4 = 0 \Rightarrow t = 2$ s. Kyun? Reversal par tangent flat hota hai (dekho Slope and tangent in calculus).
Endpoints: $x (0) = 0$ , $x (4) = 16 - 16 = 0$ . Toh $Δ x = 0$ . Yeh matter kyun karta hai? Net displacement zero hai chahe bahut move kiya ho.
$v_{a v g} = \frac{0}{4} = 0$ m/s. Zero kyun? Yeh wahan khatam hua jahan se shuru hua.
Distance: $t = 0$ se $2$ tak yeh $x (2) = 4 - 8 = - 4$ tak jaata hai ( $4$ m move kiya). $t = 2$ se $4$ tak wapas $0$ par (aur $4$ m). Total $= 8$ m. $t = 2$ par split kyun? Distance ko har unidirectional leg ki path lengths add karni chahiye.
Average speed $= \frac{8}{4} = 2$ m/s (upar di gayi definition use karke). $v_{a v g}$ se kyun alag? Speed path length use karta hai; velocity net arrow use karta hai.

Verify: $∣ v_{a v g} ∣ = 0 \leq 2 =$ avg speed ✓ (universal law $∣ v_{a v g} ∣ \leq$ avg speed ka ek instance). Turning point $t = 2$ par: $x = - 4$ parabola ka lowest point hai ✓.

Figure — parabola $x = - 4$ tak dip karti hai, $[0, 4]$ par zero-slope secant (dashed), turn $t = 2$ par flat tangent (mint), aur do equal $4$ -m legs (butter):

Example 4 — Case D: pure round trip (zero displacement)

Worked example Statement

Tum $60$ s mein $300$ m north cycle karte ho, $0$ s pause karte ho, phir $40$ s mein $300$ m south. Poore trip ke liye $v_{a v g}$ aur average speed nikalo.

Forecast: tum wapas ghar par ho. Aage padhne se pehle $v_{a v g}$ guess karo.

Net displacement $= + 300 - 300 = 0$ m. Kyun? North aur south ek vector sum ke roop mein cancel ho jaate hain.
Total time $= 60 + 40 = 100$ s. Kyun add karein? Time hamesha accumulate hota hai.
$v_{a v g} = \frac{0}{100} = 0$ m/s. Zero kyun chahe tum move kiye? Velocity sirf start-to-end dekhti hai.
Total distance $= 300 + 300 = 600$ m. Magnitudes kyun add karein? Distance direction ignore karta hai (upar average-speed definition).
Average speed $= \frac{600}{100} = 6$ m/s. Nonzero kyun? Path length real hai chahe wapas aa jao.

Verify: $v_{a v g} = 0$ , avg speed $= 6$ m/s, aur $0 \leq 6$ ✓. Yeh classic hai "averages journey chhupa lete hain."

Example 5 — Case E: degenerate interval (limit ki zaroorat kyun)

Worked example Statement

$x (t) = t^{2}$ ke liye, $[3, 3]$ interval use karke $t = 3$ par velocity paane ki koshish karo. Kya hota hai, aur limit kaise rescue karti hai?

Forecast: guess karo $\frac{Δ x}{Δ t}$ kya karta hai jab dono times same hon.

$[3, 3]$ try karo: $Δ x = x (3) - x (3) = 0$ aur $Δ t = 3 - 3 = 0$ . Toh $v_{a v g} = \frac{0}{0}$ . Yeh useless kyun? $0/0$ undefined hai — ek single instant ka koi interval nahi average karne ke liye.
Instead ek tiny window $[3, 3 + h]$ shrink karo, jahan $h$ humara chhota time gap hai ( $Δ t$ jaisi role). $x (3 + h) - x (3) = (3 + h)^{2} - 9 = 6 h + h^{2}$ . $h \neq = 0$ kyun rakhen? Tabhi hum divide kar sakte hain.
$v_{a v g} = \frac{6 h + h ^{2}}{h} = 6 + h$ . $h$ cancel kyun? Legal hai kyunki limit se pehle $h \neq = 0$ hai.
$h \to 0$ karo: $v (3) = 6$ m/s. Ab $h = 0$ set karna allowed kyun? Awkward $0/0$ gone hai; $6 + h$ $h = 0$ par perfectly defined hai.

Verify: Shortcut $v (t) = 2 t$ , toh $v (3) = 6$ ✓. Limit exactly woh tool hai jo ek forbidden $0/0$ ko ek real number mein badalta hai — yahi reason hai instantaneous velocity ek derivative hai.

Figure — parabola $x = t^{2}$ teen secants ke saath $t = 3$ se (coral $h = 1.2$ , butter $h = 0.6$ , mint $h = 0.2$ ) slope $6$ ki dashed tangent par collapse karte hue:

Example 6 — Case F: definition se seedha instantaneous velocity

Worked example Statement

$x (t) = 2 t^{3}$ (metres). Sirf limit definition use karke $v (t)$ nikalo, phir $t = 1$ par evaluate karo.

Forecast: power rule predict karta hai $6 t^{2}$ . Aao honestly prove karein. (Hum yahan $Δ t$ use karte hain; yeh exactly wahi role play karta hai jo $h$ ne Example 5 mein kiya.)

$x (t + Δ t) = 2 (t + Δ t)^{3} = 2 (t^{3} + 3 t^{2} Δ t + 3 t Δ t^{2} + Δ t^{3})$ . Poora expand kyun? Hume isolate karna hai jo division ke baad bachta hai.
$x (t) = 2 t^{3}$ subtract karo: numerator $= 6 t^{2} Δ t + 6 t Δ t^{2} + 2Δ t^{3}$ . Kyun? $2 t^{3}$ terms cancel ho jaate hain.
$Δ t$ se divide karo: $6 t^{2} + 6 t Δ t + 2Δ t^{2}$ . Legal kyun? Pre-limit $Δ t \neq = 0$ hai.
$Δ t \to 0$ : $v (t) = 6 t^{2}$ . Last do terms kyun vanish hoti hain? Har ek abhi bhi ek $Δ t$ factor carry karta hai.
$v (1) = 6 (1)^{2} = 6$ m/s.

Verify: Power rule: $\frac{d}{d t} (2 t^{3}) = 6 t^{2}$ ✓; $v (1) = 6$ ✓. Acceleration as derivative of velocity machinery se match karta hai.

Example 7 — Case G: midpoint theorem (uniform acceleration)

Worked example Statement

Constant acceleration ke under $x (t) = 2 t^{2} + t$ (toh $a$ constant hai). Dikhao ki $[2, 6]$ par $v_{a v g}$ midpoint $t = 4$ par $v$ ke barabar hai.

Forecast: parent note ne claim kiya yeh ek theorem hai, luck nahi. $v_{a v g}$ aur $v (4)$ alag alag predict karo, phir compare karo.

$x (2) = 8 + 2 = 10$ , $x (6) = 72 + 6 = 78$ . Kyun? Endpoints.
$v_{a v g} = \frac{78 - 10}{6 - 2} = \frac{68}{4} = 17$ m/s. 4 se divide kyun? $Δ t = 4$ s.
$v (t) = \frac{d}{d t} (2 t^{2} + t) = 4 t + 1$ . Linear kyun? Constant acceleration $\Rightarrow$ velocity linearly grow karti hai.
Midpoint $t_{mi d} = \frac{2 + 6}{2} = 4$ . $v (4) = 16 + 1 = 17$ m/s. Midpoint check kyun? Ek linear $v (t)$ ke liye, uska average ek interval par exactly middle time par baithta hai.

Midpoint rule generally kyun hold karta hai. Jab bhi velocity linear ho, $v (t) = m t + c$ (yahan $m = 4$ , $c = 1$ ). Linear $v$ se kyun start karein? Constant acceleration matlab $v$ equal amounts mein equal times mein change karta hai — yeh precisely ek straight line ka graph hai. Average velocity total displacement over $Δ t$ hai; displacement kyun laayein? kyunki $v_{a v g} = Δ x /Δ t$ definition se hai, aur total displacement $v (t)$ graph ke neeche ka area hai (yeh area-equals-displacement fact Equations of motion under uniform acceleration se aata hai — velocity accumulated over time hi distance covered hai). Toh $[p, q]$ par, $v_{a v g} = \frac{1}{q - p} \int_{p}^{q} (m t + c) d t = \frac{1}{q - p} [\frac{m}{2} t^{2} + c t]_{p}^{q} = \frac{m}{2} (p + q) + c .$ Yeh average-value formula kyun? Area $\int_{p}^{q} v d t$ ko width $q - p$ se divide karna exactly "total displacement over total time" hai — $v_{a v g}$ ki definition integral ke saath likhi gayi. Meanwhile $v$ midpoint par hai $v (\frac{p + q}{2}) = m \cdot \frac{p + q}{2} + c = \frac{m}{2} (p + q) + c$ — identical. Toh equality kisi bhi linear $v (t)$ ke liye, yaani kisi bhi uniform acceleration ke liye, forced hai.

Verify: $17 = 17$ ✓, aur general formula deta hai $\frac{4}{2} (2 + 6) + 1 = 16 + 1 = 17$ ✓.

Example 8 — Case H: real-world multi-leg word problem

Worked example Statement

Ek commuter $15$ min mein $12$ km east drive karta hai, ek light par $5$ min ruka rehta hai (koi motion nahi), phir $10$ min mein $8$ km west. Poore journey ke liye (a) average velocity aur (b) average speed nikalo, km/h mein.

Forecast: stop time add karta hai par koi distance nahi. Guess karo kya yeh averages ko raise ya lower karta hai.

Net displacement $= + 12 - 8 = + 4$ km (east). Signed sum kyun? East $+$ , west $-$ .
Total time $= 15 + 5 + 10 = 30$ min. Stop include kyun? Time rukne par bhi chalta rehta hai.
Units-ke-ratio method se hours mein convert karo: $30 min \times \frac{1 h}{60 min} = \frac{30}{60} h = 0.5 h$ . Yeh trick kyun? $1$ ke barabar fraction se multiply karna ( $\frac{1 h}{60 min}$ ) "min" unit cancel karta hai aur "h" bacha deta hai.
$v_{a v g} = \frac{4 km}{0.5 h} = 8$ km/h east. Direction kyun? $Δ x > 0$ matlab net eastward.
Total distance $= 12 + 8 = 20$ km. Magnitudes kyun add karein? Path length (average-speed definition).
Average speed $= \frac{20}{0.5} = 40$ km/h.

Verify: $8 \leq 40$ ✓ ( $∣ v_{a v g} ∣ \leq$ avg speed ka ek instance). $5$ -min stop ne $Δ t$ inflate karke dono averages lower kiye jabki zero distance aur zero displacement add kiya.

Example 9 — Case I: exam twist, unknown solve karo

Worked example Statement

Ek train ka constant $v_{a v g} = 25$ m/s hai ek straight track par aur $1500$ m displacement cover karta hai. Phir, doosri leg par, $600$ m par $30$ m/s average karna hai. Har leg ka time aur overall average velocity nikalo.

Forecast: overall average nahi hai $(25 + 30) /2$ . Guess karo kya true answer $27.5$ se upar ya neeche hoga.

Leg 1 time: $t_{1} = \frac{Δ x}{v _{a v g}} = \frac{1500}{25} = 60$ s. Rearrange kyun? $v = Δ x /Δ t$ se, $Δ t$ isolate karo.
Leg 2 time: $t_{2} = \frac{600}{30} = 20$ s. Same formula kyun? Same definition, second leg.
Total displacement $= 1500 + 600 = 2100$ m; total time $= 60 + 20 = 80$ s. Har ek kyun add karein? Dono legs same direction mein chalte hain, toh unke displacement arrows ek hi taraf point karte hain aur simply add ho jaate hain (koi cancellation nahi); time hamesha leg ke baad leg accumulate karta hai.
Overall $v_{a v g} = \frac{2100}{80} = 26.25$ m/s. $27.5$ kyun nahi? Velocities ko average karna galat hai; tumhe total displacement ko total time se divide karna hai, jo har leg ko time se weight karta hai, equally nahi.

Verify: $26.25$ m/s, $25$ aur $30$ ke beech hai, $25$ ke kareeb kyunki slower leg par zyada time spend hua ( $60$ s vs $20$ s) ✓. Yeh exactly wahi [!mistake] hai jiske baare mein parent note ne warn kiya tha.

Example 10 — Case J: turning point graph se read karo

Worked example Statement

Upar throw ki gayi ek ball ka $x (t) = 20 t - 5 t^{2}$ hai (metres, up ko $+$ maante hue). Kab $v = 0$ hoga (top), theek pehle aur baad ki velocity, aur launch se landing tak poore flight par $v_{a v g}$ confirm karo.

Forecast: bilkul top par ball momentarily still hoti hai. Guess karo kya poori up-and-down flight par $v_{a v g}$ zero hoga.

$v (t) = \frac{d}{d t} (20 t - 5 t^{2}) = 20 - 10 t$ . Kyun? Position parabola ka slope.
Top: $v = 0 \Rightarrow 20 - 10 t = 0 \Rightarrow t = 2$ s. Kyun? Flat tangent = reversal instant.
Theek pehle ( $t = 1.9$ ): $v = 20 - 19 = + 1$ m/s (abhi bhi upar ja raha hai). Theek baad ( $t = 2.1$ ): $v = 20 - 21 = - 1$ m/s (gir raha hai). Signs flip kyun? Direction turning point se guzarte hue reverse hoti hai, chahe $t = 2$ par exactly $v = 0$ ho.
Landing: $x = 0 \Rightarrow 20 t - 5 t^{2} = 0 \Rightarrow t (20 - 5 t) = 0 \Rightarrow t = 4$ s. Doosra root kyun? $t = 0$ launch hai; $t = 4$ landing hai.
Full-flight displacement $= x (4) - x (0) = 0 - 0 = 0$ , toh $v_{a v g} = \frac{0}{4} = 0$ m/s.

Verify: $v$ ka sign $+ \to -$ $t = 2$ par $0$ se guzarte hue change hota hai ✓ (yeh myth tod raha hai ki " $v = 0$ hone par move nahi kar sakte"). $v_{a v g} = 0$ round-trip flight par, Case D se match karta hai ✓.

Figure — height $x (t) = 20 t - 5 t^{2}$ vs time, flat top $t = 2$ par marked ( $v = 0$ ), rising leg labelled $v > 0$ , falling leg $v < 0$ , launch aur landing $x = 0$ par:

Recall Which cell trips people up most?

Case I (exam twist) ::: kyunki $(v_{1} + v_{2}) /2$ likhne ka temptation wahaan sabse zyada hota hai; fix hai total displacement over total time, jo legs ko time se weight karta hai, equally nahi. Case C aur Case J mein, turning instant par instantaneous velocity kya hoti hai? ::: exactly $0$ (flat tangent), chahe object otherwise move kar raha ho. Dono averages ko relate karne wala universal law batao. ::: $∣ v_{a v g} ∣ \leq$ average speed, kyunki path length $\geq ∣ displacement ∣$ ; equality sirf tab jab motion kabhi reverse nahi karta.

Connections

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