Average velocity vs instantaneous velocity: Exercises

1.1.14 · D4 · Physics › Measurement, Vectors & Kinematics › Average velocity vs instantaneous velocity

Shuru karne se pehle, sirf do tools ka ek-line reminder (dono parent note se):

Level 1 — Recognition

Exercise 1.1 (L1)

Ek particle $t = 1$ s par $x = 4$ m par hai, aur $t = 4$ s par $x = 10$ m par hai. Is interval par uski average velocity kya hai?

Recall Solution 1.1

KYA karte hain: $v_{a v g} = Δ x /Δ t$ seedha apply karo (1-D hai, toh signed numbers). $Δ x = 10 - 4 = 6 m, Δ t = 4 - 1 = 3 s .$ KYU divide karte hain: hum chahte hain position change per second — "kitna dur" ko "kitne time mein" se divide karna rate ke liye universal recipe hai. $v_{a v g} = \frac{6}{3} = 2 m/s (+ x direction mein, yani + 2 \overset{x}{^}).$

Exercise 1.2 (L1)

Inme se kaun si instantaneous velocity hai, aur kaun si average velocity? (A) ek instant par tangent line ki slope, (B) do instants ko join karne wale chord ki slope.

Recall Solution 1.2

(A) tangent slope $=$ instantaneous velocity — ek point, koi smearing nahi.
(B) chord (secant) slope $=$ average velocity — do points ek interval ke saath. KYU tangent instantaneous hai: tangent curve ko ek hi instant par touch karta hai, isliye uski steepness velocity us ek moment par report karti hai — koi aur time mix nahi hota. KYU chord average hai: chord ki steepness total rise $Δ x$ over total run $Δ t$ hai apne do endpoints ke beech — yeh exactly $Δ x /Δ t$ hai, yaani average. Parent note mein yahi picture hai: do chord-points ko saath slide karo aur secant tangent ban jaata hai, toh average instantaneous ban jaata hai.

Exercise 1.3 (L1)

Ek ball seedha upar phenki jaati hai. Apni flight ke exact top par, uski instantaneous velocity kya hai?

Recall Solution 1.3

KAISA dikhta hai: top par position–time graph ka flat tangent hota hai (ek horizontal peak). Flat tangent ki slope $0$ hoti hai, toh $v = \frac{d x}{d t} = 0 m/s .$ Ball momentarily velocity mein still hai, bhaale gravity abhi bhi kheench rahi ho (acceleration $\neq = 0$ ).

Level 2 — Application

Exercise 2.1 (L2)

Ek car ki position $x (t) = 2 t^{2} + 3 t$ hai (metres, seconds). $t = 1$ s se $t = 3$ s tak average velocity nikalo.

Recall Solution 2.1

KYA karte hain: dono ends par position padho, subtract karo displacement pane ke liye, phir us displacement ko rate mein convert karo. $x (1) = 2 (1) + 3 (1) = 2 + 3 = 5 m, x (3) = 2 (9) + 3 (3) = 18 + 9 = 27 m .$ $Δ x = 27 - 5 = 22 m, Δ t = 3 - 1 = 2 s .$ KYU $Δ t$ se divide karte hain: $22$ m sirf kitna car move hua hai; ise velocity (metres per second) mein badalne ke liye hume kitne time mein se divide karna hoga. Us division ke bina answer displacement hoga, rate nahi. $v_{a v g} = \frac{22}{2} = 11 m/s .$

Exercise 2.2 (L2)

Wahi $x (t) = 2 t^{2} + 3 t$ . Power rule use karke $t = 2$ s par instantaneous velocity nikalo.

Recall Solution 2.2

KYU power rule: $x$ powers of $t$ ka sum hai, aur $\frac{d}{d t} (t^{n}) = n t^{n - 1}$ har power ko seedha uski slope-formula mein badal deta hai — limit se tez. $v (t) = \frac{d x}{d t} = 2 (2 t) + 3 = 4 t + 3.$ $v (2) = 4 (2) + 3 = 11 m/s .$ Gaur karo: $[1, 3]$ ka midpoint $t = 2$ hai, aur $v (2) = v_{a v g}$ 2.1 se. Yeh constant-acceleration midpoint theorem hai, luck nahi (L4 dekho).

Exercise 2.3 (L2)

Ek drone $20$ s mein $60$ m north udata hai, phir $20$ s mein $30$ m east. (a) Average velocity ki magnitude nikalo, (b) average speed nikalo.

Neeche di gayi figure padhna (yeh genuine 2-D hai, isliye hum vectors rakhte hain): vertical axis north (m) hai, horizontal axis east (m) hai. Do black legs actual path hain — pehle seedha upar (north $60$ ), phir seedha right (east $30$ ). Red arrow displacement $Δ r$ hai: start se end tak ka ek seedha arrow. Kyunki north aur east right angles par hain, woh do legs aur red arrow ek right triangle banate hain, isliye red arrow ki length hypotenuse hai. Yahi part (a) ko chahiye, jabki do black legs ka sum part (b) ke liye path length deta hai.

Figure — Average velocity vs instantaneous velocity

Recall Solution 2.3

KYA displacement hai: north phir east perpendicular hain, toh straight-line displacement (red arrow) $60$ aur $30$ legs wale right triangle ka hypotenuse hai. $∣Δ r ∣ = 6 0^{2} + 3 0^{2} = 3600 + 900 = 4500 = 305 \approx 67.08 m .$ Total time $Δ t = 20 + 20 = 40$ s. (a) KYU displacement ko $Δ t$ se divide karte hain: red arrow ki length $305$ m sirf kitna drone effectively move hua hai; us net displacement ko velocity (net metres per second) mein badalne ke liye hum rate recipe $Δ r /Δ t$ apply karte hain — "kitna net" ko "kitne time mein" se divide karte hain. $∣ v_{a v g} ∣ = \frac{30 5}{40} = \frac{3 5}{4} \approx 1.68 m/s .$ (Direction: red arrow ke saath, north-of-east.) (b) KYU path length ko $Δ t$ se divide karte hain: average speed bhi ek rate hai, lekin distance travelled ka, net displacement ka nahi. Toh hum total path length lete hain aur phir "kitne time mein" se divide karte hain — same recipe, alag numerator. Path length (ek scalar, koi direction nahi) $= 60 + 30 = 90$ m: $avg speed = \frac{90}{40} = 2.25 m/s .$ Jaisa promise kiya, $∣ v_{a v g} ∣ \approx 1.68 \leq 2.25 =$ avg speed. Dekho Distance vs displacement.

Level 3 — Analysis

Exercise 3.1 (L3)

Neeche diya graph straight segments se bana position–time graph hai. Segment A upar jaata hai, segment B flat hai, segment C neeche jaata hai. A, B, C par instantaneous velocities ko sign ke hisaab se rank karo (positive, zero, negative).

Recall Solution 3.1

Instantaneous velocity $=$ graph ki slope (dekho Position-time graphs, Slope and tangent in calculus). KYU slope = velocity: slope rise-over-run $= Δ x /Δ t$ hai, aur position–time graph par "rise" position mein change hai aur "run" time mein change — exactly velocity recipe. Toh zyada steep tilt matlab tez velocity, aur tilt ki direction sign set karti hai.

A upar jaata hai ⇒ positive slope ⇒ $v > 0$ (position badh rahi hai, toh $+ x$ mein move kar raha hai).
B flat hai ⇒ zero slope ⇒ $v = 0$ (position change nahi ho rahi, toh rest par hai).
C neeche jaata hai ⇒ negative slope ⇒ $v < 0$ (position ghatt rahi hai, toh chhote $x$ ki taraf wapas ja raha hai). Ranking: $v_{A} > v_{B} = 0 > v_{C}$ .

Exercise 3.2 (L3)

Usi trip par object $t = 0$ par $x = 0$ se shuru hota hai, $t = 4$ s par $x = 40$ m pahunchta hai (A ka end), $t = 6$ s tak $40$ m par rehta hai (B ka end), phir $t = 9$ s par $x = 10$ m par return karta hai (C ka end). Poori trip $t = 0$ se $t = 9$ s par average velocity nikalo.

Recall Solution 3.2

Average velocity wiggles ignore karta hai; woh sirf start aur end dekhta hai. $Δ x = x (9) - x (0) = 10 - 0 = 10 m, Δ t = 9 - 0 = 9 s .$ KYU divide karte hain: $10$ m net position change hai; rate recipe $Δ x /Δ t$ ise net metres per second mein badalta hai. $v_{a v g} = \frac{10}{9} \approx 1.11 m/s .$ Geometrically yeh pehle point se aakhri point tak single chord ki slope hai — segment A se flatter, segment C se steeper (kam negative).

Exercise 3.3 (L3)

$x (t) = t^{3} - 6 t^{2} + 9 t$ (m, s) ke liye, har woh instant nikalo jab object momentarily rest par ho, aur batao position graph wahan kaisa dikhta hai.

Recall Solution 3.3

"Rest par" matlab $v = 0$ . Differentiate karo: $v (t) = \frac{d x}{d t} = 3 t^{2} - 12 t + 9 = 3 (t^{2} - 4 t + 3) = 3 (t - 1) (t - 3) .$ Zero set karo: $t = 1$ s aur $t = 3$ s. KAISA dikhta hai: dono instants par position graph ka flat tangent hota hai ( $t = 1$ par local peak, $t = 3$ par local valley) — object rukta hai aur direction reverse karta hai. Unke beech $v < 0$ hai (check: $v (2) = 3 (1) (- 1) = - 3 < 0$ ), toh wahan backward move karta hai.

Level 4 — Synthesis

Exercise 4.1 (L4)

Midpoint theorem prove karo jo tumne parent note mein dekha: constant acceleration $x (t) = x_{0} + v_{0} t + \frac{1}{2} a t^{2}$ ke liye, $[t_{1}, t_{2}]$ par average velocity midpoint $t_{m} = \frac{t _{1} + t _{2}}{2}$ par instantaneous velocity ke equal hoti hai.

Recall Solution 4.1

Instantaneous velocity: $v (t) = \frac{d x}{d t} = v_{0} + a t$ , toh midpoint par $v (t_{m}) = v_{0} + a \cdot \frac{t _{1} + t _{2}}{2} .$ Average velocity: $v_{a v g} = \frac{x ( t _{2} ) - x ( t _{1} )}{t _{2} - t _{1}} .$ Numerator compute karo. Do positions subtract karne par constant $x_{0}$ cancel ho jaata hai (dono times par same hai) aur linear aur quadratic pieces group ho jaate hain: $x (t_{2}) - x (t_{1}) = v_{0} (t_{2} - t_{1}) + \frac{1}{2} a (t_{2}^{2} - t_{1}^{2}) .$ KYU aage factor karte hain: hum $(t_{2} - t_{1})$ se divide karne wale hain, toh hum chahte hain woh exact bracket ek factor ke roop mein aaye aur cleanly cancel ho jaaye. Difference of squares humein woh deta hai: $t_{2}^{2} - t_{1}^{2} = (t_{2} - t_{1}) (t_{2} + t_{1}) .$ KYU yeh identity: $(t_{2} - t_{1}) (t_{2} + t_{1}) = t_{2}^{2} - t_{1}^{2}$ jab expand karo — cross terms $- t_{1} t_{2}$ aur $+ t_{2} t_{1}$ cancel ho jaate hain, sirf squares bachte hain. Ab har term ko $(t_{2} - t_{1})$ se divide karo; bracket jahan bhi aata hai cancel ho jaata hai: $v_{a v g} = v_{0} + \frac{1}{2} a (t_{2} + t_{1}) = v_{0} + a \cdot \frac{t _{1} + t _{2}}{2} = v (t_{m}) . ■$ KYU intuitively kaam karta hai: velocity $v_{0} + a t$ time mein ek straight line hai, aur ek straight line ka average ek interval par hamesha uski midpoint value par aata hai. Dekho Equations of motion under uniform acceleration aur Acceleration as derivative of velocity.

Exercise 4.2 (L4)

Ek particle ka $x (t) = 3 t^{2} - 2 t$ hai (parent note se). Limit definition se derivative $v (t) = 6 t - 2$ confirm karo, phir woh instant $T$ (with $T \neq = 1$ ) nikalo jab $[1, T]$ par $v_{a v g}$ , $v (T)$ ke equal ho. Comment karo.

Recall Solution 4.2

Limit se: hum ek chhoti window $[t, t + Δ t]$ par average velocity banate hain aur phir window shrink karte hain. $x (t + Δ t) = 3 (t + Δ t)^{2} - 2 (t + Δ t) = 3 t^{2} + 6 t Δ t + 3Δ t^{2} - 2 t - 2Δ t .$ KYU $x (t)$ subtract karte hain: average velocity ka numerator position mein change hai, toh hum starting position $x (t) = 3 t^{2} - 2 t$ subtract karte hain; $Δ t$ mein constant pieces ( $3 t^{2}$ aur $- 2 t$ ) cancel ho jaate hain: $numerator = 6 t Δ t + 3Δ t^{2} - 2Δ t .$ KYU $Δ t$ se divide kar sakte hain: har term mein $Δ t$ factor hai, aur jabtak $Δ t \neq = 0$ (limit se pehle) divide karna legal hai: $\frac{6 t Δ t + 3Δ t ^{2} - 2Δ t}{Δ t} = 6 t + 3Δ t - 2.$ KYU limit finish karta hai: ab $Δ t \to 0$ hone do — sirf bachne wala $Δ t$ term, $3Δ t$ , vanish ho jaata hai: $v (t) = 6 t - 2. ✓$ Ab $v_{a v g} [1, T] = v (T)$ set karo. $v_{a v g} = \frac{x ( T ) - x ( 1 )}{T - 1} = \frac{( 3 T ^{2} - 2 T ) - ( 3 - 2 )}{T - 1} = \frac{3 T ^{2} - 2 T - 1}{T - 1} .$ KYU top factor karte hain: denominator mein $(T - 1)$ cancel karna chahte hain. Factor karo $3 T^{2} - 2 T - 1 = (T - 1) (3 T + 1)$ (expand karke check karo: $3 T^{2} + T - 3 T - 1 = 3 T^{2} - 2 T - 1$ ✓): $v_{a v g} = \frac{( T - 1 ) ( 3 T + 1 )}{T - 1} = 3 T + 1 (T \neq = 1) .$ $v (T) = 6 T - 2.$ Equate karo: $3 T + 1 = 6 T - 2 \Rightarrow 3 = 3 T \Rightarrow T = 1.$ Comment: algebra $T = 1$ force karta hai, lekin woh excluded endpoint hai (woh $v_{a v g}$ ko indeterminate $0/0$ banata hai), toh koi valid $T \neq = 1$ nahi hai jahan $[1, T]$ par average endpoint velocity $v (T)$ se match kare. Clean, hamesha-sach wali match balki Exercise 4.1 ka midpoint theorem hai: is quadratic ke liye $v_{a v g} [1, T] = 3 T + 1$ aur midpoint $t = \frac{1 + T}{2}$ par $v$ hai $6 \cdot \frac{1 + T}{2} - 2 = 3 T + 1$ — har $T$ ke liye exact equality. Misal ke tor par $[1, 3]$ par: $v_{a v g} = 3 (3) + 1 = 10$ aur midpoint $t = 2$ deta hai $v (2) = 6 (2) - 2 = 10$ . ✓ Moral: average velocity midpoint par instantaneous velocity se match karta hai, endpoint par nahi.

Level 5 — Mastery

Exercise 5.1 (L5)

Ek particle $x (t) = 4 sin (\frac{π}{2} t)$ metres mein move karta hai ( $t$ seconds mein). (a) Uski instantaneous velocity function nikalo. (b) $t = 0$ se $t = 4$ s tak ek poore period par average velocity nikalo. (c) Us period par average speed nikalo aur mismatch explain karo.

Recall Solution 5.1

(a) $\frac{d}{d t} sin (k t) = k cos (k t)$ , toh $k = \frac{π}{2}$ ke saath: $v (t) = 4 \cdot \frac{π}{2} cos (\frac{π}{2} t) = 2 π cos (\frac{π}{2} t) m/s .$ (b) Ek period mein particle start par wapas aata hai: $x (4) = 4 sin (2 π) = 0 = x (0)$ . Toh, zero net displacement ke saath rate recipe apply karte hue, $v_{a v g} = \frac{x ( 4 ) - x ( 0 )}{4 - 0} = \frac{0 - 0}{4} = 0 m/s .$ (c) Path length track karo. $x = 0$ se shuru karke, particle $+ 4$ tak swing karta hai (quarter period, $t : 0 \to 1$ ), $0$ par wapas ( $t : 1 \to 2$ ), $- 4$ tak ( $t : 2 \to 3$ ), aur $0$ par wapas ( $t : 3 \to 4$ ). Har leg $4$ m ground cover karta hai, toh total distance $4 + 4 + 4 + 4 = 16$ m hai. KYU $Δ t$ se divide karte hain: average speed distance travelled per second hai, toh hum path length lete hain aur elapsed time se divide karte hain: $avg speed = \frac{16}{4} = 4 m/s .$ Mismatch explain karna: average velocity $0$ hai kyunki woh sirf net displacement dekhta hai, aur round trip net zero karta hai — particle exactly wahan finish karta hai jahan shuru hua tha. Average speed $4$ m/s hai kyunki woh har metre count karta hai jo actually travel hua, jo zero se bahut door hai. Yeh parent note ka round-trip lesson hai, ab smooth (sinusoidal) motion ke saath seedhi walk ki jagah. General rule ke saath consistent, $∣ v_{a v g} ∣ = 0 \leq 4 =$ avg speed, equality sirf tab jab motion kabhi reverse na kare — jo yahan clearly karta hai.

Exercise 5.2 (L5)

Ek stone drop ki jaati hai aur gravity ke under girta hai, $x (t) = 5 t^{2}$ m (downward positive, $g \approx 10 m/s^{2}$ ). $t = 0$ aur $t = 2$ s ke beech: (a) kya average velocity initial ya final instantaneous velocity ke closer hai, aur kyun? (b) Kis ek instant par instantaneous velocity average velocity ke equal hoti hai?

Recall Solution 5.2

$v (t) = \frac{d x}{d t} = 10 t m/s \Rightarrow v (0) = 0, v (2) = 20 m/s .$ $v_{a v g} = \frac{x ( 2 ) - x ( 0 )}{2 - 0} = \frac{5 ( 4 ) - 0}{2} = \frac{20}{2} = 10 m/s .$ (a) $v_{a v g} = 10$ exactly $0$ aur $20$ ke beech halfway hai — nah start ke closer, nah end ke. KYU: motion constant-acceleration hai, toh velocity linearly badhti hai; ek straight line ka average uski midpoint value hoti hai. (b) $v (t) = v_{a v g}$ set karo: $10 t = 10 \Rightarrow t = 1$ s — $[0, 2]$ ka midpoint, Exercise 4.1 ke theorem confirm karta hua.

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