Exercises — Average acceleration vs instantaneous acceleration

Tumhare paas do endpoints aur ek clock hai — beech ke baare mein kuch nahi. Ye bilkul average ke ingredients hain. $a_{avg}=\frac{v_f-v_i}{\Delta t}=\frac{32-8}{6}=\frac{24}{6}=4{\text{m/s}}^2$ Instantaneous acceleration ke liye ek point par slope chahiye, jiske liye $v$ ko time ki function ke roop mein jaanna zaroori hai (ek curve, na ki do dots) — isliye yeh compute nahi ho sakta yahan.

Galat. Acceleration $\frac{d\vec{v}}{dt}$ hai, aur $\vec{v}$ ek vector hai — usme direction bhi hoti hai. Speed ($\vec{v}$ ki length) fixed hai, lekin direction ghoomti rehti hai, isliye $\vec{v}$ lagaataar badal raha hai. Ek changing vector ka rate of change nonzero hota hai. Dekho Uniform circular motion. Acceleration centre ki taraf point karta hai (centripetal); yeh zero nahi hai.

Instantaneous acceleration velocity ka derivative hai (parent derivation ka Step 3). $a(t)=\frac{dv}{dt}=\frac{d}{dt}(4t^2-3t+1)=8t-3$ Term by term kyun: $\frac{d}{dt}(4t^2)=8t$, $\frac{d}{dt}(-3t)=-3$, $\frac{d}{dt}(1)=0$ (ek constant nahi badlata, isliye uska rate zero hai). $a(3)=8(3)-3=24-3=21{\text{m/s}}^2$

Average ke liye do endpoint velocities chahiye, derivatives nahi. $v(1)=4(1)-3(1)+1=2\text{m/s}v(3)=4(9)-9+1=28\text{m/s}$ $a_{avg}=\frac{v(3)-v(1)}{3-1}=\frac{28-2}{2}=\frac{26}{2}=13{\text{m/s}}^2$ Compare karo: $a_{avg}=13\ne a(3)=21$. Ye isliye alag hain kyunki yahan acceleration constant nahi hai ($a=8t-3$ time ke saath badhta hai), isliye overall chord slope ($13$) right end ke steep tangent slope ($21$) se neeche baithta hai.

Signed velocities use karo — direction sign mein hoti hai. $v_i=+12,v_f=-8\text{(west)}$ $a_{avg}=\frac{v_f-v_i}{\Delta t}=\frac{-8-(+12)}{5}=\frac{-20}{5}=-4{\text{m/s}}^2$ Magnitude $4{\text{m/s}}^2$, direction west (minus sign = $-$/east-negative direction). $\Delta v$ mein final minus initial kyun use hota hai, yeh samjhne ke liye dekho Vectors: addition and subtraction.

$v$–$t$ graph par, acceleration = slope = (velocity mein rise)/(time mein run).

• Segment A, $0\to 2$ s: $v$ $0\to 8$ jaata hai. Slope $=\frac{8-0}{2-0}=4{\text{m/s}}^2$.
• Segment B, $2\to 5$ s: $v$ $8$ par flat hai. Slope $=\frac{8-8}{5-2}=0{\text{m/s}}^2$ (cruising).
• Segment C, $5\to 8$ s: $v$ $8\to 2$ jaata hai. Slope $=\frac{2-8}{8-5}=\frac{-6}{3}=-2{\text{m/s}}^2$ (slow ho raha hai). Whole-trip average sirf trip ke do endpoints use karta hai ($v(0)=0$, $v(8)=2$): $a_{avg}=\frac{2-0}{8-0}=0.25{\text{m/s}}^2$ Average ($0.25$) ek single chord slope hai start-point se end-point tak — yeh beech ke ups and downs ko ignore karta hai, bilkul waise jaise definition ka vaada hai.

Instantaneous $a$ = tangent slope; straight segments par tangent slope segment slope ke barabar hoti hai.

• Magnitudes: $|+4|=4$ (A), $|0|=0$ (B), $|-2|=2$ (C).
• Sabse badi magnitude: Segment A, $4{\text{m/s}}^2$.
• Zero: Segment B ($2$–$5$ s) par har jagah, jahan graph flat hai.

Velocity paane ke liye har component ko ek baar differentiate karo, phir acceleration ke liye ek aur baar (parent Step 3, har axis par apply kiya gaya). $\vec{v}=\frac{d\vec{r}}{dt}=(2t)\hat{x}+(2)\hat{y}$ $\vec{a}=\frac{d\vec{v}}{dt}=(2)\hat{x}+(0)\hat{y}=2\hat{x}{\text{m/s}}^2$ Magnitude: $|\vec{a}|=\sqrt{2^2+0^2}=2{\text{m/s}}^2$, purely east point karta hai. Gaur karo ki acceleration constant hai tab bhi jab speed badhti rehti hai — dono ideas independent hain.

Speed unchanged hai (dono baar $10$ m/s), lekin direction north se east ho gayi — isliye $\Delta \vec{v}$ nonzero hai. Vectors likhte hain $\hat{x}$ = east, $\hat{y}$ = north ke saath: ${\vec{v}}_i=10\hat{y},{\vec{v}}_f=10\hat{x}$ $\Delta \vec{v}={\vec{v}}_f-{\vec{v}}_i=10\hat{x}-10\hat{y}$ Yeh figure mein red arrow hai: subtract karne ke liye, ${\vec{v}}_i$ ko reverse karo aur ${\vec{v}}_f$ mein add karo. $|\Delta \vec{v}|=\sqrt{10^2+(-10{)}^2}=\sqrt{200}=10\sqrt{2}\approx 14.14\text{m/s}$ $|{\vec{a}}_{avg}|=\frac{|\Delta \vec{v}|}{\Delta t}=\frac{10\sqrt{2}}{2}=5\sqrt{2}\approx 7.07{\text{m/s}}^2$ Direction: south-east ($\hat{x}-\hat{y}$ ke along point karta hai, yani east se $45^{\circ }$ neeche). Constant speed, real acceleration — vector nature saara kaam kar raha hai.

(a) Do baar differentiate karo. $v=\frac{dx}{dt}=3t^2-18t+24,a=\frac{dv}{dt}=6t-18$ $a=0$ set karo: $6t-18=0\Rightarrow t=3$ s. (b) $t=3$ par: $v(3)=3(9)-18(3)+24=27-54+24=-3$ m/s ($-x$ direction mein move kar raha hai). Ek body speed up karta hai jab $v$ aur $a$ ka sign same ho. $t=3$ se thodi der pehle, $a=6t-18<0$; aur $v(3)=-3<0$. Same sign ($-,-$) → $t=3$ ke paas aate waqt speeding up; bilkul $t=3$ par, $a=0$ woh instant hai jab acceleration sign switch karta hai, speed trend ka turning point.

Maano $a$ constant hai. Tab velocity linearly badhti hai: $v(t)=v_i+at$ (iska matlab hai "constant rate" — dekho Derivatives as rates of change). $[t_1,t_2]$ par average: $a_{avg}=\frac{v(t_2)-v(t_1)}{t_2-t_1}=\frac{(v_i+a{t}_2)-(v_i+a{t}_1)}{t_2-t_1}=\frac{a(t_2-t_1)}{t_2-t_1}=a$ Kisi bhi $t$ par instantaneous: $\frac{dv}{dt}=\frac{d}{dt}(v_i+at)=a$. Dono usi constant $a$ ke barabar hain, chahe tum koi bhi interval ya point chuno. $■$ Isi liye equation $v=v_i+at$ (equations of motion se) poore interval ke liye ek akele $a$ use karne ki permission deti hai — wahan average aur instantaneous values coincide karti hain.

Ex 1.1 average acceleration ::: $4{\text{m/s}}^2$ Ex 2.1 $a(3)$ for $v=4t^2-3t+1$ ::: $21{\text{m/s}}^2$ Ex 2.2 $a_{avg}$ on $[1,3]$ ::: $13{\text{m/s}}^2$ Ex 2.3 average acceleration ::: $-4{\text{m/s}}^2$ (west) Ex 3.1 whole-trip average ::: $0.25{\text{m/s}}^2$ Ex 4.1 acceleration magnitude ::: $2{\text{m/s}}^2$ (east) Ex 4.2 average acceleration magnitude ::: $5\sqrt{2}\approx 7.07{\text{m/s}}^2$ (south-east) Ex 5.1 time acceleration is zero ::: $t=3$ s