Exercises — Graphs — x-t, v-t, a-t; areas and slopes meaning

KYA karte hain: velocity $x$–$t$ graph ka slope hai — "rise over run." KYUN: $x$–$t$ ka slope $=\frac{dx}{dt}=v$ (slope ladder par $x\to v$ neeche slide karta hai). KAISA DIKHTA HAI: ek akeli seedhi line — steepness kabhi nahi badlti. $v=\frac{\Delta x}{\Delta t}=\frac{14-2}{4-0}=\frac{12}{4}=3\text{m/s}$ Seedhi line ka matlab hai slope har jagah same hai, isliye velocity constant hai $3$ m/s par (slope positive hai isliye $+x$ direction mein).

Har case mein slope padho (slope of $x$–$t$ = velocity).

• (a) flat: slope $=0\Rightarrow v=0$ ⇒ object at rest (constant velocity NAHI — neeche diya trap dekho).
• (b) upar: slope positive aur constant ⇒ $+x$ mein constant velocity.
• (c) neeche: slope negative aur constant ⇒ $-x$ mein constant velocity.
• (d) upar curve: slope badh raha hai ⇒ velocity badh rahi hai ⇒ $a>0$, $+x$ mein speeding up.

KYA: displacement $v$–$t$ graph ke neeche ka area hai. KYUN: har thin strip $dx=vdt$ hai, ek rectangle; sab jodo → area (area ladder $v\to x$ mein add up karta hai). KAISA DIKHTA HAI: ek bada rectangle, height $8$, width $5$. $\Delta x=\text{area}=8\times 5=40\text{m}$

Acceleration = slope of $v$–$t$. KYUN: acceleration define hota hai is baat se ki velocity kitni tezi se badlti hai, yaani $v$–$t$ graph ka "rise over run" ek tiny run par liya jaaye: $a=\frac{\Delta v}{\Delta t}\to \frac{dv}{dt}$. Kyunki yahan line seedhi hai, slope har jagah same hai, isliye hum poore rise ko poore run se divide kar sakte hain: $a=\frac{\Delta v}{\Delta t}=\frac{16-4}{3-0}=\frac{12}{3}=4{\text{m/s}}^2$ Displacement = $v$–$t$ ke neeche ka area. Yahan $u$ initial velocity hai (jahan graph start karta hai woh height, $u=4$ m/s) aur $v$ final velocity hai (jahan khatam hota hai, $v=16$ m/s). Area ek trapezium hai (upar define kiya gaya): donon parallel heights ka average times width. $\Delta x=\frac{u+v}{2}t=\frac{4+16}{2}\times 3=10\times 3=30\text{m}$ (Same as rectangle $4\times 3=12$ plus triangle $\frac{1}{2}\times 3\times 12=18$; $12+18=30$ ✓.)

(a) Accelerations = har seedhe segment ke slopes. (Slope of $v$–$t$ $=\frac{dv}{dt}=a$.)

• Phase 1: $a_1=\frac{20-0}{4}=5{\text{m/s}}^2$.
• Phase 2: flat line ⇒ $a_2=0$.
• Phase 3: $a_3=\frac{0-20}{5}=-4{\text{m/s}}^2$ (negative = slow ho raha hai, kyunki $v>0$ lekin $a<0$).

(b) Displacement = total area (yahan sab axis ke upar hai, isliye sab positive).

• Phase 1 (triangle): $\frac{1}{2}\times 4\times 20=40$ m.
• Phase 2 (rectangle): $20\times 6=120$ m.
• Phase 3 (triangle): $\frac{1}{2}\times 5\times 20=50$ m.
• Total $=40+120+50=210$ m.

(c) Average velocity. KYUN yeh formula: average velocity define hota hai total displacement divided by total time se — "agar tum ek steady speed par chalte, to same displacement same time mein cover karne ke liye kya speed chahiye?" Graph par woh displacement hai hi $v$–$t$ ke neeche ka total area, aur time poori width hai, isliye: $\bar{v}=\frac{\text{total displacement}}{\text{total time}}=\frac{\text{area under whole }v\text{–}t}{\text{total width}}=\frac{210}{4+6+5}=\frac{210}{15}=14\text{m/s}.$ Dekho Instantaneous vs Average Velocity: average velocity total-displacement-over-total-time hai, naa ki phase speeds ka average.

Signed areas mein split karo. Humare sign convention se, time axis ke upar wala region $+$ hai ($+x$ mein motion), neeche wala $-$ hai ($-x$ mein motion).

• $0$–$2$ s (rectangle above): $+6\times 2=+12$ m.
• $2$–$4$ s (triangle above, $v$ from $+6$ to $0$): $+\frac{1}{2}\times 2\times 6=+6$ m.
• $4$–$6$ s (triangle below, $v$ from $0$ to $-6$): $-\frac{1}{2}\times 2\times 6=-6$ m.
• $6$–$8$ s (rectangle below): $-6\times 2=-12$ m.

(a) Displacement = net signed area: $\Delta x=+12+6-6-12=0\text{m}.$ Particle wahan khatam hota hai jahan se shuru hua tha. (b) Distance = magnitudes ka sum (below-axis area ko positive count karo): $\text{distance}=|12|+|6|+|6|+|12|=36\text{m}.$ Dekho Distance vs Displacement: yahi exactly woh jagah hai jahan donon alag hote hain kyunki velocity ka sign badl gaya.

Area under $a$–$t$ use karke ladder $a\to v$ chadho ($\Delta v=\int adt$).

• $t=4$ tak: $\Delta v=3\times 4=12\Rightarrow v(4)=2+12=14$ m/s.
• $4$–$7$: $a=0\Rightarrow$ koi change nahi $\Rightarrow v(7)=14$ m/s.
• $7$–$10$: $\Delta v=-2\times 3=-6\Rightarrow v(10)=14-6=8$ m/s.

Ab area under $v$–$t$ use karke $v\to x$ chadho (har phase ek trapezium hai; start aur end heights ka average times width).

• Phase 1 ($v:2\to 14$ over $4$ s): $\frac{2+14}{2}\times 4=8\times 4=32$ m.
• Phase 2 ($v=14$ constant, $3$ s): $14\times 3=42$ m.
• Phase 3 ($v:14\to 8$ over $3$ s): $\frac{14+8}{2}\times 3=11\times 3=33$ m.
• Total displacement $=32+42+33=107$ m.

Saari velocities positive rahi, isliye distance $=$ displacement $=107$ m. Uses Equations of Uniformly Accelerated Motion aur Differentiation and Integration in Physics.

Acceleration = slope ($a=\frac{dv}{dt}$): $a=\frac{25-5}{8}=\frac{20}{8}=2.5$ m/s². Displacement = trapezium area: $\Delta x=\frac{u+v}{2}\times t=\frac{5+25}{2}\times 8=15\times 8=120$ m. $v^2=u^2+2ax$ se check (yeh khud graph se derivable hai — parent note dekho): $v^2=5^2+2(2.5)(120)=25+600=625=25^2✓$ Third kinematic equation sirf trapezium area hai $t$ ke bina likha hua.

(a) Shape: $a=-10$ constant hai, isliye $v$–$t$ ek seedhi line hai slope $-10$ ke saath, $+20$ m/s se start karti hai. Yeh zero cross karti hai (top par) aur negative $v$ mein continue karti hai (gir rahi hai $-x$ mein, yaani neeche). $v(t)=20-10t$ (b) Top tak time = jahan $v=0$: $0=20-10t\Rightarrow t=2$ s. (Top par ball momentarily rest mein hai — $v=0$, lekin $a$ abhi bhi $-10$ hai; $a$ ke sign mein kabhi pause nahi hota.) (c) Total flight time: symmetry se down-trip up-trip ki mirror hai, isliye total $=4$ s. Check: start par wapas aane ka matlab hai net displacement $0$, isliye $\Delta x=20t-\frac{1}{2}(10)t^2=0$ solve karne par $t(20-5t)=0\Rightarrow t=4$ s ✓. (d) Max height = $0$ se $2$ s tak $v$–$t$ ke neeche ka area (triangle above axis): $\frac{1}{2}\times 2\times 20=20$ m. (e) Net displacement = $0$ se $4$ s tak net signed area: triangle above ($+20$ m) plus equal triangle below ($-20$ m) $=0$ m — haath mein wapas. Total distance $=|+20|+|-20|=40$ m $=2\times$ max height. Dekho Free Fall and Projectile Motion aur Vectors — Components and Signs ki "up $=+$" automatically har sign kaise fix karta hai.

(a) Displacement = curved $v$–$t$ graph ke neeche ka area = integral: $\Delta x={\int }_0^3{t}^{2}dt={\left[\frac{t^3}{3}\right]}_0^3=\frac{27}{3}=9\text{m}.$ (b) Triangle/trapezium kyun nahi: woh formulas assume karte hain ki top edge ek seedhi line hai (constant acceleration). Yahan $a=\frac{dv}{dt}=2t$ time ke saath badlta hai, isliye top edge curve karti hai. Ek curved edge ka reliable "area" sirf integral hai — infinitely many thin rectangles $vdt$ add karne ki limit (yahi exactly area idea hai parent note se, apni limit tak le jaaya gaya). Dekho Differentiation and Integration in Physics. Sanity check bounding se: curve ko straight chord se replace karo $(0,0)$ se $(3,9)$ tak, jo area $\frac{1}{2}\times 3\times 9=13.5$ m ka triangle deta hai. Kyunki actual curve $v=t^2$ is interval mein us chord ke neeche rehti hai, yeh straight-chord triangle real area ko over-estimate karta hai. Isliye exact answer $13.5$ m se chota hona chahiye — aur sach mein $9<13.5$ ✓. (Yahan koi under-estimate nahi hai; chord area ko upar se bound karta hai.)