Motion in a Plane: Class 11 Physics Chapter 3 Hinglish mein

Physics Part 1 · Hinglish notes, simulations aur practice · Official NCERT chapter PDF ↗

Published 18 Oct 2026 · Updated 18 Oct 2026

95 min
1 simulation
28 practice questions
29 MCQs

Is chapter mein aap kya seekhenge

Scalars aur vectors mein fark karna, aur position aur displacement ko vectors ki tarah represent karna
Vectors ko graphically aur components se add, subtract aur scale karna, aur kisi vector ko di gayi directions mein resolve karna
Resultant ka magnitude aur direction nikalne ke liye law of cosines aur law of sines use karna
Plane mein average aur instantaneous velocity aur acceleration define karna, aur unhe r(t) ko differentiate karke nikalna
Constant-acceleration equations ko vector aur component form mein lagana, aur perpendicular motions ko independently treat karna
Projectile ka path, time of flight, maximum height aur range derive karna, aur projectile problems solve karna
Uniform circular motion ka centripetal acceleration derive karna, aur v, ω, T, ν aur a_c ko aapas mein jodna

NCERT Physics Class 11, Chapter 3, pp. 27–48 par based original companion text. Yeh NCERT se juda nahi hai aur NCERT ne ise endorse nahi kiya hai.

3.1 Introduction

Crossing par idhar-udhar ghoomta auto-rickshaw, Makar Sankranti par chhaton ke upar tairti patang, board par tirchha fisalta carrom striker: inmein se koi bhi ek line par nahi chalta. Chapter 2 mein direction batane ke liye plus ya minus sign kaafi tha, kyunki line par sirf do directions hoti hain. Plane mein anginat directions hoti hain, aur ab ek sign yeh nahi bata sakta ki kaun si.

Isliye hum pehle vectors ki bhasha seekhte hain, yaani aisi quantities jo apne size ke saath ek direction bhi rakhti hain: unhe kaise add, subtract aur scale karte hain, aur axes ke along kaise todte hain. Phir hum plane mein velocity aur acceleration describe karte hain, constant acceleration wali motion solve karte hain, aur ise hawa mein udti ball aur circle mein ek jaisi speed se ghoomti cheez par lagate hain. Yeh sab ek extra coordinate jodkar three dimensions tak badh jaata hai.

3.2 Scalars aur vectors

Kuch quantities ko poori tarah describe karne ke liye ek number aur uski unit kaafi hoti hai. Tumhare school bag ka mass, aakhri bell bajne ka time, chai ke cup ka temperature aur packet mein doodh ka volume kisi direction mein point nahi karte. Aisi quantities scalars (अदिश) kehlati hain. Yeh normal arithmetic se judti hain: 4.5 kg ka bag aur 1.2 kg ka lunchbox milkar 5.7 kg hote hain, aur 40 minute ki class ke baad 45 minute ki class kul 85 minute leti hai. Do scalars ko multiply ya divide karna hamesha meaningful hai, chahe unki units kuch bhi hon: 120 km ko 2.0 h se divide karne par 60 km/h ki speed milti hai, jo ek naya scalar hai. Lekin unhe add ya subtract karna tabhi meaningful hai jab units same hon.

Kuch doosri quantities direction ke bina koi matlab nahi rakhtin. “300 m chalo” kehne par tum 300 m radius ke circle par kahin bhi pahunch sakte ho; “mandir ke gate ki taraf 300 m chalo” ek hi jagah fix kar deta hai. Ek vector (सदिश) ka magnitude aur direction hota hai, aur woh apni kism ke doosre vectors se Section 3.4 ke triangle (ya parallelogram) rule se judta hai. Yeh aakhri shart zaroori hai. Wire mein electric current ki wire ke along ek disha hoti hai, lekin junction par milne wale do currents seedhe numbers ki tarah add hote hain, isliye current ek scalar hai. Displacement, velocity, acceleration aur force triangle rule follow karte hain; yeh vectors hain.

Print mein hum vector ko bold mein likhte hain, $𝐯$ , aur uska magnitude normal italic type mein, $v$ , ya $∣ 𝐯 ∣$ ki tarah. Haath se likhte waqt letter ke upar ek chhota arrow banao. Magnitude kabhi negative nahi hota: yeh kisi chune hue scale par arrow ki length hai.

Compare

Scalars aur vectors

Scalars

Sirf magnitude: unit ke saath ek number
Normal algebra se judte hain
Jaise mass, time, temperature, speed, energy, current

Vectors

Magnitude aur direction
Triangle law se judte hain
Jaise displacement, velocity, acceleration, force

3.2.1 Position aur displacement vectors

Plane mein koi cheez kahan hai, yeh batane ke liye pehle ek reference point chuno, origin O. Phir kisi bhi point P tak O se P tak bane ek arrow se pahuncha jaata hai. Woh arrow P ka position vector $𝐫$ hai. Yeh O se P ki taraf point karta hai, aur iski length seedhi line wali distance OP hai. Origin badlo to har position vector badal jaata hai, isliye hamesha batao ki O kahan hai.

Ab maan lo object time $t$ par P se chalkar baad ke time $t^{'}$ par P′ tak pahunchta hai. P se P′ tak ka arrow uska displacement (विस्थापन) hai. Yeh sirf position ka badlaav record karta hai, yaani object kahan se shuru hua aur kahan khatam hua, aur beech ke raaste ke baare mein kuch nahi batata. Fig. 3.1 ek housing colony mein ghar H se school S tak teen raaston ko follow karta hai. Raaston ki lengths alag hain, phir bhi teeno ka displacement ek hi hai: H se S tak seedha arrow.

4/4Har raaste ka displacement same hai, H se S tak 500 m. Iska magnitude kabhi path length se zyada nahi hota.

Fig. 3.1Teen raaste, teen path lengths, ek displacement.NCERT Fig. 3.1 · p. 28

Upar se dikhta square blocks wala street grid, jismein ghar H neeche left mein hai aur school S usse chaar blocks east aur teen blocks north mein. Ek scale bar har block ko 100 m dikhata hai aur ek arrow north dikhata hai. Step 1: ek solid blue raasta neeche wali street par east jaata hai aur phir north mein S tak, label 700 m. Step 2: ek dashed orange raasta ek block east aage nikal jaata hai, north jaata hai aur phir west mein wapas S par aata hai, label 900 m. Step 3: ek dotted green raasta blocks ke beech se mudta hua jaata hai, label lagbhag 590 m. Step 4: teeno raaste halke ho jaate hain aur ek mota vermillion arrow H se seedha S tak jaata hai, label sab: 500 m.

Isse ek kaam ki inequality milti hai. Displacement ka magnitude kabhi path length se zyada nahi ho sakta, aur dono sirf tab barabar hote hain jab motion seedhi line mein ho aur wapas na mude. Isi tarah, kisi bhi trip mein average speed (path length ÷ time) kam se kam average velocity ke magnitude (displacement ÷ time) jitni badi hoti hai. Din khatam hone par shop par wapas aane wali delivery rider ne shayad 60 km chalaya ho, lekin us din ka uska displacement zero hai.

Ise bahar try karo

Kadmon se naapo

Tumhe chahiye

aasani se dikhne wale do landmarks, jaise ek gate aur ek nal
ek measuring tape (optional)
ek notebook

Pehle guess karo

Tum gate se nal tak teen alag raaston se chalte ho. Kaun sa statement sahi hai?

Sabse lamba raasta sabse bada displacement deta haiTeeno raaste same displacement dete hainDisplacement sabse chhote raaste ke steps ki ginti ke barabar hai

Apne school ground ya park mein do fixed jagah chuno, jaise gate (G) aur paani ka nal (T).
G se T tak apni pasand ke teen alag raaston se chalo aur har baar apne steps gino.
Ab G se T tak seedhi line mein chalo aur phir se gino.
Tape se apne 10 steps ki length naapo, aur har ginti ko metres mein badlo.

Table 3.1

Route	Steps	Metres
Route 1
Route 2
Route 3
G se T tak seedhi line

Tumhari entries apne aap save hoti hain.

Abhi nahi kar sakte? Simulation try karo · Fig. 3.10

3.2.2 Vectors ki equality

Do vectors tab equal hote hain jab unke magnitudes match karein aur unki directions bhi match karein; page par woh kahan bane hain, isse fark nahi padta. Ek quick test: ek arrow ko bina ghumaye tab tak khiskao jab tak uski tail doosre ki tail par na aa jaaye. Agar heads bhi mil jaayein, to vectors equal hain (Fig. 3.2). Same length ke do arrows jo alag directions mein point karte hain, yeh test fail karte hain, isliye magnitudes match hone par bhi woh alag vectors hain.

3/3 $𝐂$ ka magnitude same hai lekin direction alag, isliye uska head nahi milta: $𝐂 \neq 𝐀$ .

Fig. 3.2Bina ghumaye khiskao: equal vectors head se head milte hain.NCERT Fig. 3.2 · p. 29

Barabar length ke teen arrows. Picture ke left mein ek blue arrow A upar right ki taraf jaata hai, ek dashed blue arrow B usi direction mein thoda aage right mein point karta hai, aur neeche ek orange arrow C horizontally right ki taraf point karta hai. Step 1: teeno arrows apni-apni jagah par hain. Step 2: B bina ghume khisakta hai jab tak uski tail A ki tail par na aa jaaye; uska head theek A ke head par aata hai aur ek note kehta hai heads mile: equal. Step 3: B wapas apni jagah par hai aur ab C khisak kar A ki tail par aata hai; heads alag jagah par khatam hote hain aur note kehta hai heads alag: unequal.

Ek bhaari darwaaze ko hinges ke paas dhakka do, phir utni hi zor se handle ke paas dhakka do. Force vector bilkul same hai, phir bhi pehli baar darwaaza mushkil se hilta hai. Aise force ke liye point of application matter karta hai, aur kisi point ya line of action se bandhe vector ko localised vector kehte hain; torque ke liye Chapter 6 mein tumhe yeh idea chahiye hoga. Is chapter ke vectors free vectors hain: sirf magnitude aur direction maayne rakhte hain, isiliye unhe idhar-udhar khiskana allowed hai.

3.3 Vectors ka real numbers se multiplication

Kisi vector ko ek normal number λ se multiply karne par woh khinch jaata hai ya sikud jaata hai. Agar λ positive hai, to $λ 𝐀$ usi taraf point karta hai jidhar $𝐀$ , aur λ guna lamba hota hai. Agar λ negative hai, to arrow ghoom kar ulti direction mein bhi point karne lagta hai. Dono cases mein naya magnitude hai

∣ λ 𝐀 ∣ = ∣ λ ∣ A

−1 se multiply karne par vector bas ulta ho jaata hai: $- 𝐀$ ka magnitude $𝐀$ jitna hi hai aur direction ulti. Fig. 3.3 $𝐀$ , $2 𝐀$ , $- 𝐀$ aur $- 1.5 𝐀$ ko ek common tail se line mein dikhata hai taaki tum grid par unki lengths compare kar sako.

4/4 $- 1.5 𝐀$ : ulta aur 1.5 guna lamba.

Fig. 3.3Positive multipliers khinchte hain; negative wale ulta bhi kar dete hain.NCERT Fig. 3.3 · p. 29

Rows mein chaar arrows, sab ek vertical dashed line se shuru hote hain, halki vertical grid lines ke upar jo A ki ek length ki doori par hain. Step 1: sirf blue arrow A, do grid squares lamba, right ki taraf point karta hua. Step 2: uske neeche ek green arrow 2A aata hai, right ki taraf aur double lamba. Step 3: ek orange arrow minus A aata hai, left ki taraf aur A jitna hi lamba. Step 4: ek purple arrow minus 1.5A aata hai, left ki taraf aur A se dedh guna lamba.

Multipliers ki apni units bhi ho sakti hain. 15 m/s due north ki steady velocity ko 4.0 s ke time se multiply karo to 60 m due north milta hai: yeh ek displacement hai, kyunki $(m / s) \times s = m$ . Chapter 4 mein, acceleration ko mass se multiply karne par isi tarah force milega.

Quick check

2 mein se 1

𝐀

4 units lamba hai aur east ki taraf point karta hai.

- 2.5 𝐀

kya hai?

10 units, east ki taraf10 units, west ki taraf1.6 units, west ki taraf6.5 units, west ki taraf

Answer dekho

Answer (B) 10 units, west ki taraf

Magnitude

∣ - 2.5 ∣ \times 4 = 10

units hai, aur minus sign direction ko ulta kar deta hai, isliye result west ki taraf point karta hai. Magnitude kabhi negative nahi hota; sign direction ke badlaav ke roop mein dikhta hai.

3.4 Vectors ka addition aur subtraction — graphical method

4 m east chalo aur phir 3 m north, to tum apne starting point se 5 m door, east aur north ke beech ki kisi direction mein pahunchte ho. Start se finish tak ek seedhi walk wahi kaam kar deti jo dono walks milkar karti hain. Vectors ko add karne ka yahi matlab hai. Paper par $𝐀$ mein $𝐁$ add karne ke liye, $𝐀$ ko scale se banao, phir $𝐁$ wahan se shuru karo jahan $𝐀$ khatam hota hai. $𝐀$ ke starting point ko $𝐁$ ke end point se jodo: yeh arrow sum hai, yaani resultant (परिणामी), $𝐑 = 𝐀 + 𝐁$ (Fig. 3.4). Arrows ko nose se tail milakar banane par ek triangle band hota hai, isliye is method ke do naam hain: head-to-tail ya triangle method.

4/4Pehle $𝐁$ aur phir $𝐀$ add karne par bhi same point par pahunchte hain: addition commutative hai, aur dono raaste milkar ek parallelogram banate hain.

Fig. 3.4Triangle method: head se tail, phir tail se head.NCERT Fig. 3.4 · p. 30

Step 1: ek blue arrow A point O se shuru hokar right aur thoda upar point karta hai; ek orange arrow B, jo upar aur right ki taraf point karta hai, usse alag neeche rakha hai. Step 2: B bina ghume tab tak khisakta hai jab tak uski tail A ke head par na aa jaaye. Step 3: ek mota vermillion arrow R, O ko B ke door wale end se jodta hai aur triangle band karta hai, label R = A + B. Step 4: dashed copies dikhati hain ki B pehle O se aur A uske baad bana hai; dono same head par pahunchte hain, isliye dono raaste ek parallelogram banate hain jiska diagonal R hai, label A + B = B + A.

Pehle $𝐁$ aur phir $𝐀$ banao, to tum same final point par pahunchte ho: dono orders ek parallelogram ke do halves banate hain jiska diagonal $𝐑$ hai. Isliye vector addition commutative hai. Teen ya zyada vectors ke saath grouping se bhi fark nahi padta; yeh associative law hai.

𝐀 + 𝐁 = 𝐁 + 𝐀

(3.1)

\begin{aligned} (𝐀 + 𝐁) + 𝐂 \\ = 𝐀 + (𝐁 + 𝐂) \end{aligned}

(3.2)

Kisi vector ko uske ulte vector se add karo to tum wahin wapas aa jaate ho jahan se shuru kiya tha. Result null vector (zero vector) $𝟎$ hai, yaani zero length ka arrow. Length na hone ki wajah se iski koi fixed direction nahi hoti.

\begin{array}{c} 𝐀 + (- 𝐀) = 𝟎 \\ ∣ 𝟎 ∣ = 0 \end{array}

(3.3)

Vectors mein null vector waise hi behave karta hai jaise numbers mein zero:

\begin{array}{c} 𝐀 + 𝟎 = 𝐀 \\ λ 𝟎 = 𝟎, 0 𝐀 = 𝟎 \end{array}

(3.4)

Physically, null vector kisi bhi round trip ka displacement hai. Ek school bus jo apne depot se nikalti hai, apna route chalati hai aur usi bay mein park hoti hai, usne shayad 40 km cover kiye hon, phir bhi us din ka uska displacement $𝟎$ hai.

Subtraction ko addition se define kiya jaata hai: $𝐁$ ko subtract karne ke liye uska ulta vector add karo.

𝐀 - 𝐁 = 𝐀 + (- 𝐁)

(3.5)

Fig. 3.5 isi tarah $𝐀 - 𝐁$ banata hai aur use $𝐀 + 𝐁$ ke saath rakhta hai. Agar dono vectors ek point se banakar parallelogram poora karo, to ek diagonal sum hai, aur doosra, jo $𝐁$ ke head se $𝐀$ ke head tak jaata hai, difference hai.

4/4Parallelogram poora karne par: ek diagonal $𝐀 + 𝐁$ hai, aur doosra, $𝐁$ ke head se $𝐀$ ke head tak, $𝐀 - 𝐁$ ke barabar hai.

Fig. 3.5Parallelogram ke do diagonals: sum aur difference.NCERT Fig. 3.5 · p. 30

Ek common point O se do arrows shuru hote hain: ek blue arrow A jo right aur thoda upar point karta hai, aur ek orange arrow B jo tezi se upar aur right ki taraf point karta hai. Step 1: sirf A aur B dikhte hain. Step 2: A ke head se ek dashed orange arrow minus B banta hai, B jitna lamba lekin ulta, jo neeche aur left ki taraf point karta hai. Step 3: ek purple arrow A minus B, O se minus B ke head tak jaata hai. Step 4: A aur B par parallelogram dashed sides se poora hota hai; ek vermillion arrow A plus B, O se door wale corner tak jaata hai, aur ek dashed purple diagonal B ke head ko A ke head se jodta hai, jo A minus B ke parallel aur barabar hai.

Yahi picture ek doosra graphical tareeka deti hai, parallelogram method. $𝐀$ aur $𝐁$ ki tails ko ek point O par milao, har arrow ke head se doosre arrow ke parallel ek line kheencho, aur O se parallelogram ke door wale corner tak ka diagonal $𝐀 + 𝐁$ hai. Parallelogram ka aadha hissa theek head-to-tail triangle hai, isliye dono methods hamesha same answer dete hain. Tum dono ko vector lab (Fig. 3.8) mein try kar sakte ho.

Example 3.1NCERT Example 3.1 · p. 31

River ke paar tairna

Meera 60 m chaudi river ke paar tairti hai. Woh paani ke relative 1.20 m/s se tairti hai, hamesha seedha saamne (due north) face karte hue, jabki current paani ko due east 0.50 m/s se le jaata hai. (a) Bank ke relative uski velocity kya hai? (b) Paar karne mein use kitna time lagta hai, aur woh kitni door downstream pahunchti hai?

Pehle khud try karo. Phir solution ek-ek step karke kholo.

3.5 Vectors ka resolution

Addition do vectors ko ek bana deta hai. Ulta kaam, yaani ek vector ko chuni hui directions mein tukdon mein todna, resolution kehlata hai, aur isse hum vector sums ruler ke bajaye arithmetic se kar paate hain. Plane mein do directions chuno, jo vectors $𝐚$ aur $𝐛$ se di gayi hain aur parallel nahi hain, aur vector $𝐀$ ko O se P tak banao. Ab do guide lines kheencho: ek tail O se $𝐚$ ke parallel, aur ek head P se $𝐛$ ke parallel. Jahan yeh milti hain, wahan Q mark karo (Fig. 3.6). Ab $𝐀$ , arrows OQ aur QP ka sum hai. Pehla $𝐚$ ke along hai, isliye yeh $𝐚$ ka koi multiple λ hai; doosra $𝐛$ ka multiple μ hai:

𝐀 = λ 𝐚 + μ 𝐛

(3.8)

3/3OQ $= λ 𝐚$ aur QP $= μ 𝐛$ , isliye $𝐀 = λ 𝐚 + μ 𝐛$ (yahan λ ≈ 2.0, μ ≈ 2.4).

Fig. 3.6Tail aur head se kheenchi gayi parallels kisi vector ko kinhi bhi do directions mein tod deti hain.NCERT Fig. 3.8 · p. 31

Upar left mein do chhote reference arrows hain: blue mein a, jo right aur thoda upar point karta hai, aur orange mein b, jo tezi se upar point karta hai. Ek lamba vermillion arrow A neeche point O se upar right mein point P tak jaata hai. Step 1: sirf a, b aur A. Step 2: O se a ke parallel ek dashed line aur P se b ke parallel ek dashed line banti hai; yeh ek marked point Q par cross karti hain. Step 3: A thoda halka hota hai aur do mote arrows uski jagah lete hain, O se Q tak ek blue arrow jiska label lambda a hai aur Q se P tak ek orange arrow jiska label mu b hai, equation A = lambda a + mu b ke saath.

Vectors $λ 𝐚$ aur $μ 𝐛$ ko $𝐚$ aur $𝐛$ ke along $𝐀$ ke components kehte hain. Plane mein non-parallel directions ke kisi bhi pair ke liye yeh split exist karta hai aur unique hota hai.

Unit vectors

Directions ko handle karna unit vectors (एकांक सदिश) se sabse aasaan hai: yeh theek 1 magnitude wale vectors hain, bina unit aur bina dimension ke, jinka kaam sirf point karna hai. Right-handed axes ke set mein, $+ x$ , $+ y$ aur $+ z$ ke along point karne wale teen unit vectors ko $\hat{𝐢}$ , $\hat{𝐣}$ aur $\hat{𝐤}$ likhte hain. Yeh ek doosre ke perpendicular hain, aur

∣ \hat{𝐢} ∣ = ∣ \hat{𝐣} ∣ = ∣ \hat{𝐤} ∣ = 1

(3.9)

Har vector apne magnitude aur apne along unit vector $\hat{𝐧}$ ka product hai, isliye $\hat{𝐧} = 𝐀 / A$ . Jaise, north-east ki taraf 20 m/s ki velocity, $20 m / s$ guna north-east ki taraf point karne wala unit vector hai.

𝐀 = A \hat{𝐧}

(3.10)

Rectangular components

Ab khud axes ke along resolve karo. Maan lo $𝐀$ , $x$ – $y$ plane mein positive $x$ -axis se θ angle par hai, aur uske head se dono axes par perpendiculars giraao (Fig. 3.7). $x$ -axis ke along wala tukda $\hat{𝐢}$ ka multiple hai, aur $y$ -axis ke along wala tukda $\hat{𝐣}$ ka multiple hai:

𝐀 = A_{x} \hat{𝐢} + A_{y} \hat{𝐣}

(3.12)

Numbers $A_{x}$ aur $A_{y}$ , $𝐀$ ke $x$ - aur $y$ -components (घटक) hain. Yeh fark saaf rakho: $A_{x}$ unit ke saath ek number hai, jabki $A_{x} \hat{𝐢}$ ek vector hai. Right-angled trigonometry magnitude aur angle se components deti hai:

\begin{array}{c} A_{x} = A \cos θ \\ A_{y} = A \sin θ \end{array}

(3.13)

3/4 $A_{x} = A \cos θ$ aur $A_{y} = A \sin θ$ ; dono milkar, $𝐀 = A_{x} \hat{𝐢} + A_{y} \hat{𝐣}$ .

Fig. 3.7Components axes par vector ki parchhaiyan hain, aur unke signs hote hain.NCERT Fig. 3.9 · p. 32

Origin O ke saath coordinate axes. Step 1: fixed length ka ek vermillion arrow A, O se positive x-axis ke lagbhag 35 degrees upar angle theta par nikalta hai, angle ek arc se marked hai. Step 2: A ke head se dono axes par dashed perpendiculars girte hain. Step 3: x-axis ke along ek blue arrow, label A_x = A cos theta, aur y-axis par upar ek green arrow, label A_y = A sin theta, dono components dikhate hain. Step 4: A ko 140 degrees par, second quadrant mein, dobara banaya jaata hai; ab uska x-component negative x-axis ke along left point karta hai, label A_x less than 0, jabki uska y-component ab bhi upar point karta hai.

Component positive, negative ya zero ho sakta hai; kaun sa, yeh θ par depend karta hai. $x$ -axis se 150° par vector ke liye $\cos 150^{\circ} < 0$ hai, isliye uska $x$ -component negative hai: vector $- x$ ki taraf jhukta hai. Isliye plane mein ek vector do equivalent tareekon se diya ja sakta hai: uske magnitude aur angle $(A, θ)$ se, ya uske components $(A_{x}, A_{y})$ se. Components se wapas magnitude aur angle par jaane ke liye Pythagoras aur tangent use hote hain:

A = \sqrt{A_{x}^{2} + A_{y}^{2}}

(3.14)

\tan θ = \frac{A_{y}}{A_{x}}

(3.15)

Three dimensions mein vectors

Space mein ek vector ke teen components hote hain. Agar $𝐀$ , $x$ -, $y$ - aur $z$ -axes se angles α, β aur γ banata hai, to

\begin{array}{c} A_{x} = A \cos α \\ A_{y} = A \cos β \\ A_{z} = A \cos γ \end{array}

(3.16a)

\begin{aligned} 𝐀 & = A_{x} \hat{𝐢} + A_{y} \hat{𝐣} + A_{z} \hat{𝐤} \\ A & = \sqrt{A_{x}^{2} + A_{y}^{2} + A_{z}^{2}} \end{aligned}

(3.16b)

Inmein se har angle us plane mein naapa jaata hai jismein $𝐀$ aur woh axis dono hain, isliye teeno angles ek flat drawing mein nahi aate. Coordinates $(x, y, z)$ wale point ka position vector hai

𝐫 = x \hat{𝐢} + y \hat{𝐣} + z \hat{𝐤}

(3.17)

Ek delivery drone jo apne launch pad se 40 m east aur 20 m north wali jagah ke 30 m upar hover kar raha hai, uska $𝐫 = (40 \hat{𝐢} + 20 \hat{𝐣} + 30 \hat{𝐤}) m$ hai, jahan $x$ east, $y$ north aur $z$ upar hai. Pad se uski distance $\sqrt{40^{2} + 20^{2} + 30^{2}} \approx 53.9 m$ hai.

Quick check

3 mein se 1

60 N ka ek force

+ x

-axis se 30° upar lagta hai. Newtons mein uska

y

-component kitna hai?

Answer dekho

Answer 30 N

F_{y} = F \sin 30^{\circ} = 60 \times 0.5 = 30 N

x

-component

60 \cos 30^{\circ} \approx 52 N

hai.

3.6 Vectors ka addition — analytical method

Ruler aur protractor se tum kinhi bhi do vectors ko add kar sakte ho, lekin har sum mein time lagta hai aur usmein tumhari drawing ki galtiyan aa jaati hain. Components ke saath kaam karna zyada fast hai, aur utna hi exact jitna tumhara arithmetic.

DerivationComponents se vectors add karna

Dono vectors ko unke components ke terms mein likho:

\begin{array}{c} 𝐀 = A_{x} \hat{𝐢} + A_{y} \hat{𝐣} \\ 𝐁 = B_{x} \hat{𝐢} + B_{y} \hat{𝐣} \end{array}

(3.18)

Inhe add karo, aur commutative aur associative laws use karke $\hat{𝐢}$ wale terms ko ek saath aur $\hat{𝐣}$ wale terms ko ek saath rakho:

\begin{aligned} 𝐑 & = (A_{x} + B_{x}) \hat{𝐢} \\ + (A_{y} + B_{y}) \hat{𝐣} \end{aligned}

Ise $𝐑 = R_{x} \hat{𝐢} + R_{y} \hat{𝐣}$ se compare karne par milta hai

\begin{array}{c} R_{x} = A_{x} + B_{x} \\ R_{y} = A_{y} + B_{y} \end{array}

(3.21)

Yaani vectors component by component add hote hain: $x$ ke saath $x$ , $y$ ke saath $y$ . Yeh rule turant aage badh jaata hai. Three dimensions mein $z$ -components bhi add karo ( $R_{z} = A_{z} + B_{z}$ ). Kitne bhi vectors ke liye, subtraction ke saath bhi, components ko unhi signs ke saath jodo jo vectors ke hain. $𝐓 = 𝐚 + 𝐛 - 𝐜$ ke liye,

\begin{array}{c} T_{x} = a_{x} + b_{x} - c_{x} \\ T_{y} = a_{y} + b_{y} - c_{y} \\ T_{z} = a_{z} + b_{z} - c_{z} \end{array}

(3.23)

Vector lab mein tum is rule ko, aur Section 3.4 ke graphical methods ko, apni pasand ke kinhi bhi do vectors par check kar sakte ho. Arrow heads ko drag karo aur components ki table ko drawing se compare karo.

InteractiveVector addition lab

Fig. 3.8

𝐀

aur

𝐁

ke heads drag karo. Triangle method, parallelogram method aur difference

𝐀 - 𝐁

ke beech switch karo; table har component deti hai.

Example

Delivery run ke teen hisse

Ek delivery rider 4.0 km east jaati hai, phir east se 60° north ki direction mein 3.0 km, phir north-west ki taraf 2.0 km. Uska kul displacement nikalo.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 3.2NCERT Example 3.2 · p. 33

Resultant ka magnitude aur direction

Do vectors $𝐀$ aur $𝐁$ ke magnitudes $A$ aur $B$ hain, aur unke beech ka angle θ hai. $𝐑 = 𝐀 + 𝐁$ ka magnitude $R$ aur $𝐑$ ka $𝐀$ ke saath angle α nikalo. $A = 3$ , $B = 4$ aur $θ = 60 °$ ke liye apna answer check karo.

Fig. 3.9Head to tail bana

𝐑 = 𝐀 + 𝐁

; SN triangle ki height hai.

Ek blue arrow A, horizontal line ke along O se P tak jaata hai. P se ek orange arrow B, OP ke extension se theta angle upar S tak jaata hai. Ek vermillion arrow R, O ko S se jodta hai aur OP ke saath alpha angle banata hai. Ek dashed line OP ko N tak badhaati hai, jo S ke theek neeche hai, aur S se N tak ek dashed vertical line right-angle mark ke saath girti hai. Labels PN = B cos theta aur SN = B sin theta dete hain.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

In do results ke naam hain. Fig. 3.9 ke triangle ke angles ke saath (β, $𝐑$ aur $𝐁$ ke beech ka angle hai), yeh law of cosines aur law of sines hain:

Law of cosines

\begin{aligned} R^{2} & = A^{2} + B^{2} \\ + 2 A B \cos θ \end{aligned}

(3.24a)

Law of sines

\frac{R}{\sin θ} = \frac{A}{\sin β} = \frac{B}{\sin α}

(3.24d)

\tan α = \frac{B \sin θ}{A + B \cos θ}

(3.24f)

Agar tum trigonometry se cosine rule ko $c^{2} = a^{2} + b^{2} - 2 a b \cos C$ ke roop mein jaante ho, to yahan plus sign koi contradiction nahi hai: P par triangle ka interior angle $180^{\circ} - θ$ hai, aur $\cos (180^{\circ} - θ) = - \cos θ$ . Extremes check karo: θ = 0 se $R = A + B$ milta hai, θ = 180° se $R = ∣ A - B ∣$ , aur θ = 90° se Pythagoras.

Example 3.3NCERT Example 3.3 · p. 34

Crosswind mein ek aircraft

Ek light aircraft 240 km/h ki airspeed se due east ja raha hai. 60 km/h ki steady hawa south-east se chal rahi hai, yaani north-west ki taraf. Zameen ke relative aircraft ki velocity nikalo.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Quick check

3 mein se 1

𝐀 = 3 \hat{𝐢} - 2 \hat{𝐣}

aur

𝐁 = - \hat{𝐢} + 6 \hat{𝐣}

, units mein.

𝐀 + 𝐁

ka magnitude kya hai?

units

Answer dekho

Answer 4.47 units

𝐀 + 𝐁 = 2 \hat{𝐢} + 4 \hat{𝐣}

, isliye magnitude

\sqrt{4 + 16} = \sqrt{20} \approx 4.47

hai.

3.7 Plane mein motion

Vectors haath mein aane ke baad hum aisi motion describe kar sakte hain jo ek line tak simit nahi hai: board par fisalta carrom striker, mandir ke gumbad ke chakkar lagata kabootar, mod leta scooter. Chapter 2 ke teen ideas, displacement, velocity aur acceleration, yahan bhi kaam aate hain, bas signed numbers ki jagah vectors aa jaate hain.

3.7.1 Position vector aur displacement

Kisi bhi suvidhajanak point par origin rakhkar $x$ – $y$ axes banao. Coordinates $(x, y)$ wale point P par ek particle ka position vector $𝐫 = x \hat{𝐢} + y \hat{𝐣}$ hai; kisi point ke coordinates bas uske position vector ke components hain. Agar particle time $t$ par P par hai aur baad ke time $t^{'}$ par P′ par, to uska displacement hai

Δ 𝐫 = 𝐫^{'} - 𝐫

(3.25)

Component by component subtract karne par,

Δ 𝐫 = Δ x \hat{𝐢} + Δ y \hat{𝐣}

(3.26)

jahan

$Δ x$: $x^{'} - x$ , $x$ -coordinate mein badlaav
$Δ y$: $y^{'} - y$ , $y$ -coordinate mein badlaav

Simulation mein O fixed hai jabki P aur Q ko drag kiya ja sakta hai. Segments OP aur OQ ki lengths dono position vectors ki lengths hain, PQ ki length P se Q tak ke displacement ki length hai, aur har segment par right-triangle ki legs uske $x$ - aur $y$ -components dikhati hain.

SimulationPosition vectors aur displacement

Dono axes par −10 se 10 tak ka square coordinate grid, jismein ek fixed origin O aur do draggable points hain, P (−4, 2) par aur Q (4, 8) par. Segments O ko P se, O ko Q se aur P ko Q se jodte hain, har ek par uski length ka label hai, aur right-triangle ki legs har segment ke x- aur y-components dikhati hain. Kisi point ko drag karne par uske coordinates aur har length update hoti hai.

Fig. 3.10P aur Q ko drag karo. OP aur OQ position vectors hain; PQ displacement hai, legs Δx aur Δy ke saath.NCERT Fig. 3.12 · p. 35

Try karo

Q ko tab tak drag karo jab tak displacement PQ 5 units lamba na ho jaaye aur Δx = 3 ho. Kitne grid points kaam karte hain?
P aur Q ko same direction mein same amount se khiskao. Kaun si lengths badalti hain, aur kaun si fixed rehti hai?
OQ ko OP se chhota karo jabki PQ 10 units lamba rahe.

Velocity

Interval $Δ t = t^{'} - t$ par average velocity displacement ko liye gaye time se divide karne par milti hai:

\begin{aligned} \overset{ˉ}{𝐯} & = \frac{Δ 𝐫}{Δ t} \\ = \frac{Δ x}{Δ t} \hat{𝐢} + \frac{Δ y}{Δ t} \hat{𝐣} \end{aligned}

(3.27)

Positive scalar se divide karne par direction nahi badalti, isliye $\overset{ˉ}{𝐯}$ , $Δ 𝐫$ ke along point karta hai, yaani P se P′ tak ki chord ke along. Velocity (instantaneous velocity) average ki limit hai jab interval zero tak chhota ho jaata hai:

𝐯 = \lim_{Δ t \to 0} \frac{Δ 𝐫}{Δ t} = \frac{d 𝐫}{d t}

(3.28)

Limit direction ke saath kya karti hai? Fig. 3.11 P ko fixed rakhta hai aur baad wale point ko path ke along uski taraf wapas khiskne deta hai. Chord ghoomti hai, aur limit mein woh P par path ki tangent ke along aa jaati hai. Isliye path ke har point par velocity path ki tangent hoti hai aur us taraf point karti hai jidhar object chal raha hai. Grinding wheel se udti chingaariyan theek isi wajah se tangents ke along nikalti hain.

4/4Limit mein direction P par tangent hai: yahi $𝐯$ ki direction hai.

Fig. 3.11Jaise Δt → 0, chord PP′, P par tangent ban jaati hai.NCERT Fig. 3.13 · p. 36

Ek curved path neeche left se uthta hai, arch banata hai aur right mein neeche aata hai, uske neeche left end ke paas ek fixed point P hai. Step 1: ek blue arrow, yaani displacement, P se path par kaafi aage arch ke us paar ek point P1 tak jaata hai. Step 2: pehli chord halki hoti hai aur doosri chord top ke paas ek nazdeek ke point P2 tak jaati hai; yeh zyada tirchhi hai. Step 3: teesri chord aur bhi paas ke point tak jaati hai. Step 4: chords halki hain aur ek mota vermillion arrow v, P se path ki tangent ke along nikalta hai, wahi direction jiski taraf chords badh rahi thin.

Components mein, har term ki limit alag-alag lo:

v_{x} = \frac{d x}{d t}, v_{y} = \frac{d y}{d t}

(3.30a)

Isliye agar tumhe $x (t)$ aur $y (t)$ pata hain, to differentiate karne se velocity milti hai. Iska magnitude, yaani speed, aur $x$ -axis ke saath iska angle θ kisi bhi vector ki tarah nikalte hain:

\begin{array}{c} v = \sqrt{v_{x}^{2} + v_{y}^{2}} \\ \tan θ = \frac{v_{y}}{v_{x}} \end{array}

(3.30b)

Acceleration

Acceleration velocity se theek waise hi banta hai jaise velocity position se bani thi. Interval $Δ t$ par average acceleration hai

\begin{aligned} \overset{ˉ}{𝐚} & = \frac{Δ 𝐯}{Δ t} \\ = \frac{Δ v_{x}}{Δ t} \hat{𝐢} + \frac{Δ v_{y}}{Δ t} \hat{𝐣} \end{aligned}

(3.31)

aur acceleration (instantaneous acceleration) uski limit hai:

\begin{array}{c} 𝐚 = \lim_{Δ t \to 0} \frac{Δ 𝐯}{Δ t} = \frac{d 𝐯}{d t} \\ a_{x} = \frac{d v_{x}}{d t} = \frac{d^{2} x}{d t^{2}} \\ a_{y} = \frac{d v_{y}}{d t} = \frac{d^{2} y}{d t^{2}} \end{array}

(3.32)

Average acceleration $Δ 𝐯$ ke along point karta hai, jo tum triangle rule se nikalte ho (Fig. 3.12). Zaroori nahi ki yeh motion ki direction ho, aur yeh nayi baat hai. Line ke along, $𝐯$ aur $𝐚$ sirf same direction (speed badhna) ya ulti directions (speed ghatna) mein point kar sakte hain. Plane mein woh 0° aur 180° ke beech koi bhi angle bana sakte hain. Jab yeh theek 90° hota hai, to speed steady rehti hai aur direction mudti hai, jaise Section 3.10 mein.

4/4 $\overset{ˉ}{𝐚} = Δ 𝐯 / Δ t$ , $Δ 𝐯$ ke along point karta hai, yahan path ke upar aur left ki taraf.

Fig. 3.12Acceleration Δv ko follow karta hai, motion ki direction ko nahi.NCERT Fig. 3.15 · p. 37

Ek curved path neeche left se upar ki taraf mudta hai. Step 1: point P par ek blue arrow v path ke along point karta hai. Step 2: baad ke point P-prime par ek lamba green arrow v-prime path ke along ek nayi, zyada tirchhi direction mein point karta hai. Step 3: right mein dono velocities ek common tail se dobara banti hain, aur ek vermillion arrow delta v, v ke head ko v-prime ke head se jodta hai. Step 4: P aur P-prime ke beech path par ek purple arrow, yaani average acceleration, banta hai jo delta v ki direction mein, upar aur left, point karta hai, path ke along nahi.

Example 3.4NCERT Example 3.4 · p. 37

Position vector se velocity aur acceleration

Ek particle aise chalta hai ki $𝐫 = (2.0 t^{2} \hat{𝐢} + 6.0 t \hat{𝐣} + 4.0 \hat{𝐤}) m$ , jahan $t$ seconds mein hai aur har coefficient ki zaroori unit hai. (a) $𝐯 (t)$ aur $𝐚 (t)$ nikalo. (b) $t = 2.0 s$ par speed aur motion ki direction nikalo. (c) Us pal $𝐯$ aur $𝐚$ ke beech angle kya hai?

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Quick check

3 mein se 1

Ek car ek curve par mudti hai. Kisi bhi pal uski velocity point karti hai

curve ke centre ki tarafcurve ke end tak ki chord ke alongcurve ke start se line ke alongcurve ki tangent ke along, motion ki direction mein

Answer dekho

Answer (D) curve ki tangent ke along, motion ki direction mein

Velocity Δr/Δt ki limit hai, aur limit mein chord tangent ban jaati hai. Curve ka centre woh jagah hai jidhar acceleration (velocity nahi) point karta hai jab speed steady ho.

3.8 Constant acceleration ke saath plane mein motion

Aksar acceleration kuch der tak size aur direction mein fixed rehta hai: hawa mein udti ball $𝐠$ se accelerate karti hai, jhuki hui air table par puck slope ke neeche ek steady push mehsoos karta hai. Maan lo $𝐚$ constant hai aur $t = 0$ par particle ki velocity $𝐯_{0}$ aur position $𝐫_{0}$ hai.

DerivationConstant acceleration ki equations

Constant acceleration mein kisi bhi interval par average acceleration $𝐚$ ke barabar hota hai. $0$ aur $t$ ke beech, $𝐚 = (𝐯 - 𝐯_{0}) / t$ , isliye

𝐯 = 𝐯_{0} + 𝐚 t

(3.33a)

Velocity time ke saath linearly badalti hai, isliye interval par uska average uski end values ka mean hai, $(𝐯_{0} + 𝐯) / 2$ . Displacement $Δ 𝐫 = 𝐫 - 𝐫_{0}$ , is average ko $t$ se multiply karke milta hai:

\begin{aligned} Δ 𝐫 & = \frac{𝐯_{0} + 𝐯}{2} t \\ = \frac{2 𝐯_{0} + 𝐚 t}{2} t \end{aligned}

𝐫 = 𝐫_{0} + 𝐯_{0} t + \frac{1}{2} 𝐚 t^{2}

(3.34a)

Check: differentiate karne par $d 𝐫 / d t = 𝐯_{0} + 𝐚 t$ milta hai, jo (3.33a) hai, aur $t = 0$ par $𝐫 = 𝐫_{0}$ milta hai.

Components mein likhne par do vector equations chaar normal equations ban jaati hain:

\begin{array}{c} v_{x} = v_{0 x} + a_{x} t \\ v_{y} = v_{0 y} + a_{y} t \end{array}

(3.33b)

\begin{array}{c} x = x_{0} + v_{0 x} t + \frac{1}{2} a_{x} t^{2} \\ y = y_{0} + v_{0 y} t + \frac{1}{2} a_{y} t^{2} \end{array}

(3.34b)

Dekho kya missing hai. $x$ -equations mein sirf $x$ -quantities hain aur $y$ -equations mein sirf $y$ -quantities. Constant acceleration ke saath plane mein motion, perpendicular axes ke along do independent straight-line motions hain jo ek hi samay par ho rahi hain, aur har ek Chapter 2 ki equations follow karti hai. Fig. 3.13 ise “shadows” se dikhata hai: har axis par particle ka projection ek alag one-dimensional object ki tarah chalta hai. Three dimensions mein teesra axis ek teesri independent motion jodta hai.

4/4Akele dekhne par har shadow ek one-dimensional equation follow karta hai; saath milkar woh curve banate hain.

Fig. 3.13Do straight-line motions, har axis par ek, milkar curved path banati hain.NCERT p. 38

Neeche left mein origin wale axes aur ek dotted curve jo origin se x ke along shuru hokar upar mudne wale particle ka path dikhata hai. Ek vermillion dot particle ko t = 0, 1, 2 aur 3 s par dikhata hai, har step mein ek, aur pehle ki positions halki reh jaati hain. Particle se har axis par dashed lines girti hain, jahan ek blue dot uska x-shadow aur ek green dot uska y-shadow dikhata hai. Step by step x-shadow x-axis par barabar time mein badhte hue steps se chalta hai, aur y-shadow y-axis par badhte steps se chadhta hai, har ek constant acceleration wali ek alag straight-line motion ki tarah.

Vector forms (3.33a) aur (3.34a) mein kisi axes ka zikr hi nahi hai. Axes tum sirf tab chunte ho jab numbers chahiye, aur tum woh pair chun sakte ho jisse algebra sabse aasaan ho. Projectile ke liye yeh horizontal aur vertical honge.

Example 3.5NCERT Example 3.5 · p. 38

Steady push wali model boat

Talaab par ek model boat $t = 0$ par origin se $6.0 \hat{𝐢} m / s$ velocity ke saath shuru hoti hai. Uski motor aur ek steady breeze milkar use constant acceleration $(2.0 \hat{𝐢} + 4.0 \hat{𝐣}) m / s^{2}$ dete hain. (a) Jab uska $y$ -coordinate 18 m ho, tab boat kahan hai? (b) Tab uski velocity kya hai?

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Quick check

3 mein se 1

Ek puck ka

+ x

ke along constant acceleration hai aur woh

+ y

ke along velocity se shuru hota hai. Kaun sa statement sahi hai?

v_{x}

aur

v_{y}

dono badhte hain

v_{y}

constant rehta hai jabki

v_{x}

badhta hai

v_{y}

dheere-dheere ghatkar zero ho jaata haiPuck seedhi line mein chalta hai

Answer dekho

Answer (B) $v_{y}$ constant rehta hai jabki $v_{x}$ badhta hai

y

ke along koi acceleration nahi hai, isliye

v_{y}

kabhi nahi badalta;

v_{x}

zero se badhta hai. Path mudta hai, kyunki velocity lagataar

+ x

ki taraf ghoomti rehti hai.

3.9 Projectile motion

Ball phenko, football ko kick karo, ya garden hose se paani ka arc dekho: chhodne ke baad har ek projectile (प्रक्षेप्य) hai, yaani hawa mein udta aisa object jis par sirf gravity lagti hai. Hum poore section mein air resistance ko neglect karte hain, jo chhoti doori par compact, kaafi dheemi cheezon ke liye theek hai. Tab acceleration constant $𝐠$ hai, seedha neeche, isliye Section 3.8 seedhe lagu hota hai. $x$ ko horizontal aur $y$ ko vertically upar lo. Horizontal motion mein koi acceleration hi nahi hai, jabki vertical motion normal free fall hai, aur dono saath-saath chalti hain bina ek doosre par asar daale.

Try karke dekho

Ek click ya do?

Tumhe chahiye

do milte-julte coins
30 cm ka ruler
sakht farsh par ek table

Pehle guess karo

Ek coin ko table se horizontally giraaya jaata hai aur usi pal doosra seedha neeche girta hai. Farsh par pehle kaun pahunchta hai?

Seedha neeche girne wala coinSide mein dhakka khaane wala coinDono ek saath

Ek coin ko table ke bilkul kinaare par flat rakho.
Uske paas table par ek ruler flat rakho, jiska ek end kinaare se bahar nikla ho, aur us end par doosra coin rakho.
Ruler ko beech mein pivot ki tarah pakad kar, andar wale end ko tezi se side mein flick karo taaki ruler kinaare wale coin ko horizontally table se gira de jabki ruler wala coin lagbhag seedha neeche gire.
Dhyaan se suno, phir zyada zor ke flick ke saath dobara karo.

Projectile ko origin se $v_{0}$ speed ke saath horizontal se $θ_{0}$ angle upar launch karo. Uska acceleration aur uski initial velocity ke components hain

a_{x} = 0, a_{y} = - g

(3.35)

\begin{array}{c} v_{0 x} = v_{0} \cos θ_{0} \\ v_{0 y} = v_{0} \sin θ_{0} \end{array}

(3.36)

Inhe $x_{0} = y_{0} = 0$ ke saath constant-acceleration equations mein daalne par kisi bhi time $t$ par position aur velocity milti hai:

\begin{array}{c} x = (v_{0} \cos θ_{0}) t \\ y = (v_{0} \sin θ_{0}) t - \frac{1}{2} g t^{2} \end{array}

(3.37)

\begin{array}{c} v_{x} = v_{0} \cos θ_{0} \\ v_{y} = v_{0} \sin θ_{0} - g t \end{array}

(3.38)

Horizontal velocity kabhi nahi badalti. Sirf vertical velocity badalti hai, har second $g$ se ghatti hai: top par yeh zero ho jaati hai aur phir neeche point karte hue badhti hai (Fig. 3.14). Sabse upar wale point par velocity horizontal hoti hai, zero nahi, kyunki ball ab bhi $v_{0} \cos θ_{0}$ se aage badh rahi hai.

3/5Top par $v_{y} = 0$ aur velocity horizontal hai; yahan height $h_{m}$ hai.

Fig. 3.14Horizontal velocity same rehti hai; vertical velocity lagataar badalti hai.NCERT Fig. 3.17 · p. 39

Ek dashed parabola level ground line ke left mein launch point se uthta hai aur right mein land karta hai. Har step path ke ek point par ek yellow ball dikhata hai, horizontal velocity ke liye ek blue arrow jo hamesha same length ka hai, vertical velocity ke liye ek green arrow aur total velocity ke liye ek dashed vermillion arrow. Step 1, launch: ek lamba green arrow upar point karta hai aur total velocity v0 launch angle par tirchhi upar jaati hai. Step 2, upar jaate hue: green arrow aadha lamba hai. Step 3, top: koi green arrow nahi, sirf blue wala, aur ek dashed line ground tak girti hai jis par h_m mark hai. Step 4, girte hue: green arrow neeche point karta hai, aadhi length ka. Step 5, landing: green arrow neeche point karta hai aur launch jitna lamba hai, aur ground ke along ek double-headed arrow range R mark karta hai.

Projectile ke path ki equation

DerivationPath ek parabola hai

Time ko eliminate karo. (3.37) ki pehli equation se $t = x / (v_{0} \cos θ_{0})$ . Ise doosri mein substitute karo:

\begin{aligned} y & = x \tan θ_{0} \\ - \frac{g x^{2}}{2 v_{0}^{2} \cos^{2} θ_{0}} \end{aligned}

(3.39)

Ek di gayi launch ke liye $g$ , $v_{0}$ aur $θ_{0}$ constants hain, isliye iska form $y = a x - b x^{2}$ hai jahan $b > 0$ : origin se guzarne wala parabola jo neeche ki taraf khulta hai.

Isliye har projectile ka path, yaani trajectory, ek parabola hota hai, jab tak air resistance negligible ho.

Maximum height ka time aur time of flight

DerivationProjectile kitni der upar jaata hai aur kitni der udta hai

Top par $v_{y} = 0$ . (3.38) se, $v_{0} \sin θ_{0} - g t_{m} = 0$ , isliye

t_{m} = \frac{v_{0} \sin θ_{0}}{g}

(3.40a)

Projectile apni launch height par tab wapas aata hai jab phir se $y = 0$ ho. (3.37) mein $y = 0$ rakhne par $t (v_{0} \sin θ_{0} - \frac{1}{2} g t) = 0$ milta hai, isliye ya to $t = 0$ (launch) ya

T_{f} = \frac{2 v_{0} \sin θ_{0}}{g}

(3.40b)

Yeh time of flight theek $2 t_{m}$ hai: level ground par utarna chadhne ki mirror image hai.

Maximum height

$y (t)$ mein $t = t_{m}$ rakho, ya top par $v_{y} = 0$ ke saath $v_{y}^{2} = v_{0 y}^{2} - 2 g y$ use karo. Dono tareekon se pahunchi gayi sabse badi height hai

h_{m} = \frac{(v_{0} \sin θ_{0})^{2}}{2 g}

(3.41)

Horizontal range

Horizontal range $R$ woh horizontal doori hai jo projectile apni launch height par wapas aane tak tay karta hai. Steady horizontal velocity ko time of flight se multiply karo, aur $2 \sin θ_{0} \cos θ_{0} = \sin 2 θ_{0}$ use karo:

\begin{aligned} R & = (v_{0} \cos θ_{0}) T_{f} \\ = \frac{v_{0}^{2} \sin 2 θ_{0}}{g} \end{aligned}

(3.42a)

Fixed launch speed ke liye $R$ sabse bada tab hota hai jab $\sin 2 θ_{0} = 1$ , yaani jab $θ_{0} = 45^{\circ}$ :

R_{m} = \frac{v_{0}^{2}}{g}

(3.42b)

45° se barabar upar aur neeche ke angles same range dete hain, jaisa Fig. 3.15 dikhata hai aur Example 3.6 prove karta hai. Zyada steep launch zyada upar jaata hai aur zyada der hawa mein rehta hai. Tum yeh sab projectile lab mein test kar sakte ho.

3/345° par range sabse zyada possible hai, $v_{0}^{2} / g$ .

Fig. 3.15Complementary angles ki range same hoti hai; 45° sabse door jaata hai.NCERT p. 40

Level ground line ke left mein ek point se launch hue parabolic paths, sab same speed par. Step 1: 30 degrees par ek solid blue path aur 60 degrees par ek dashed blue path; 60 degree wala path teen guna upar jaata hai, phir bhi dono same marked point par land karte hain, label same range. Step 2: woh paths halke ho jaate hain aur 15 aur 75 degrees ka ek orange pair ek nazdeek ke common point par land karta hai. Step 3: 45 degrees par ek mota vermillion path sabse door land karta hai, label maximum range, aur comparison ke liye baaki pairs halke hain.

InteractiveProjectile lab

Level ground ka side view jismein left mein ek adjustable platform par launcher hai. Sliders launch speed 5 se 30 m/s, launch angle 0 se 90 degrees aur launch height 0 se 30 m tak set karte hain. Play karne se pehle predicted parabola dashed bana hota hai; Play dabane par ek ball uske along jaati hai, har dasve hisse second par dots chhodte hue, horizontal aur vertical velocity ke arrows ke saath. Readouts time of flight, maximum height, range aur landing speed dikhate hain. 90 degrees minus angle par launch hone wali ek optional twin ball dikhati hai ki launch height zero hone par complementary angles ki range same hoti hai.

Fig. 3.16Launch speed, angle aur height set karo, phir Play dabao. Complementary angles compare karne ke liye 90° − θ wala twin on karo.

Example 3.6NCERT Example 3.6 · p. 40

Complementary angles, barabar ranges

(a) Dikhao ki same launch speed ke liye $45^{\circ} + α$ aur $45^{\circ} - α$ ke launch angles same horizontal range dete hain. (b) Ek ball 25 m/s se phenki jaati hai, ek baar 30° par aur ek baar 60° par. Ranges, sabse badi heights aur flight times compare karo. $g = 9.8 m / s^{2}$ lo.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 3.7NCERT Example 3.7 · p. 40

Pul se phenka gaya parcel

River se 44.1 m upar ek pul se, ek ladka boat mein baithe apne dost ki taraf ek parcel 12 m/s se horizontally phenkta hai. Air resistance neglect karo aur $g = 9.8 m / s^{2}$ lo. (a) Parcel kitni der hawa mein rehta hai? (b) Pul ke neeche wale point se horizontally kitni door woh land karta hai? (c) Woh kis velocity se paani se takraata hai?

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 3.8NCERT Example 3.8 · p. 40

Angle par kick ki gayi football

Ek footballer ground se ball ko 22 m/s se, horizontal se 37° upar kick karta hai. Nikalo (a) ball ki sabse badi height, (b) level ground par land hone tak ka time, aur (c) woh kitni door land karti hai. $g = 9.8 m / s^{2}$ lo aur air resistance ignore karo.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Quick check

3 mein se 1

Apne path ke sabse upar wale point par projectile ke paas hota hai

ek horizontal velocity, aur neeche ki taraf acceleration gzero velocity, aur neeche ki taraf acceleration gzero velocity aur zero accelerationek horizontal velocity aur zero acceleration

Answer dekho

Answer (A) ek horizontal velocity, aur neeche ki taraf acceleration g

Top par sirf

v_{y}

zero hota hai;

v_{x} = v_{0} \cos θ_{0}

chalta rehta hai. Gravity kabhi band nahi hoti, isliye poori flight mein acceleration neeche ki taraf

g

hai.

3.10 Uniform circular motion

Dhaage par ghumaya gaya patthar, ceiling-fan ke blade ki tip, circular orbit mein satellite, steady speedometer reading ke saath roundabout ka chakkar lagati car: har ek constant speed se circle par chalta hai. Yeh uniform circular motion hai. Sirf speed steady hai: motion ki direction har pal badalti hai, isliye velocity badalti hai, aur object ko accelerate karna hi padta hai bhale hi uski speed kabhi na badle. Yeh acceleration kis taraf point karta hai, aur kitna bada hai?

Maan lo object centre C ke around radius $R$ ke circle par speed $v$ se chal raha hai, aur time $t$ par P par hai aur thode time $Δ t$ baad P′ par. Uski velocities $𝐯$ aur $𝐯^{'}$ tangents hain, isliye har ek apne point tak bane radius ke perpendicular hai. Dono ki length $v$ hai, isliye $Δ 𝐯 = 𝐯^{'} - 𝐯$ ek isosceles triangle ka base hai. Fig. 3.17 yeh triangle circle ke paas banata hai. Jab P aur P′ symmetrically rakhe hon, to $Δ 𝐯$ seedha andar ki taraf point karta hai, us line ke along jo angle PCP′ ko bisect karti hai, isliye average acceleration $Δ 𝐯 / Δ t$ bhi andar ki taraf point karta hai. Jaise-jaise $Δ t$ chhota hota hai, P′, P ke paas aata jaata hai, aur limit mein P par acceleration theek centre ki taraf point karta hai.

3/3Limit mein P par acceleration C ki taraf point karta hai, magnitude $v^{2} / R$ ke saath.

Fig. 3.17Interval kitna bhi chhota ho, Δv centre ki taraf point karta hai.NCERT Fig. 3.18 · p. 41

Left mein centre C wala ek circle. Step 1: points P aur P-prime circle ke right mein 60 degrees ki doori par hain, radii se C se jude hain aur angle delta theta marked hai; P par ek blue velocity arrow v aur P-prime par ek green arrow v-prime tangents ke along bane hain. Right mein dono arrows ek common tail se dobara bane hain, aur ek vermillion arrow delta v unke heads ko jodta hai, jo left ki taraf point karta hai, yaani arc se C ki taraf. Step 2: wahi construction, lekin points sirf 30 degrees ki doori par; delta v chhota hai aur ab bhi centre ki taraf point karta hai. Step 3: limit mein ek hi point P hai jismein blue tangent velocity v aur ek mota vermillion acceleration arrow a seedha C ki taraf point karta hai, label a_c = v squared over R.

DerivationCentripetal acceleration ka magnitude

Radii CP aur CP′, har ek ki length $R$ , angle $Δ θ$ par milti hain. Har velocity apne radius se 90° ghumi hui hai, isliye $𝐯$ aur $𝐯^{'}$ ke beech ka angle bhi $Δ θ$ hai. Isliye position vectors ka triangle (sides $R$ , $R$ , $∣ Δ 𝐫 ∣$ ) aur velocities ka triangle (sides $v$ , $v$ , $∣ Δ 𝐯 ∣$ ) similar isosceles triangles hain, aur

\frac{∣ Δ 𝐯 ∣}{v} = \frac{∣ Δ 𝐫 ∣}{R}

$Δ t$ se divide karo aur limit lo. Bahut chhote interval ke liye chord $∣ Δ 𝐫 ∣$ arc $v Δ t$ ke barabar hoti hai, isliye $∣ Δ 𝐫 ∣ / Δ t \to v$ :

\begin{aligned} a & = \lim_{Δ t \to 0} \frac{∣ Δ 𝐯 ∣}{Δ t} \\ = \frac{v}{R} \lim_{Δ t \to 0} \frac{∣ Δ 𝐫 ∣}{Δ t} \\ = \frac{v}{R} \cdot v \end{aligned}

a_{c} = \frac{v^{2}}{R}

(3.43)

Yeh centripetal acceleration (अभिकेंद्र त्वरण) poore chakkar mein same size, $v^{2} / R$ , rakhta hai, lekin hamesha centre ki taraf aim karta hai: jab object circle ke top par ho to neeche ki taraf, jab bottom par ho to upar ki taraf. Isliye $𝐚_{c}$ constant vector nahi hai, aur Section 3.8 ki constant-acceleration equations circular motion ke liye kabhi use nahi ki ja saktin.

Angular speed

Circular motion ko aksar distance ke bajaye angle se describe karna aasaan hota hai. Jab object P se P′ tak jaata hai, to radius angle $Δ θ$ sweep karta hai, jo uska angular displacement hai. Angular speed ω woh rate hai jisse yeh angle badalta hai:

ω = \frac{Δ θ}{Δ t}

(3.44)

Ise radians per second mein naapte hain. $Δ t$ mein tay kiya gaya arc $Δ s = R Δ θ$ hai (θ radians mein), isliye speed aur centripetal acceleration hain

v = \frac{Δ s}{Δ t} = R ω

(3.45)

a_{c} = \frac{v^{2}}{R} = ω^{2} R

(3.46)

Ek poora revolution period $T$ leta hai aur $2 π R$ distance cover karta hai. Frequency $ν = 1 / T$ per second revolutions ki ginti hai. Isliye

v = \frac{2 π R}{T} = 2 π R ν

(3.47)

\begin{array}{c} ω = 2 π ν \\ a_{c} = 4 π^{2} ν^{2} R \end{array}

(3.48)

Try karke dekho

Chhod do!

Tumhe chahiye

ek soft ball ya gaanth lagi kapde ki potli
lagbhag 1 m mazboot dhaaga
ek khula playground

Pehle guess karo

Ball tumhare sir ke upar anticlockwise ghoom rahi hai. Tum use tab chhodte ho jab woh theek tumhare saamne hai aur tumhare left ki taraf ja rahi hai. Woh kis taraf udti hai?

Seedha aage, tumse doorTumhare left ki taraf, us pal ki motion ki direction ke alongWapas tumhari taraf

Ball ko dhaage se mazbooti se baandho.
Khuli jagah mein ball ko apne sir ke upar horizontal circle mein dheere-dheere ghumao.
Jab ball theek tumhare saamne ho, us pal dhaaga chhodo, aur dekho woh kis taraf udti hai.
Circle ke alag-alag points par chhodte hue kuch baar dobara karo.

Example 3.9NCERT Example 3.9 · p. 42

Circular track par toy train

Ek toy train 0.75 m radius ke circular track par steady chalti hai aur 60 s mein 12 laps poore karti hai. (a) Uski angular speed aur speed nikalo. (b) Kya uska acceleration ek constant vector hai? Uska magnitude nikalo.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Recap

Summary aur key points

Scalars ka sirf magnitude hota hai aur woh normal algebra follow karte hain; vectors ka magnitude aur direction hota hai aur woh triangle law se add hote hain (Section 3.2).
Displacement sirf end points par depend karta hai; iska magnitude kabhi path length se zyada nahi hota (Section 3.2.1).
Equal vectors ka magnitude aur direction same hota hai; $λ 𝐀$ ka magnitude $∣ λ ∣ A$ hai aur λ < 0 hone par yeh ulta ho jaata hai (Section 3.2.2, Section 3.3).
Vectors head to tail ya parallelogram rule se add hote hain; addition commutative aur associative hai, aur $𝐀 - 𝐁 = 𝐀 + (- 𝐁)$ (Section 3.4).
Null vector ka magnitude zero hai aur koi direction nahi: yeh round trip ka displacement hai ((3.3)).
Plane mein koi bhi vector do non-parallel directions ke along uniquely tootta hai: $𝐀 = λ 𝐚 + μ 𝐛$ (Section 3.5).
$𝐀 = A_{x} \hat{𝐢} + A_{y} \hat{𝐣}$ jahan $A_{x} = A \cos θ$ , $A_{y} = A \sin θ$ , $A = \sqrt{A_{x}^{2} + A_{y}^{2}}$ aur $\tan θ = A_{y} / A_{x}$ (Rectangular components).
Sum ke components, components ka sum hote hain; $𝐀 + 𝐁$ ka magnitude $\sqrt{A^{2} + B^{2} + 2 A B \cos θ}$ hai (Section 3.6).
$𝐯 = d 𝐫 / d t$ path ki tangent hai; $𝐚 = d 𝐯 / d t$ , $Δ 𝐯$ ke along point karta hai aur $𝐯$ ke saath koi bhi angle bana sakta hai (Section 3.7).
Constant $𝐚$ ke saath: $𝐯 = 𝐯_{0} + 𝐚 t$ aur $𝐫 = 𝐫_{0} + 𝐯_{0} t + \frac{1}{2} 𝐚 t^{2}$ , do independent 1D motions (Section 3.8).
Projectile girte hue side mein uniformly chalta hai; uska path ek parabola hai (Section 3.9).
$t_{m} = v_{0} \sin θ_{0} / g$ , $T_{f} = 2 t_{m}$ , $h_{m} = (v_{0} \sin θ_{0})^{2} / 2 g$ , $R = v_{0}^{2} \sin 2 θ_{0} / g$ , 45° par sabse zyada (Horizontal range).
Uniform circular motion mein $a_{c} = v^{2} / R = ω^{2} R$ , hamesha centre ki taraf, isliye yeh constant vector nahi hai (Section 3.10).
$v = R ω$ , $ω = 2 π ν = 2 π / T$ aur $a_{c} = 4 π^{2} ν^{2} R$ (Angular speed).

Flash cards · 9

Formula revision

1 / 9

Apna NCERT exercise question dhoondo

Hamare practice questions ncert.ai ke liye likhe gaye hain aur NCERT exercises ko follow karte hain. Apni textbook ke question se match karne wala question dhoondo; uska answer uske saath hai.

Tumhari NCERT textbook mein	Yahan practice question
NCERT Exercise 3.1	Practice Q1
NCERT Exercise 3.2	Practice Q2
NCERT Exercise 3.3	Practice Q3
NCERT Exercise 3.4	Practice Q4
NCERT Exercise 3.5	Practice Q5
NCERT Exercise 3.6	Practice Q6
NCERT Exercise 3.7	Practice Q7
NCERT Exercise 3.8	Practice Q8
NCERT Exercise 3.9	Practice Q9
NCERT Exercise 3.10	Practice Q10
NCERT Exercise 3.11	Practice Q11
NCERT Exercise 3.12	Practice Q12
NCERT Exercise 3.13	Practice Q13
NCERT Exercise 3.14	Practice Q14
NCERT Exercise 3.15	Practice Q15
NCERT Exercise 3.16	Practice Q16
NCERT Exercise 3.17	Practice Q17
NCERT Exercise 3.18	Practice Q18
NCERT Exercise 3.19	Practice Q19
NCERT Exercise 3.20	Practice Q20
NCERT Exercise 3.21	Practice Q21
NCERT Exercise 3.22	Practice Q22

Quick check: MCQs

3.2.2 Vectors ki equality4 questions
3.3 Vectors ka real numbers se multiplication2 questions
3.4 Vectors ka addition aur subtraction — graphical method4 questions
Three dimensions mein vectors3 questions
3.6 Vectors ka addition — analytical method3 questions
Acceleration3 questions
3.8 Constant acceleration ke saath plane mein motion3 questions
Horizontal range3 questions
Angular speed4 questions

Glossary · 15

Key terms (English + Hindi)

angular speedकोणीय चालkoniya chaal: Radius dwara sweep kiye gaye angle ke badlaav ka rate, $ω = Δ θ / Δ t$ , rad/s mein.Chapter mein pehli baar yahan aaya →

centripetal accelerationअभिकेंद्र त्वरणabhikendra tvaran: Uniform circular motion ka acceleration $v^{2} / R$ , jo hamesha centre ki taraf point karta hai.Chapter mein pehli baar yahan aaya →
componentघटकghatak: Chune hue axis ke along vector ka hissa; $A_{x} = A \cos θ$ positive, negative ya zero ho sakta hai.Chapter mein pehli baar yahan aaya →

displacementविस्थापनvisthapan: Position mein badlaav, $Δ 𝐫 = 𝐫^{'} - 𝐫$ , initial se final point tak, raasta chahe jo bhi ho.Chapter mein pehli baar yahan aaya →

horizontal rangeक्षैतिज परासkshaitij paras: Launch height par wapas aane se pehle projectile ki tay ki gayi horizontal doori, $v_{0}^{2} \sin 2 θ_{0} / g$ .Chapter mein pehli baar yahan aaya →

null vectorशून्य सदिशshunya sadish: Zero magnitude wala vector, $𝟎$ , jiski koi fixed direction nahi hoti.Chapter mein pehli baar yahan aaya →

position vectorस्थिति सदिशsthiti sadish: Chune hue origin se us point tak ka arrow jahan object hai: $𝐫 = x \hat{𝐢} + y \hat{𝐣}$ .Chapter mein pehli baar yahan aaya →
projectileप्रक्षेप्यprakshepya: Phenke ya launch kiye jaane ke baad hawa mein udta object, jo sirf gravity ke asar mein chalta hai.Chapter mein pehli baar yahan aaya →

resultantपरिणामीparinami: Woh ek vector jo do ya zyada vectors ke sum ke barabar ho.Chapter mein pehli baar yahan aaya →

scalarअदिशadish: Aisi quantity jo ek number aur unit se poori tarah describe ho jaati hai, jaise mass, time ya temperature.Chapter mein pehli baar yahan aaya →

time of flightउड्डयन कालuddayan kaal: Launch height par wapas aane se pehle projectile ka hawa mein bitaya time, $2 v_{0} \sin θ_{0} / g$ .Chapter mein pehli baar yahan aaya →
trajectoryप्रक्षेप पथprakshep path: Chalte object ka path; air drag ke bina projectile ke liye ek parabola.Chapter mein pehli baar yahan aaya →

uniform circular motionएकसमान वृत्तीय गतिeksaman vrittiya gati: Constant speed se circle ke along motion; velocity ki direction lagataar badalti rehti hai.Chapter mein pehli baar yahan aaya →
unit vectorएकांक सदिशekank sadish: 1 magnitude wala dimensionless vector jo sirf direction dikhata hai, jaise $\hat{𝐢}$ , $\hat{𝐣}$ , $\hat{𝐤}$ .Chapter mein pehli baar yahan aaya →

vectorसदिशsadish: Magnitude aur direction wali quantity jo triangle law se add hoti hai, jaise displacement ya force.Chapter mein pehli baar yahan aaya →

Motion in a Plane: Class 11 Physics Chapter 3 Hinglish mein

Is chapter mein aap kya seekhenge

3.1 Introduction

3.2 Scalars aur vectors

3.2.1 Position aur displacement vectors

3.2.2 Vectors ki equality

3.3 Vectors ka real numbers se multiplication

3.4 Vectors ka addition aur subtraction — graphical method

3.5 Vectors ka resolution

Unit vectors

Rectangular components

Three dimensions mein vectors

3.6 Vectors ka addition — analytical method

3.7 Plane mein motion

3.7.1 Position vector aur displacement

Velocity

Acceleration

3.8 Constant acceleration ke saath plane mein motion

3.9 Projectile motion

Projectile ke path ki equation

Maximum height ka time aur time of flight

Maximum height

Horizontal range

3.10 Uniform circular motion

Angular speed

Summary aur key points

Practice questions answers ke saath (NCERT “Exercises”)

Apna NCERT exercise question dhoondo

Quick check: MCQs

Key terms (English + Hindi)

Tum chapter ke end tak pahunch gaye