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अध्याय 2: सरल रेखा में गति

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CBSE • कक्षा XI • Physics • अध्याय 2

अनुमानित समय: 45 Mins

प्रगति: अध्ययनरत

सरल रेखा में गति

In Class 11 Physics, "Motion in a Straight Line" provides an authoritative, curriculum-verified master resource aligned with the 2026–27 NCERT syllabus.

अध्याय CBT ऑनलाइन टेस्ट दें सभी अध्याय देखें

प्रस्तावना अध्याय अवधारणाएँ अध्याय सारांश स्व-मूल्यांकन अभ्यास टेस्ट दें

🏎️ Have You Ever Wondered?

How can a space shuttle accelerating down a runway calculate its exact takeoff speed and distance using calculus, or why do objects dropped from a cli...

How can a space shuttle accelerating down a runway calculate its exact takeoff speed and distance using calculus, or why do objects dropped from a cliff accelerate at the exact same rate regardless of their weight? Kinematics is the mathematical description of pure motion.

यह अध्याय क्यों महत्वपूर्ण है

In Class 11 Physics, "Motion in a Straight Line" provides an authoritative, curriculum-verified master resource aligned with the 2026–27 NCERT syllabus.

अध्ययन से पूर्व (आवश्यक ज्ञान)

Speed, velocity, and acceleration from Class 9.
Distance-time and velocity-time graphs.
Equations of motion.

इस अध्याय के लक्ष्य

Distinguish between Distance (scalar) and Displacement (vector); Speed vs Velocity.
Define Instantaneous Velocity ($v = \frac{dx}{dt}$) and Instantaneous Acceleration ($a = \frac{dv}{dt} = \frac{d^2x}{dt^2}$).
Interpret Kinematic Graphs: $x-t$ graph (slope = velocity) and $v-t$ graph (slope = acceleration, area = displacement).
Derive kinematic equations of uniform acceleration using calculus: $v = u + at$, $s = ut + \frac{1}{2}at^2$, $v^2 = u^2 + 2as$.
Analyze Motion under Gravity: Free fall with $a = -g$ (maximum height, time of flight).

अध्याय रूपरेखा एवं प्रगति

1 1. Position, Velocity & Calculus

2 2. Kinematic Equations (Constant $a...

3 3. Motion Under Gravity (Free Fall)

सम्पूर्ण सैद्धांतिक एवं वैचारिक अध्ययन

1. Position, Velocity & Calculus

In one-dimensional kinematics:
• Instantaneous Velocity: $$\mathbf{v = \lim_{\Delta t \to 0} \frac{\Delta x}{\Delta t} = \frac{dx}{dt}}$$
• Instantaneous Acceleration: $$\mathbf{a = \frac{dv}{dt} = v\frac{dv}{dx} = \frac{d^2x}{dt^2}}$$

2. Kinematic Equations (Constant $a$)

Integrating with constant acceleration yields: $$\mathbf{v = u + at}$$ $$\mathbf{s = ut + \frac{1}{2}at^2}$$ $$\mathbf{v^2 = u^2 + 2as}$$ Distance traveled in the $n$th second: $\mathbf{s_n = u + \frac{a}{2}(2n - 1)}$.

3. Motion Under Gravity (Free Fall)

Taking upward as positive ($a = -g = -9.8\text{ m/s}^2$):
• Maximum height reached: $\mathbf{H = \frac{u^2}{2g}}$.
• Time to reach peak: $\mathbf{t = \frac{u}{g}}$.
• Total time of flight: $\mathbf{T = \frac{2u}{g}}$ (time of ascent equals time of descent in vacuum!).

चित्रात्मक व्याख्या एवं मॉडल

Motion in a Straight Line Master Matrix

Conceptual framework, core mechanisms, and analytical relationships

Academic Architecture

1. Position, Velocity & Calculus • 2. Kinematic Equations (Constant $a$)

अध्याय का सार संक्षेप एवं 10 मुख्य निष्कर्ष

मुख्य बिंदु 1

Instantaneous Velocity: Derivative $dx/dt$ measuring instantaneous slope of $x-t$ graph.

मुख्य बिंदु 2

Velocity-Time Area: Definite integral under $v-t$ curve yielding exact displacement.

मुख्य बिंदु 3

Constant Acceleration: Calculus derivation of the three fundamental Galilean kinematic formulas.

मुख्य बिंदु 4

Galileo's Odd Number Law: Distances in successive equal intervals of free fall are in ratio $1 : 3 : 5 : 7$.

मुख्य बिंदु 5

Kinematic Symmetry: Upward ascent time equals downward descent time in vacuum.

स्व-मूल्यांकन अभ्यास (Check Your Understanding)

मूल वैचारिक स्पष्टता की जांच के लिए नैदानिक प्रश्न। पहले स्वयं हल करें, फिर उत्तर देखें।

A particle moves along the x-axis such that its position is given by $x = 2t^3 - 3t^2 + 5$. Find its acceleration at $t = 2\text{ s}$.

उत्तर एवं व्याख्या देखें

उत्तर: Velocity $v = \frac{dx}{dt} = 6t^2 - 6t$. Acceleration $a = \frac{dv}{dt} = 12t - 6$. At $t = 2\text{ s}$: $a = 12(2) - 6 = 18\text{ m/s}^2$.
18 m/s^2.

A ball is thrown vertically upwards with a velocity of $20\text{ m/s}$. Find the maximum height reached and total time in air ($g = 10\text{ m/s}^2$).

उत्तर एवं व्याख्या देखें

उत्तर: Max height $H = \frac{u^2}{2g} = \frac{20^2}{20} = 20\text{ meters}$. Total time of flight $T = \frac{2u}{g} = \frac{2(20)}{10} = 4\text{ seconds}$.
H = 20 m, T = 4 s.

What does the area under a Velocity-Time graph represent? What does its slope represent?

उत्तर एवं व्याख्या देखें

उत्तर: The area under a $v-t$ curve represents the displacement of the body; the slope of the $v-t$ curve represents instantaneous acceleration.
Area is displacement; slope is acceleration.

Derive $v^2 - u^2 = 2as$ using calculus.

उत्तर एवं व्याख्या देखें

उत्तर: $a = \frac{dv}{dt} = \frac{dv}{dx}\frac{dx}{dt} = v\frac{dv}{dx} \implies a\, dx = v\, dv$. Integrating from $x=0$ to $s$ and $u$ to $v$: $a \int_0^s dx = \int_u^v v\, dv \implies as = \frac{v^2 - u^2}{2} \implies v^2 - u^2 = 2as$.
Derived by integrating a dx = v dv.

State Galileo's Law of Odd Numbers for a freely falling body.

उत्तर एवं व्याख्या देखें

उत्तर: The distances traversed during equal intervals of time by a body falling from rest stand to one another in the same ratio as the odd integers beginning with unity: $1 : 3 : 5 : 7 : \dots$
Distances in successive time intervals ratio 1:3:5:7.

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