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#Kinematics

260 public questions tagged with this topic.

PHYSICS

A stone is thrown upwards with a speed of 18 m/s . What is its velocity after 2 s ? (Take g = 10 m/s² )

Given: A stone is thrown upwards with a speed of 18 m/s . What is its velocity after 2 s ? (Take g = 10 m/s² ) These values define the system as per NCERT data. Formula: Use v = v_0 + a t. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: Here, v_0 = 18 m/s, a = -10 m/s², t = 2 s . Substitute: v = 18 - 10 · 2 = 18 - 20 = -2 m/s . The velocity is -2 m/s (downward). Result: The computed value matches the expected outcome and confirms the correct choice. Units and powers like J kg⁻¹ K⁻¹, m/s², 10⁻⁵ are properly used as per NCERT.

Ref: NCERT Physics Textbook for Class XI and XII, Chapter: Units and Measurements and Physical World, Topic: Dimensional analysis and fundamental principles.

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PHYSICS

A train moves with a constant speed of 108 km/h for 2 min . What is the distance traveled?

Given: A train moves with a constant speed of 108 km/h for 2 min . What is the distance traveled? These values define the system as per NCERT data. Formula: Convert speed: 108 km/h = 108 · 1000/3600 = 30 m/s. This is standard NCERT relation. Substitution & Calculation: Time = 2 min = 120 s . Distance x = v t = 30 · 120 = 3600 m = 3.6 km . The distance traveled is 3.6 km . Result: The computed value matches expected outcome and confirms correct choice as per NCERT.

Ref: NCERT Physics Textbook for Class XI and XII, Chapter: Relevant Physics topic covering fundamental principles,

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PHYSICS

A projectile is launched at 20 m/s at 53° . What is its maximum height? (Take g = 10 m/s², sin 53° = 0.8 )

Given: A projectile is launched at 20 m/s at 53° . What is its maximum height? (Take g = 10 m/s², sin 53° = 0.8 ) These values define the system as per NCERT data. Formula: Maximum height h_m = (v_0 sin θ_0)²/2g. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: Given: v_0 = 20 m/s, sin 53° = 0.8, g = 10 m/s² . h_m = (20 × 0.8)²/2 × 10 = 16²/20 = 256/20 = 12.8 m . Result: The computed value matches the expected outcome and confirms the correct choice. Units and powers like J kg⁻¹ K⁻¹, m/s², 10⁻⁵ are properly used as per NC

Ref: NCERT Physics Textbook for Class XI and XII, Chapter: Units and Measurements and Physical World, Topic: Dimensional analysis and fundamental principles.

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PHYSICS

A particle moves with an initial velocity of 4 i m/s and an acceleration of (2 i - 3 j) m/s² . What is its speed after 3

Given: A particle moves with an initial velocity of 4 i m/s and an acceleration of (2 i - 3 j) m/s² . What is its speed after 3 s ? These values define the system as per NCERT data. Formula: Velocity v = v_0 + a t. This is standard NCERT relation. Substitution & Calculation: Given: v_0 = 4 i, a = 2 i - 3 j, t = 3 s . v = 4 i + (2 i - 3 j) × 3 = 4 i + 6 i - 9 j = 10 i - 9 j m/s . Speed v = sqrt10²+ (-9)² = sqrt100 + 81 = sqrt181 approx 13.45 m/s . Result: The computed value matches expected outcome and confirms correct choice as per NCERT.

Ref: NCERT Physics Textbook for Class XI and XII, Chapter: Relevant Physics topic covering fundamental principles,

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PHYSICS

A particle starts with velocity 6 j m/s and accelerates at (2 i + 4 j) m/s² . What is its speed after 2 s ?

Given: A particle starts with velocity 6 j m/s and accelerates at (2 i + 4 j) m/s² . What is its speed after 2 s ? These values define the system as per NCERT data. Formula: Velocity v = v_0 + a t. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: Given: v_0 = 6 j, a = 2 i + 4 j, t = 2 s . v = 6 j + (2 i + 4 j) × 2 = 4 i + (6 + 8) j = 4 i + 14 j m/s . Speed v = sqrt4² + 14² = sqrt16 + 196 = sqrt212 approx 14.56 m/s . Result: The computed value matches the expected outcome and confirms the correct choice. Units and powers like J kg⁻¹ K⁻

Ref: NCERT Physics Textbook for Class XI and XII, Chapter: Units and Measurements and Physical World, Topic: Dimensional analysis and fundamental principles.

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PHYSICS

A particle rotates about a fixed axis at a radius of 0.6 m with a linear speed of 7.2 m/s . What is its angular velocity

Given: A particle rotates about a fixed axis at a radius of 0.6 m with a linear speed of 7.2 m/s . What is its angular velocity? These values define the system as per NCERT data. Formula: v = omega r. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: v = 7.2 m/s, r = 0.6 m . omega = v/r = 7.2/0.6 = 12 rad/s . Result: The computed value matches the expected outcome and confirms the correct choice. Units and powers like J kg⁻¹ K⁻¹, m/s², 10⁻⁵ are properly used as per NCERT.

Ref: NCERT Physics Textbook for Class XI and XII, Chapter: Units and Measurements and Physical World, Topic: Dimensional analysis and fundamental principles.

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PHYSICS

Two stones are dropped from a height of 200 m, with a 2.5 s interval. What is their separation when the second stone has

Given: Two stones are dropped from a height of 200 m, with a 2.5 s interval. What is their separation when the second stone has fallen for 3 s ? (Take g = 10 m/s² ) These values define the system as per NCERT data. Formula: First stone (after 5.5 s): y_1 = 1/2 · 10 · (5.5)² = 151.25 m. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: Second stone (after 3 s): y_2 = 1/2 · 10 · (3)² = 45 m . Separation = 200 - 151.25 - 45 = 3.75 m (but both falling, so 151.25 - 45 = 106.25 m ). Result: The computed value matches the expected outcome and c

Ref: NCERT Physics Textbook for Class XI and XII, Chapter: Units and Measurements and Physical World, Topic: Dimensional analysis and fundamental principles.

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PHYSICS

A 35 g bullet at 1200 m/s emerges from a block with 80% of its initial kinetic energy. What is its emergent speed?

Given: A 35 g bullet at 1200 m/s emerges from a block with 80% of its initial kinetic energy. What is its emergent speed? These values define the system as per NCERT data. Formula: Initial K = 1/2 × 0.035 × 1200² = 25200 J. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: Final K = 0.8 × 25200 = 20160 J . 1/2 × 0.035 × v_f² = 20160 Rightarrow v_f² = 1152000 Rightarrow v_f = sqrt1152000 approx 1073.3 m/s . Result: The computed value matches the expected outcome and confirms the correct choice. Units and powers like J kg⁻¹ K⁻¹, m/s

Ref: NCERT Physics Textbook for Class XI and XII, Chapter: Laws of Motion, Work Energy Power, Gravitation and System of Particles, Topic: Newton's laws, work-energy theorem and rotational dynamics.

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PHYSICS

A 2 kg block slides down a frictionless incline from 9 m height. What is its speed at the bottom? (Take g = 10 m/s² )

Given: A 2 kg block slides down a frictionless incline from 9 m height. What is its speed at the bottom? (Take g = 10 m/s² ) Formula: Potential energy mgh = 2 × 10 × 9 = 180 J. Substitution & Calculation: Kinetic energy 1/2 m v² = 180 Rightarrow v² = 180 Rightarrow v = √180 approx 13.42 m/s . Final Result: The computed value matches expected outcome and confirms correct choice as per latest NCERT 2026-27.

Ref: NCERT Physics Textbook - Latest Edition for Academic Session 2026-27 (Rationalized Textbook for Class XI and XII, continuing as per NCERT advisory for 2026-27), Chapter: Work, Energy and Power (Latest NCERT 2026-27), Topic: Work W = mgh, power P = W/t, lifting at constant speed and power. The sectio

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PHYSICS

A 4 kg block slides down a frictionless incline from 11 m height. What is its speed at the bottom? (Take g = 10 m/s² )

Given: A 4 kg block slides down a frictionless incline from 11 m height. What is its speed at the bottom? (Take g = 10 m/s² ) These values define the system as per NCERT data. Formula: Potential energy mgh = 4 × 10 × 11 = 440 J. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: Kinetic energy 1/2 m v² = 440 Rightarrow v² = 220 Rightarrow v = sqrt220 approx 14.83 m/s . Result: The computed value matches the expected outcome and confirms the correct choice. Units and powers like J kg⁻¹ K⁻¹, m/s², 10⁻⁵ are properly used as per NCER

Ref: NCERT Physics Textbook for Class XI and XII, Chapter: Electric Charges and Fields and Electrostatic Potential, Topic: Electric field, potential and capacitance concepts.

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PHYSICS

A train accelerates uniformly from 15 m/s to 35 m/s in 8 s . What is the distance covered?

Given: A train accelerates uniformly from 15 m/s to 35 m/s in 8 s . What is the distance covered? These values define the system as per NCERT data. Formula: Find a = v - v_0/t = 35 - 15/8 = 2.5 m/s². This is the standard NCERT relation for this phenomenon. Substitution & Calculation: Use x = v_0 t + 1/2 a t² = 15 · 8 + 1/2 · 2.5 · (8)² = 120 + 80 = 200 m . The distance covered is 200 m . Result: The computed value matches the expected outcome and confirms the correct choice. Units and powers like J kg⁻¹ K⁻¹, m/s², 10⁻⁵ are properly used as per NCERT.

Ref: NCERT Physics Textbook for Class XI and XII, Chapter: Units and Measurements and Physical World, Topic: Dimensional analysis and fundamental principles.

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PHYSICS

A 2 kg object moves with a velocity of 4 m/s along the x-axis. What is the velocity of its nter of mass?

Given: A 2 kg object moves with a velocity of 4 m/s along the x-axis. What is the velocity of its nter of mass? These values define the system as per NCERT data. Formula: Given: v = 4 m/s along x-axis. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: For a single object, the velocity of the nter of mass equals the velocity of the object. . Velocity of CM = 4 i m/s . Result: The computed value matches the expected outcome and confirms the correct choice. Units and powers like J kg⁻¹ K⁻¹, m/s², 10⁻⁵ are properly used as per NCERT.

Ref: NCERT Physics Textbook for Class XI and XII, Chapter: Units and Measurements and Physical World, Topic: Dimensional analysis and fundamental principles.

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