Kinematics Formulas are the fundamental mathematical expressions used to describe the motion of objects in one, two, and three dimensions, forming a core part of NCERT Class 11 Physics Chapter 3 and Chapter 4. These formulas relate displacement, velocity, acceleration, and time without considering the cause of motion. For CBSE board students, mastering kinematics is essential for scoring full marks in Unit 1. For JEE Main and NEET aspirants, kinematics questions appear in almost every paper. This article covers every important kinematics formula, a complete reference sheet, three progressive solved examples, CBSE exam tips, common mistakes, and JEE/NEET application strategies.
Key Kinematics Formulas at a Glance
Quick reference for the most important kinematics formulas used in CBSE and competitive exams.
- First equation of motion: \( v = u + at \)
- Second equation of motion: \( s = ut + \frac{1}{2}at^2 \)
- Third equation of motion: \( v^2 = u^2 + 2as \)
- Average velocity: \( v_{avg} = \frac{u + v}{2} \)
- Displacement in nth second: \( s_n = u + \frac{a}{2}(2n – 1) \)
- Projectile range: \( R = \frac{u^2 \sin 2\theta}{g} \)
- Maximum height: \( H = \frac{u^2 \sin^2 \theta}{2g} \)
What is Kinematics?
Kinematics is the branch of classical mechanics that studies the motion of points, objects, and systems of bodies without considering the forces that cause such motion. The Kinematics Formulas describe relationships between displacement, velocity, acceleration, and time. Scientists and engineers often call kinematics the “geometry of motion” because it focuses purely on describing how objects move through space.
In the NCERT curriculum, kinematics is introduced in Class 11 Physics. Chapter 2 covers motion in a straight line (1D kinematics). Chapter 4 extends these ideas to motion in a plane, including projectile motion and circular motion. The three equations of motion derived in these chapters form the backbone of all kinematics problems.
Kinematics applies to everyday phenomena. A ball thrown upward, a car accelerating on a highway, and a satellite orbiting Earth all obey kinematics equations. Understanding these formulas helps students build a strong foundation for dynamics, work-energy theorems, and rotational motion studied later in Class 11 and Class 12.
Kinematics Formulas — Expressions and Variables
The three standard equations of motion (also called equations of kinematics) apply to uniform acceleration in a straight line.
First Equation of Motion:
\[ v = u + at \]
Second Equation of Motion:
\[ s = ut + \frac{1}{2}at^2 \]
Third Equation of Motion:
\[ v^2 = u^2 + 2as \]
Displacement in the nth Second:
\[ s_n = u + \frac{a}{2}(2n – 1) \]
| Symbol | Quantity | SI Unit |
|---|---|---|
| \( u \) | Initial velocity | m/s |
| \( v \) | Final velocity | m/s |
| \( a \) | Uniform acceleration | m/s² |
| \( t \) | Time elapsed | s (seconds) |
| \( s \) | Displacement | m (metres) |
| \( n \) | nth second (integer) | dimensionless |
| \( g \) | Acceleration due to gravity | m/s² (9.8 m/s²) |
| \( \theta \) | Angle of projection | degrees or radians |
Derivation of the Three Equations of Motion
All three equations are derived from the definitions of velocity and acceleration.
Step 1: Acceleration is defined as \( a = \frac{v – u}{t} \). Rearranging gives the first equation: \( v = u + at \).
Step 2: Displacement equals the area under a velocity-time graph. For uniform acceleration, this area is a trapezium. Therefore \( s = \frac{(u + v)}{2} \times t \). Substituting \( v = u + at \) into this expression yields the second equation: \( s = ut + \frac{1}{2}at^2 \).
Step 3: Eliminating \( t \) from the first equation and the average-velocity expression gives the third equation: \( v^2 = u^2 + 2as \).
These three steps follow the NCERT Class 11 Physics derivation exactly, making them directly examinable in CBSE board papers.
Complete Kinematics Formula Sheet
Use this table as a one-stop reference for all kinematics formulas covered in NCERT Class 11 Physics and tested in JEE Main, JEE Advanced, and NEET.
| Formula Name | Expression | Variables | SI Units | NCERT Chapter |
|---|---|---|---|---|
| First Equation of Motion | \( v = u + at \) | u=initial vel., v=final vel., a=acceleration, t=time | m/s | Class 11, Ch 2 |
| Second Equation of Motion | \( s = ut + \frac{1}{2}at^2 \) | s=displacement, u=initial vel., a=acceleration, t=time | m | Class 11, Ch 2 |
| Third Equation of Motion | \( v^2 = u^2 + 2as \) | v=final vel., u=initial vel., a=acceleration, s=displacement | m²/s² | Class 11, Ch 2 |
| Displacement in nth Second | \( s_n = u + \frac{a}{2}(2n-1) \) | s_n=displacement in nth second, u=initial vel., a=acceleration | m | Class 11, Ch 2 |
| Average Velocity | \( v_{avg} = \frac{u + v}{2} \) | u=initial vel., v=final vel. | m/s | Class 11, Ch 2 |
| Projectile Time of Flight | \( T = \frac{2u \sin\theta}{g} \) | u=initial speed, θ=angle, g=9.8 m/s² | s | Class 11, Ch 4 |
| Projectile Maximum Height | \( H = \frac{u^2 \sin^2\theta}{2g} \) | u=initial speed, θ=angle of projection, g=gravity | m | Class 11, Ch 4 |
| Projectile Horizontal Range | \( R = \frac{u^2 \sin 2\theta}{g} \) | u=initial speed, θ=angle, g=gravity | m | Class 11, Ch 4 |
| Relative Velocity (1D) | \( v_{AB} = v_A – v_B \) | v_A=velocity of A, v_B=velocity of B | m/s | Class 11, Ch 2 |
| Velocity Components (Projectile) | \( v_x = u\cos\theta,\; v_y = u\sin\theta – gt \) | v_x=horizontal vel., v_y=vertical vel., t=time | m/s | Class 11, Ch 4 |
| Free Fall Displacement | \( h = \frac{1}{2}gt^2 \) | h=height fallen, g=9.8 m/s², t=time | m | Class 11, Ch 2 |
| Uniform Circular Motion Speed | \( v = \frac{2\pi r}{T} \) | r=radius, T=time period | m/s | Class 11, Ch 4 |
Kinematics Formulas — Solved Examples
Example 1 (Class 9-10 Level — Uniform Acceleration)
Problem: A car starts from rest and accelerates uniformly at 4 m/s². Find its velocity and displacement after 6 seconds.
Given: Initial velocity \( u = 0 \) m/s, acceleration \( a = 4 \) m/s², time \( t = 6 \) s.
Step 1: Apply the first equation of motion: \( v = u + at \)
Step 2: Substitute values: \( v = 0 + 4 \times 6 = 24 \) m/s
Step 3: Apply the second equation of motion: \( s = ut + \frac{1}{2}at^2 \)
Step 4: Substitute values: \( s = 0 \times 6 + \frac{1}{2} \times 4 \times 6^2 = 0 + 2 \times 36 = 72 \) m
Answer
Final velocity = 24 m/s; Displacement = 72 m
Example 2 (Class 11-12 Level — Projectile Motion)
Problem: A ball is projected at an angle of 30° with the horizontal at an initial speed of 40 m/s. Calculate the maximum height reached and the horizontal range. Take \( g = 10 \) m/s².
Given: \( u = 40 \) m/s, \( \theta = 30^\circ \), \( g = 10 \) m/s².
Step 1: Find maximum height using \( H = \frac{u^2 \sin^2\theta}{2g} \)
Step 2: Calculate \( \sin 30^\circ = 0.5 \), so \( \sin^2 30^\circ = 0.25 \)
Step 3: Substitute: \( H = \frac{40^2 \times 0.25}{2 \times 10} = \frac{1600 \times 0.25}{20} = \frac{400}{20} = 20 \) m
Step 4: Find horizontal range using \( R = \frac{u^2 \sin 2\theta}{g} \)
Step 5: Calculate \( \sin 2 \times 30^\circ = \sin 60^\circ = \frac{\sqrt{3}}{2} \approx 0.866 \)
Step 6: Substitute: \( R = \frac{1600 \times 0.866}{10} = \frac{1385.6}{10} \approx 138.6 \) m
Answer
Maximum Height = 20 m; Horizontal Range ≈ 138.6 m
Example 3 (JEE/NEET Level — Relative Motion and Displacement)
Problem: Two trains A and B move along parallel tracks. Train A moves at 72 km/h and Train B moves at 54 km/h in the same direction. Train A decelerates at 1 m/s². Find the time after which Train A and Train B have the same velocity. Also find the distance covered by Train A in this time.
Given: \( u_A = 72 \) km/h \( = 20 \) m/s, \( u_B = 54 \) km/h \( = 15 \) m/s (constant), \( a_A = -1 \) m/s².
Step 1: Train B moves at constant velocity, so \( v_B = 15 \) m/s throughout.
Step 2: Apply the first equation to Train A: \( v_A = u_A + a_A t = 20 – t \)
Step 3: Set \( v_A = v_B \): \( 20 – t = 15 \Rightarrow t = 5 \) s
Step 4: Find displacement of Train A using \( s = ut + \frac{1}{2}at^2 \):
Step 5: \( s_A = 20 \times 5 + \frac{1}{2} \times (-1) \times 5^2 = 100 – 12.5 = 87.5 \) m
Step 6: Relative displacement of A with respect to B = \( 87.5 – (15 \times 5) = 87.5 – 75 = 12.5 \) m. Train A is still 12.5 m ahead of Train B when they reach the same speed.
Answer
Time = 5 s; Displacement of Train A = 87.5 m; Train A is 12.5 m ahead of Train B at that instant.
CBSE Exam Tips 2025-26
- Memorise all three equations of motion with their conditions. The equations apply only under uniform acceleration. Write this condition explicitly in your answer to earn full marks.
- Convert units before substituting. CBSE examiners frequently give speed in km/h. Always convert to m/s by multiplying by \( \frac{5}{18} \). Failing to convert is the single most common error in board papers.
- Draw a velocity-time graph for every problem where possible. CBSE 2025-26 marking schemes award 1 mark for a correct diagram. This also helps you verify your answer visually.
- Use the nth-second formula for tricky 3-mark questions. The expression \( s_n = u + \frac{a}{2}(2n-1) \) is frequently tested. We recommend practising at least five problems using this formula before your board exam.
- For projectile problems, split motion into horizontal and vertical components. Treat each component independently. Horizontal: uniform velocity. Vertical: uniformly accelerated motion under gravity.
- State all given quantities at the start of your solution. CBSE 2025-26 step-marking awards 1 mark for identifying given data correctly, even if the final answer is wrong.
Common Mistakes to Avoid
- Using equations without checking conditions: The three equations of motion are valid only for uniform (constant) acceleration. Never apply them to problems involving variable acceleration, such as a body moving under a variable force.
- Ignoring sign conventions: Always define a positive direction at the start. If upward is positive, then \( g = -9.8 \) m/s² for a body thrown upward. Many students use \( g = +9.8 \) m/s² regardless of direction, leading to sign errors.
- Confusing distance and displacement: Displacement is a vector quantity and can be negative. Distance is always positive. When a ball is thrown upward and returns, its displacement is zero but the distance covered is twice the maximum height.
- Forgetting that \( s_n \) is not total displacement: The formula \( s_n = u + \frac{a}{2}(2n-1) \) gives displacement in the nth second only, not the total displacement after n seconds. Students frequently substitute this into the wrong context.
- Applying maximum range angle blindly: Maximum horizontal range occurs at \( \theta = 45^\circ \) only on flat ground with no air resistance. If the question involves a cliff or elevated launch point, the angle for maximum range changes.
JEE/NEET Application of Kinematics Formulas
In our experience, JEE aspirants who master Kinematics Formulas early gain a significant advantage. Kinematics is the starting point of mechanics, and almost every JEE Main paper includes 2–3 questions from this chapter. JEE Advanced tests kinematics through multi-concept problems combining relative motion, projectile motion, and graphs.
Application Pattern 1: Velocity-Time Graph Analysis
JEE Main frequently presents a velocity-time graph and asks for displacement, acceleration, or the time at which two bodies meet. The area under a v-t graph gives displacement. The slope gives acceleration. Practise identifying these quantities quickly from graphs. NEET also tests this concept in its physics section.
Application Pattern 2: Projectile Motion with Constraints
JEE Advanced problems often involve a projectile that must pass through a specific point or land on an inclined plane. Use the parametric equations \( x = u\cos\theta \cdot t \) and \( y = u\sin\theta \cdot t – \frac{1}{2}gt^2 \) simultaneously. Eliminate \( t \) to get the equation of trajectory: \( y = x\tan\theta – \frac{gx^2}{2u^2\cos^2\theta} \). This trajectory equation appears directly in JEE Advanced problems.
Application Pattern 3: Relative Motion in Two Dimensions
NEET and JEE Main test relative velocity in river-boat problems and rain-man problems. If a boat heads perpendicular to a river current, the resultant velocity is \( v_{resultant} = \sqrt{v_{boat}^2 + v_{river}^2} \). The drift along the river bank is \( d = v_{river} \times t \), where \( t = \frac{width}{v_{boat}} \). Our experts suggest drawing a vector diagram for every relative motion problem to avoid direction errors.
We also recommend reviewing the Friction Force Formula alongside kinematics, as JEE problems often combine uniform acceleration on rough surfaces with these equations of motion.
FAQs on Kinematics Formulas
Explore More Physics Formulas
Kinematics is just the beginning of your physics journey. Once you are confident with these formulas, explore related topics to build a complete understanding of mechanics and beyond.
- Visit our Complete Physics Formulas Hub for a full list of Class 11 and Class 12 formula sheets.
- Study the Friction Force Formula to understand how kinematics combines with surface forces in real-world problems.
- Review the Heat Capacity Formula for Class 11 thermodynamics, which follows kinematics in the NCERT syllabus sequence.
- Check the Electrical Formulas page for Class 12 physics preparation.
For the official NCERT Physics textbook and syllabus, refer to the NCERT official website.