Kinematics Formulas For JEE 2026
Kinematics is one of the most basic and important chapters in JEE Physics, as it helps students understand the fundamentals of motion. In this topic, students learn about key concepts such as displacement, velocity, acceleration, and the equations of motion, which explain how objects move in different situations. The chapter also includes topics like projectile motion and relative motion, which are frequently used in numerical problems. When students clearly understand these concepts, it becomes much easier to approach and solve more advanced topics in mechanics.
In the JEE Mains exam, Kinematics usually contributes around 2–3 questions every year, making it a reliable scoring chapter for students. Since many mechanics problems are built on the concepts of motion, having a strong understanding of this topic is very helpful during problem solving. Practicing different types of questions and revising important formulas regularly can improve both speed and accuracy. For quick and effective revision, students can also refer to a well-organized JEE Mains Physics Formula PDF, which helps them remember key formulas and concepts during their preparation.
What Is Kinematics?
Kinematics is the study of motion — how objects move through space and time — without worrying about why they move (that comes later in Newton's laws). If you have ever watched a ball flying through the air or a car speeding down a highway, you were observing kinematics in action.
In this chapter we will build up the language of motion step by step: position, distance, displacement, speed, velocity, and acceleration. Then we will derive the key equations that let you predict where an object will be at any future time.
Distance and Displacement Formulas
When an object moves from one point to another, there are two different ways to describe "how far" it went. Understanding the difference is crucial because many JEE problems test exactly this distinction.
Distance: The total length of the path actually travelled by an object. Distance is always positive (or zero). It is a scalar — it has magnitude only, no direction.
Displacement: The shortest straight-line distance from the starting point to the ending point, along with the direction. Displacement is a vector — it has both magnitude and direction. It can be positive, negative, or zero.
Distance vs Displacement
- Distance $$\geq |\text{Displacement}|$$ always.
- Distance $$= |\text{Displacement}|$$ only when the object moves in a straight line without changing direction.
- If an object returns to its starting point, displacement $$= 0$$, but distance $$\neq 0$$.
Worked Example
A person walks 4 m east and then 3 m north. Find the distance and displacement.
Distance $$= 4 + 3 = 7$$ m (total path length).
Displacement $$= \sqrt{4^2 + 3^2} = \sqrt{16 + 9} = \sqrt{25} = 5$$ m, directed at an angle $$\theta = \tan^{-1}(3/4) \approx 36.87°$$ north of east.
Notice: distance (7 m) $$>$$ displacement (5 m).
Speed and Velocity Formulas
Speed and velocity both describe how fast an object is moving, but they differ in the same way that distance and displacement differ.
Speed: The rate at which distance is covered. Speed is a scalar. It is always positive or zero.
Velocity: The rate of change of displacement. Velocity is a vector. It can be positive, negative, or zero.
Speed and Velocity Formulas
Average speed:
$$$\text{Average speed} = \frac{\text{Total distance}}{\text{Total time}}$$$Average velocity:
$$$\text{Average velocity} = \frac{\text{Displacement}}{\text{Total time}} = \frac{\Delta x}{\Delta t}$$$Instantaneous velocity (velocity at a specific instant):
$$$v = \lim_{\Delta t \to 0} \frac{\Delta x}{\Delta t} = \frac{dx}{dt}$$$Worked Example
A car travels 100 km in 2 hours, then returns 60 km in 1 hour. Find average speed and average velocity.
Total distance $$= 100 + 60 = 160$$ km. Total time $$= 2 + 1 = 3$$ hours.
Average speed $$= \dfrac{160}{3} = 53.3$$ km/h.
Net displacement $$= 100 - 60 = 40$$ km (in the original direction).
Average velocity $$= \dfrac{40}{3} = 13.3$$ km/h (in the original direction).
Tip: When an object travels equal distances at speeds $$v_1$$ and $$v_2$$, the average speed is the harmonic mean: $$v_{\text{avg}} = \dfrac{2 v_1 v_2}{v_1 + v_2}$$. When it travels for equal times, the average speed is the arithmetic mean: $$v_{\text{avg}} = \dfrac{v_1 + v_2}{2}$$. JEE loves testing this!
Acceleration Formulas
Acceleration tells us how quickly the velocity is changing. If velocity is constant, acceleration is zero. If velocity increases, acceleration is in the direction of motion (positive). If velocity decreases, acceleration is opposite to motion (called deceleration or retardation).
Acceleration: The rate of change of velocity with respect to time. It is a vector quantity. SI unit: m/s$$^2$$.
Acceleration Formulas
Average acceleration:
$$$a_{\text{avg}} = \frac{\Delta v}{\Delta t} = \frac{v - u}{t}$$$where $$u$$ = initial velocity, $$v$$ = final velocity, $$t$$ = time taken.
Instantaneous acceleration:
$$$a = \frac{dv}{dt} = \frac{d^2 x}{dt^2}$$$Worked Example
A bike accelerates from 10 m/s to 30 m/s in 5 seconds. Find the acceleration.
$$a = \dfrac{v - u}{t} = \dfrac{30 - 10}{5} = \dfrac{20}{5} = 4 \text{ m/s}^2$$
This means the velocity increases by 4 m/s every second.
Equations of Motion (SUVAT)
When an object moves in a straight line with constant (uniform) acceleration, its motion is completely described by five variables: $$s$$ (displacement), $$u$$ (initial velocity), $$v$$ (final velocity), $$a$$ (acceleration), and $$t$$ (time). Any three of these determine the other two. The following three equations connect them.
Three Equations of Motion
First equation (relates $$v$$, $$u$$, $$a$$, $$t$$):
$$$v = u + at$$$Second equation (relates $$s$$, $$u$$, $$a$$, $$t$$):
$$$s = ut + \frac{1}{2}at^2$$$Third equation (relates $$v$$, $$u$$, $$a$$, $$s$$ — eliminates time):
$$$v^2 = u^2 + 2as$$$where $$u$$ = initial velocity (m/s), $$v$$ = final velocity (m/s), $$a$$ = acceleration (m/s$$^2$$), $$t$$ = time (s), $$s$$ = displacement (m).
Note: These equations are valid only when acceleration is constant. If acceleration varies with time, you must use calculus ($$v = \int a\,dt$$, $$s = \int v\,dt$$).
Worked Example
A car starts from rest and accelerates uniformly at 3 m/s$$^2$$. Find the distance covered in 10 seconds.
Given: $$u = 0$$ (starts from rest), $$a = 3$$ m/s$$^2$$, $$t = 10$$ s.
Using $$s = ut + \frac{1}{2}at^2$$:
$$s = 0 \times 10 + \frac{1}{2} \times 3 \times 10^2 = 0 + \frac{1}{2} \times 3 \times 100 = 150$$ m.
The car covers 150 m in 10 seconds.
Worked Example
A train moving at 20 m/s applies brakes and decelerates at 2 m/s$$^2$$. How far does it travel before stopping?
Given: $$u = 20$$ m/s, $$v = 0$$ (stops), $$a = -2$$ m/s$$^2$$ (deceleration).
Using $$v^2 = u^2 + 2as$$:
$$0 = 20^2 + 2(-2)s$$
$$0 = 400 - 4s \Rightarrow s = 100$$ m.
The train travels 100 m before stopping.
Distance Covered in the nth Second
Sometimes JEE asks for the distance covered specifically in the $$n$$th second (not in $$n$$ seconds). There is a direct formula for this.
Distance in the nth Second
$$$s_n = u + \frac{a}{2}(2n - 1)$$$where $$s_n$$ = displacement during the $$n$$th second, $$u$$ = initial velocity, $$a$$ = acceleration.
Worked Example
An object starts from rest with acceleration 4 m/s$$^2$$. Find the distance covered in the 3rd second.
$$s_3 = 0 + \dfrac{4}{2}(2 \times 3 - 1) = 2 \times 5 = 10$$ m.
Free Fall Formulas
When an object is dropped (or thrown) near the Earth's surface and air resistance is negligible, it experiences a constant acceleration due to gravity. This special case of uniformly accelerated motion is called free fall.
Acceleration due to gravity ($$g$$): The constant acceleration that all freely falling objects experience near Earth's surface. $$g \approx 9.8$$ m/s$$^2$$ (often approximated as 10 m/s$$^2$$ in JEE).
Free Fall Equations
Take downward as positive (common convention):
Object dropped from rest ($$u = 0$$):
- $$v = gt$$
- $$h = \frac{1}{2}gt^2$$
- $$v^2 = 2gh$$
Object thrown upward with initial velocity $$u$$ (take upward positive, so $$a = -g$$):
- $$v = u - gt$$
- $$h = ut - \frac{1}{2}gt^2$$
- $$v^2 = u^2 - 2gh$$
- Time to reach maximum height: $$t_{\text{up}} = \dfrac{u}{g}$$
- Maximum height: $$H = \dfrac{u^2}{2g}$$
- Total time of flight (return to same level): $$T = \dfrac{2u}{g}$$
Worked Example
A ball is thrown vertically upward at 20 m/s. Find (a) the maximum height, (b) the time of flight. Take $$g = 10$$ m/s$$^2$$.
(a) $$H = \dfrac{u^2}{2g} = \dfrac{20^2}{2 \times 10} = \dfrac{400}{20} = 20$$ m.
(b) $$T = \dfrac{2u}{g} = \dfrac{2 \times 20}{10} = 4$$ s.
The ball rises for 2 s, reaches 20 m, and falls back in another 2 s.
Tip: For a body thrown upward: (1) speed at any height during ascent = speed at the same height during descent, (2) time of ascent = time of descent (when returning to the same level).
Relative Motion Formulas
Motion is always measured with respect to some reference point (called a frame of reference). When two objects A and B are both moving, the velocity of A as seen by B (or "relative to B") is different from the velocity of A as seen by a stationary observer.
Relative Velocity: The velocity of object A as observed from object B.
Relative Velocity Formula
$$$\vec{v}_{AB} = \vec{v}_A - \vec{v}_B$$$where $$\vec{v}_{AB}$$ = velocity of A relative to B, $$\vec{v}_A$$ = velocity of A, $$\vec{v}_B$$ = velocity of B.
Special cases (1D motion):
- Objects moving in the same direction: $$v_{\text{rel}} = v_A - v_B$$
- Objects moving in opposite directions: $$v_{\text{rel}} = v_A + v_B$$
Worked Example
Two trains A and B move on parallel tracks. A moves at 60 km/h east and B at 40 km/h east. Find the velocity of A relative to B.
$$v_{AB} = v_A - v_B = 60 - 40 = 20$$ km/h (east).
Passenger in B sees train A moving ahead at 20 km/h.
If B were moving west at 40 km/h instead:
$$v_{AB} = 60 - (-40) = 100$$ km/h (east).
They approach each other at 100 km/h.
Projectile Motion Formulas
When an object is launched into the air at an angle (or horizontally) and moves under the influence of gravity alone (no air resistance), it follows a curved path called a projectile. Examples: a ball kicked at an angle, a bullet fired from a gun, water from a hose.
The key insight is that projectile motion can be broken into two independent components:
- Horizontal: No acceleration (constant velocity), since gravity acts only vertically.
- Vertical: Constant acceleration $$g$$ downward (same as free fall).
Projectile Launched at Angle $$\theta$$
Consider an object launched from the ground with initial speed $$u$$ at angle $$\theta$$ above the horizontal.
Projectile Motion Formulas
Components of initial velocity:
- Horizontal: $$u_x = u\cos\theta$$
- Vertical: $$u_y = u\sin\theta$$
At any time $$t$$:
- Horizontal position: $$x = (u\cos\theta)\,t$$
- Vertical position: $$y = (u\sin\theta)\,t - \frac{1}{2}gt^2$$
- Horizontal velocity: $$v_x = u\cos\theta$$ (constant)
- Vertical velocity: $$v_y = u\sin\theta - gt$$
Key quantities:
- Time of flight: $$T = \dfrac{2u\sin\theta}{g}$$
- Maximum height: $$H = \dfrac{u^2 \sin^2\theta}{2g}$$
- Range (horizontal distance): $$R = \dfrac{u^2 \sin 2\theta}{g}$$
Equation of Trajectory
Eliminating time from the $$x$$ and $$y$$ equations gives the path equation:
$$$y = x\tan\theta - \frac{g x^2}{2u^2 \cos^2\theta}$$$This is a parabola opening downward.
Worked Example
A ball is thrown at 40 m/s at $$30°$$ above the horizontal. Find the time of flight, maximum height, and range. ($$g = 10$$ m/s$$^2$$)
$$u = 40$$ m/s, $$\theta = 30°$$, so $$\sin 30° = 0.5$$, $$\cos 30° = \frac{\sqrt{3}}{2}$$, $$\sin 60° = \frac{\sqrt{3}}{2}$$.
Time of flight: $$T = \dfrac{2 \times 40 \times 0.5}{10} = \dfrac{40}{10} = 4$$ s.
Maximum height: $$H = \dfrac{40^2 \times (0.5)^2}{2 \times 10} = \dfrac{1600 \times 0.25}{20} = \dfrac{400}{20} = 20$$ m.
Range: $$R = \dfrac{40^2 \times \sin 60°}{10} = \dfrac{1600 \times \frac{\sqrt{3}}{2}}{10} = \dfrac{800\sqrt{3}}{10} = 80\sqrt{3} \approx 138.6$$ m.
Note: The range is maximum when $$\theta = 45°$$, giving $$R_{\max} = \dfrac{u^2}{g}$$. Two complementary angles ($$\theta$$ and $$90° - \theta$$) give the same range but different maximum heights.
Horizontal Projectile
When an object is thrown horizontally from a height (e.g., a ball rolling off a table), it is a special case with $$\theta = 0$$.
Horizontal Projectile ($$\theta = 0$$)
- $$u_x = u$$, $$u_y = 0$$
- $$x = ut$$, $$y = \frac{1}{2}gt^2$$
- Time to reach ground (from height $$h$$): $$t = \sqrt{\dfrac{2h}{g}}$$
- Range: $$R = u\sqrt{\dfrac{2h}{g}}$$
- Speed on hitting ground: $$v = \sqrt{u^2 + 2gh}$$
Worked Example
A stone is thrown horizontally at 10 m/s from the top of a 45 m high tower. Find where it lands and its speed on hitting the ground. ($$g = 10$$ m/s$$^2$$)
Time to fall: $$t = \sqrt{\dfrac{2 \times 45}{10}} = \sqrt{9} = 3$$ s.
Horizontal distance: $$x = 10 \times 3 = 30$$ m from the base of the tower.
Speed: $$v = \sqrt{10^2 + 2 \times 10 \times 45} = \sqrt{100 + 900} = \sqrt{1000} = 10\sqrt{10} \approx 31.6$$ m/s.
Motion Graphs — How to Read and Interpret
Graphs are a powerful way to visualise and analyse motion. JEE frequently tests your ability to read and interpret three types of graphs. Understanding what each graph's slope and area represent is the key.
Position-Time ($$x$$-$$t$$) Graph
Reading an $$x$$-$$t$$ Graph
- Slope of the $$x$$-$$t$$ graph $$= \dfrac{\Delta x}{\Delta t} =$$ velocity.
- Straight line $$\Rightarrow$$ constant velocity (uniform motion).
- Curve (concave up) $$\Rightarrow$$ increasing velocity (acceleration).
- Curve (concave down) $$\Rightarrow$$ decreasing velocity (deceleration).
- Horizontal line $$\Rightarrow$$ object is at rest (velocity $$= 0$$).
- A steeper slope means faster motion.
Velocity-Time ($$v$$-$$t$$) Graph
Reading a $$v$$-$$t$$ Graph
- Slope of the $$v$$-$$t$$ graph $$= \dfrac{\Delta v}{\Delta t} =$$ acceleration.
- Area under the $$v$$-$$t$$ graph $$=$$ displacement.
- Horizontal line $$\Rightarrow$$ constant velocity (zero acceleration).
- Straight line with positive slope $$\Rightarrow$$ uniform acceleration.
- Straight line with negative slope $$\Rightarrow$$ uniform deceleration.
- Area above time axis $$=$$ positive displacement; area below $$=$$ negative displacement.
Tip: For distance (not displacement) from a $$v$$-$$t$$ graph, take the absolute value of each area segment and add them. This matters when the velocity goes negative (object reverses direction).
Acceleration-Time ($$a$$-$$t$$) Graph
Reading an $$a$$-$$t$$ Graph
- Area under the $$a$$-$$t$$ graph $$=$$ change in velocity ($$\Delta v$$).
- Horizontal line at $$a = 0$$ $$\Rightarrow$$ constant velocity.
- Horizontal line at $$a = \text{constant}$$ $$\Rightarrow$$ uniformly accelerated motion.
Worked Example
A $$v$$-$$t$$ graph shows a straight line from $$v = 0$$ at $$t = 0$$ to $$v = 20$$ m/s at $$t = 5$$ s, then horizontal at $$v = 20$$ m/s from $$t = 5$$ s to $$t = 10$$ s. Find the total displacement.
Phase 1 ($$t = 0$$ to $$5$$ s): The graph is a triangle.
Area $$= \frac{1}{2} \times 5 \times 20 = 50$$ m.
Acceleration $$= \text{slope} = \dfrac{20 - 0}{5 - 0} = 4$$ m/s$$^2$$.
Phase 2 ($$t = 5$$ to $$10$$ s): The graph is a rectangle.
Area $$= 20 \times 5 = 100$$ m.
Acceleration $$= 0$$ (constant velocity).
Total displacement $$= 50 + 100 = 150$$ m.
Summary of Key Formulas
Kinematics Formula Sheet
| Quantity / Situation | Formula |
|---|---|
| Average speed | Total distance / Total time |
| Average velocity | $$\Delta x / \Delta t$$ |
| Equation 1 | $$v = u + at$$ |
| Equation 2 | $$s = ut + \frac{1}{2}at^2$$ |
| Equation 3 | $$v^2 = u^2 + 2as$$ |
| $$n$$th second distance | $$s_n = u + \frac{a}{2}(2n-1)$$ |
| Free fall (from rest) | $$h = \frac{1}{2}gt^2$$, $$v = gt$$ |
| Max height (thrown up) | $$H = u^2 / (2g)$$ |
| Projectile time of flight | $$T = 2u\sin\theta / g$$ |
| Projectile max height | $$H = u^2\sin^2\theta / (2g)$$ |
| Projectile range | $$R = u^2\sin 2\theta / g$$ |
| Relative velocity | $$\vec{v}_{AB} = \vec{v}_A - \vec{v}_B$$ |
Tip: In JEE problems, always start by listing the known quantities ($$u$$, $$v$$, $$a$$, $$s$$, $$t$$), identify which three you know, and pick the SUVAT equation that contains those three plus the unknown. This avoids unnecessary steps.
Kinematics Formulas For JEE 2026: Conclusion
Kinematics is one of the most fundamental chapters in JEE Physics, as it introduces students to the basic concepts of motion such as displacement, velocity, acceleration, and the equations of motion. A strong understanding of Kinematics Formulas For JEE 2026 makes it easier to solve numerical problems and understand advanced mechanics topics. Since this chapter frequently appears in the JEE exam, mastering the formulas and concepts can significantly improve a student's problem-solving ability.
For effective revision, students should regularly practice problems and keep a quick reference sheet of important formulas. Using a well-organized JEE Mains Physics Formula PDF can help students revise key equations quickly before exams and strengthen their preparation. By understanding the concepts clearly and practicing regularly, students can score well in Kinematics and build a strong foundation for the rest of JEE Physics.
