Physics
Kinematics
Kinematic Equations

Kinematics · JEE & NEET Physics

Kinematic Equations: notes and previous year questions

The four equations of motion for constant acceleration: where they come from, how to pick one, and how to use them for falling and thrown objects.

43 JEE Main questions (2002–2026)
13 NEET questions (2001–2022)

Kinematic Equations in short

The kinematic equations work only when the acceleration is constant; otherwise integrate.
Five quantities: u, v, a, t and s; knowing any three gives the other two.
u, v, a and s carry signs; choose the positive direction first; only t is never negative.
v = u + at because the velocity grows by a every second.

1Constant acceleration and the five variables

Throw a ball straight up: it slows down, stops for an instant and falls back. How high does it go, how long is it in the air, how fast does it land? The kinematic equations answer such questions whenever the acceleration is constant.

Constant acceleration means the velocity changes by the same amount every second. With $a = 2\ \text{m/s}^2$ from rest, the velocity goes 0, 2, 4, 6, 8 m/s. Gravity near the ground is like this: a thrown ball loses 10 m/s every second on the way up and gains 10 m/s every second on the way down.

Symbol	Meaning
$u$	initial velocity
$v$	final velocity
$a$	acceleration (constant)
$t$	time taken
$s$	displacement

Know any three of the five, and the equations give the other two. $u$ , $v$ , $a$ and $s$ have directions, so first choose which way is positive; only $t$ can never be negative. When something slows down, $a$ has the opposite sign to $u$ .

2The four equations and where they come from

Equation 1. Start with velocity $u$ and add $a$ every second. After $t$ seconds you have added $at$ :

v = u + at

A car from rest at 2 m/s² for 10 s: v = 0 + 2 × 10 = 20 m/s.

Equation 4. Because the velocity grows steadily, the average velocity is halfway between $u$ and $v$ . Displacement is average velocity times time:

s = \frac{(u + v)}{2}\,t

The same car: (0 + 20)/2 × 10 = 100 m.

Equation 2. Put $v = u + at$ into equation 4: $s = (2u + at)t/2$ . Here $ut$ is how far you would go at the starting speed, and $\tfrac{1}{2}at^2$ is the extra from speeding up:

s = ut + \tfrac{1}{2}at^2

The same car: 0 + ½ × 2 × 10² = 100 m.

Equation 3. Take $t = (v - u)/a$ from equation 1 and put it into equation 4: $s = (v + u)(v - u)/2a = (v^2 - u^2)/2a$ . The time disappears:

v^2 = u^2 + 2as

The most used equation, because many problems never give the time.

3Picking the right equation

Each equation leaves out exactly one of the five variables. Write down what is given and what is wanted; the one variable that is neither tells you which equation to use.

Variable not involved	Use
$s$	$v = u + at$
$v$	$s = ut + \tfrac{1}{2}at^2$
$t$	$v^2 = u^2 + 2as$
$a$	$s = (u + v)t/2$

4Free fall

In free fall only gravity acts (no air resistance). Then every object, heavy or light, has the same acceleration $g \approx 9.8\ \text{m/s}^2$ , which we round to $10\ \text{m/s}^2$ . Without air, a feather and a hammer land together.

For an object dropped from rest, $u = 0$ and $a = g$ :

What you need	Formula	Example
Velocity after time $t$	$v = gt$	after 2 s: 20 m/s
Height fallen in time $t$	$h = \tfrac{1}{2}gt^2$	after 2 s: 20 m
Velocity after falling $h$	$v = \sqrt{2gh}$	after 20 m: 20 m/s

Second by second: after 1 s it moves at 10 m/s and has fallen 5 m; after 2 s, 20 m/s and 20 m; after 3 s, 30 m/s and 45 m. The speed grows like $t$ , the distance like $t^2$ . A ball dropped from 45 m lands after 3 s; after 2 s it is still $45 - 20 = 25$ m above the ground.

5Thrown straight up

Choose up as positive. A ball thrown up at 20 m/s has $u = +20$ m/s and $a = -g = -10\ \text{m/s}^2$ for the whole flight: going up, at the top and coming down.

At the top $v = 0$ . So $0 = 20^2 - 2(10)H$ gives $H = 20$ m, and $0 = 20 - 10t$ gives 2 s to the top. The way down takes another 2 s.

Thrown up at speed u	Formula
Maximum height	$H = \dfrac{u^2}{2g}$
Time to reach the top	$t = \dfrac{u}{g}$
Total time in the air	$T = \dfrac{2u}{g}$
Speed at height $h$	$v = \sqrt{u^2 - 2gh}$

6Patterns and tricks

For motion starting from rest, the distances in successive seconds follow the odd numbers, and the totals follow the squares:

S_1 : S_2 : S_3 : S_4 = 1 : 3 : 5 : 7

A dropped ball falls 5, 15, 25, 35 m in its first four seconds.

s_1 : s_2 : s_3 : s_4 = 1 : 4 : 9 : 16

Totals 5, 20, 45, 80 m, because s = ½at² grows as t².

The distance in the nth second alone is the total after $n$ seconds minus the total after $n - 1$ seconds:

S_n = u + \frac{a}{2}(2n - 1)

From rest this gives 1, 3, 5, 7 … times a/2: the odd-number pattern.

7Solving a problem, step by step

List what is given and what is wanted.
Pick the equation by the missing variable.
Check the signs: which way is positive, and is $a$ positive or negative?
Use SI units: m, s, m/s, m/s².
Ask whether the answer makes sense.

Example: a car at 30 m/s brakes at $6\ \text{m/s}^2$ . No time is given, so $0 = 30^2 - 2(6)s$ and $s = 900/12 = 75$ m.

Summary

Key ideas

The kinematic equations work only when the acceleration is constant; otherwise integrate.
Five quantities: u, v, a, t and s; knowing any three gives the other two.
u, v, a and s carry signs; choose the positive direction first; only t is never negative.
v = u + at because the velocity grows by a every second.
With steady change, the average velocity is (u + v)/2, which gives s = (u + v)t/2 and then s = ut + ½at².
Eliminating t gives v² = u² + 2as, the most used equation.
Pick the equation that does not contain the variable you neither know nor want.
In free fall every object has the same acceleration g ≈ 10 m/s² downward.
Thrown up: take up as positive so a = −g all the way; at the top v = 0 but a is still g downward.
The upward flight mirrors the downward one: equal times and equal speeds at each height.
From rest, distances in successive seconds go 1 : 3 : 5 : 7 and totals go 1 : 4 : 9 : 16.
A ball dropped from H and one thrown up at u from below meet after H/u.

Every equation

Velocity: $v = u + at$
Displacement: $s = ut + \tfrac{1}{2}at^2$
No time: $v^2 = u^2 + 2as$
Average velocity: $s = \frac{(u + v)}{2}\,t$
Distance in the nth second: $S_n = u + \frac{a}{2}(2n - 1)$
Dropped: velocity: $v = gt$
Dropped: height fallen: $h = \tfrac{1}{2}gt^2$
Dropped: speed after falling h: $v = \sqrt{2gh}$
Thrown up: maximum height: $H = \frac{u^2}{2g}$
Thrown up: time to the top: $t = \frac{u}{g}$
Thrown up: time of flight: $T = \frac{2u}{g}$
Thrown up: speed at height h: $v = \sqrt{u^2 - 2gh}$
From rest, successive seconds: $1 : 3 : 5 : 7$
From rest, totals: $1 : 4 : 9 : 16$
Dropped and thrown balls meet: $t = \frac{H}{u}$

Previous year questions with solutions

Real JEE and NEET questions on kinematic equations. Try each one before you open the solution.

Q1JEE Main 2026Numerical answer

From $18\ \mathrm{m}$ height above the ground a ball is dropped from rest. The height above the ground at which the magnitude of velocity is equal to the magnitude of acceleration due to gravity (in the same set of units) is __________ $\mathrm{m}$ . (Take $g=10\ \mathrm{m/s^2}$ and neglect the air resistance.)

Show answer and solution

Answer: 13

The question asks where the speed reaches the number $10$ , since $g=10$ in these units. From $v^{2}=u^{2}+2gy$ with $u=0$ : $100=2(10)y$ , so $y=5\ \mathrm{m}$ of falling. The ball started at $18\ \mathrm{m}$ , so it is then $18-5=13\ \mathrm{m}$ above the ground.

Q2JEE Main 2026One correct option

Water drops fall from a tap on the floor, $5\ \mathrm{m}$ below, at regular intervals of time, the first drop striking the floor when the sixth drop begins to fall. The height at which the fourth drop will be from the ground, at the instant when the first drop strikes the ground, is: ( $g=10\ \mathrm{m/s^2}$ )

A $3.8\ \mathrm{m}$
B $4.0\ \mathrm{m}$
C $4.2\ \mathrm{m}$
D $2.5\ \mathrm{m}$

Show answer and solution

Answer: Option C

Six drops means five equal intervals $T$ , and the first drop's whole fall takes $5T$ . From $5=\tfrac{1}{2}(10)(5T)^{2}$ comes $T=0.2\ \mathrm{s}$ . The fourth drop started three intervals later, so at that instant it has been falling for $5T-3T=0.4\ \mathrm{s}$ and has dropped $\tfrac{1}{2}(10)(0.4)^{2}=0.8\ \mathrm{m}$ , leaving it $5-0.8=4.2\ \mathrm{m}$ up.

Q3JEE Main 2026One correct option

A gas balloon is going up with a constant velocity of $10\ \mathrm{m/s}$ . When this balloon reaches a height of $75\ \mathrm{m}$ , a stone is dropped from it and the balloon keeps moving up with the same velocity. The height of the balloon when the stone hits the ground is __________ $\mathrm{m}$ . (Take $g=10\ \mathrm{m/s^2}$ .)

A $85$
B $150$
C $129$
D $125$

Show answer and solution

Answer: Option D

"Dropped" from a rising balloon does not mean $u=0$ : the stone leaves with the balloon's own $10\ \mathrm{m/s}$ upwards. Taking up as positive, $-75=10t-5t^{2}$ gives $t^{2}-2t-15=0$ , so $t=5\ \mathrm{s}$ . In those $5\ \mathrm{s}$ the balloon climbs a further $50\ \mathrm{m}$ , reaching $75+50=125\ \mathrm{m}$ .

Practice questions, easy to hard

Three questions from the kinematic equations practice ladder: one easy, one medium, one hard.

Q4Numerical answer

A car already moving at $10\ \mathrm{m/s}$ accelerates steadily at $2\ \mathrm{m/s^{2}}$ for $5\ \mathrm{s}$ .

How far does it travel in those $5\ \mathrm{s}$ , in $\mathrm{m}$ ?

Show answer and solution

Answer: 75

The final speed is neither given nor asked for, so reach for the equation with no $v$ : $s = ut + \frac{1}{2}at^{2} = 10 \times 5 + \frac{1}{2} \times 2 \times 25 = 50 + 25 = 75\ \mathrm{m}$ . The $50\ \mathrm{m}$ is what it would have covered anyway; the $25\ \mathrm{m}$ is what the acceleration bought.