Kinematic Equations

Prerequisites

Before diving in, make sure you understand:

Motion in a Straight Line — displacement, velocity, acceleration basics
Frame of Reference — coordinate systems and sign conventions

The Setup: Why This Matters

Connect: Real Life → Physics

Imagine Virat Kohli hitting a six. The ball leaves the bat, rises, slows down, stops momentarily, then falls back.

Can you predict:

How HIGH will it go?
How LONG will it stay in air?
How FAST when it lands?

That’s EXACTLY what kinematic equations do — predict motion when acceleration is constant!

The Core: Four Magical Equations

When acceleration is constant, we have 5 variables and 4 equations. Master these, and you master 30% of Kinematics!

Symbol	What it means	Memory Trick
$u$	Initial velocity	U = “You start here”
$v$	Final velocity	V = “Velocity at the end”
$a$	Acceleration	A = “Accelerating forward”
$t$	Time	T = “Time elapsed”
$s$	Displacement	S = “Space covered”

Interactive: Visualize the Equations

Adjust the sliders to see how initial velocity and acceleration affect the s-t, v-t, and a-t graphs:

The “SUVAT” Equations

Equation 1: When you know time, need final velocity

$$\boxed{v = u + at}$$

Memory Trick: “Velocity = U + Acceleration × Time” → v = u + at

When to use: Given $u$, $a$, $t$ → Find $v$

Equation 2: When you know time, need displacement

$$\boxed{s = ut + \frac{1}{2}at^2}$$

Memory Trick: “Start with U×T, then add ½A×T²” → s = ut + ½at²

When to use: Given $u$, $a$, $t$ → Find $s$

Equation 3: When time is NOT given

$$\boxed{v^2 = u^2 + 2as}$$

Memory Trick: “V-squared minus U-squared equals 2AS” (like a chant!)

When to use: Given $u$, $v$, $a$ (no time!) → Find $s$

Pro Tip: This is the MOST USED equation in JEE because many problems don’t give time!

Equation 4: Average method (backup equation)

$$\boxed{s = \frac{(u + v)}{2} \times t}$$

When to use: When you know both initial AND final velocity

The Shortcut: Which Equation to Pick?

5-Second Selection Rule

Step 1: List what’s GIVEN and what’s NEEDED

Step 2: Find the equation that has EXACTLY those variables

Missing Variable	Use Equation
$v$ not in question	$s = ut + \frac{1}{2}at^2$
$t$ not in question	$v^2 = u^2 + 2as$
$s$ not in question	$v = u + at$
$a$ not in question	$s = \frac{(u+v)}{2}t$

“The variable you DON’T need should NOT be in the equation you choose!”

Special Weapon: nth Second Formula

For displacement in the nth second only (not total):

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

Memory Trick: “u + (a/2) × (2n - 1)” — just remember the odd number pattern!

Why is this powerful?

In JEE, they often ask: “Distance in 3rd second = ?” or “Distance in 5th second = ?”

Without this formula, you’d calculate $s_5 - s_4$. With it, direct answer!

Common Mistakes to Avoid

Trap #1: Sign Convention Errors

Problem: Ball thrown up with 20 m/s, find max height.

WRONG approach: $v^2 = u^2 + 2as$ $0 = 400 + 2(10)s$ → $s = -20$ m (negative?? 😱)

CORRECT approach: When going UP, $a = -g = -10$ m/s² $0 = 400 + 2(-10)s$ → $s = +20$ m (correct!)

Rule: Going UP = acceleration is NEGATIVE

Trap #2: Displacement vs Distance

Ball thrown up returns to hand.

Displacement = 0 (back to start)
Distance = 2 × max height

Many students write displacement = 2h. WRONG!

Trap #3: Using equations for variable acceleration

If acceleration is NOT constant (like $a = 2t$), these equations DON’T work!

Use calculus: $v = \int a \, dt$ and $s = \int v \, dt$

For variable acceleration problems, see Differentiation basics and Integration.

Free Fall: Ball Physics

Free fall = acceleration due to gravity = $g ≈ 10$ m/s² (or 9.8 m/s²)

Object Dropped (Released from Rest)

What you need	Formula	Quick Result
Velocity after $t$ sec	$v = gt$	After 2s → 20 m/s
Height fallen in $t$ sec	$h = \frac{1}{2}gt^2$	After 2s → 20 m
Velocity after falling $h$	$v = \sqrt{2gh}$	After 20m → 20 m/s

Object Thrown Upward

What you need	Formula
Max height	$H = \frac{u^2}{2g}$
Time to reach top	$t = \frac{u}{g}$
Total flight time	$T = \frac{2u}{g}$
Speed at height $h$	$v = \sqrt{u^2 - 2gh}$

Golden Rule

Speed going UP at height h = Speed coming DOWN at height h

If thrown at 30 m/s, it returns at 30 m/s (ignoring air resistance)

Pattern Recognition: JEE Favorites

Pattern 1: “Ratio of distances in successive seconds”

For object starting from rest ($u = 0$):

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

Memory: Odd numbers! 1, 3, 5, 7… — distances in successive seconds form an odd number sequence!

Why? Using $S_n = \frac{a}{2}(2n-1)$, for consecutive seconds you get $1a/2$, $3a/2$, $5a/2$…

Pattern 2: “Ratio of total distances”

For object starting from rest:

$$s_1 : s_2 : s_3 : s_4 = 1 : 4 : 9 : 16 = 1^2 : 2^2 : 3^2 : 4^2$$

Memory: Perfect squares! 1, 4, 9, 16… — total distances follow the square number pattern!

Pattern 3: “Two objects, one dropped, one thrown”

Object A dropped, Object B thrown up simultaneously from same height.

They meet at: Height $= H - \frac{1}{2}gt^2$ (distance A has fallen)

Time to meet: $t = \frac{H}{u}$ (where u is throwing speed of B)

Practice Problems (Level-wise)

Level 1: Foundation (NCERT Type)

Problem 1

A car accelerates from rest at 2 m/s² for 10 s. Find final velocity and distance.

Quick Solution:

$v = u + at = 0 + 2(10) =$ 20 m/s
$s = ut + \frac{1}{2}at² = 0 + \frac{1}{2}(2)(100) =$ 100 m

Level 2: JEE Main Type

Problem 2

A ball is dropped. Find ratio of distances in 1st, 2nd, 3rd seconds.

Instant Answer: $1 : 3 : 5$ (odd number pattern!)

Problem 3

A stone is thrown up with 40 m/s. Find: (a) max height, (b) time of flight