Motion in a Straight Line JEE Notes 2026 | Formulas, PYQs
Covers core kinematics concepts—position, displacement, distance, speed, velocity, and acceleration—along with differences between scalar and vector quantities; explains uniform and non-uniform motion, average vs instantaneous values, and key kinematic equations (v = u + at, s = ut + ½at², v² = u² + 2as, nth-second formula); includes interpretation of x–t and v–t graphs (slope and area meaning), calculus-based motion for variable acceleration, relative motion formulas, and free fall equations under gravity; provides solved JEE-level examples and highlights that this chapter consistently contributes 2–4 questions annually in JEE Main.
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Table of Contents
- Key Concepts
- Types of Straight-Line Motion
- All Formulas — Motion in a Straight Line
- Position–Time and Velocity–Time Graphs
- How to Use Calculus in Kinematics (JEE Advanced)
- Relative Motion in a Straight Line
- Free Fall and Motion Under Gravity
- Solved Examples — JEE Main Level
- Quick Formula Sheet — Paste in Your Notes
- JEE Main Weightage and Previous Year Trends
eSaral › JEE › JEE Main ›Motion in a Straight Line JEE Notes 2026
Key Concepts
Before diving into equations, you must be solid on these six physical quantities. A surprising number of JEE errors come from confusing scalar and vector quantities.
Position
Location of an object w.r.t. a reference origin. Measured in metres (m). Can be positive or negative.
Displacement
Change in position. Vector quantity. $\Delta x = x_f - x_i$. Can be zero even if the distance is non-zero.
Distance
Total path length covered. Scalar quantity. Always ≥ |displacement|.
Velocity
Rate of change of displacement. Vector. $v = \frac{dx}{dt}$. SI unit: m/s.
Speed
Rate of distance covered. Scalar. Speed ≥ |velocity| always.
Acceleration
Rate of change of velocity. Vector. $a = \frac{dv}{dt}$. SI unit: m/s².
In JEE, most "speed vs velocity" traps appear when an object reverses direction. A ball thrown upward and caught back has zero displacement — so its average velocity is zero — but non-zero average speed. Always ask: Is the path in one direction only?
Types of Straight-Line Motion
There are two fundamental categories of straight-line motion. Almost every JEE numerical fits into one of these.
| Type | Speed | Acceleration | v–t Graph | JEE Example |
|---|---|---|---|---|
| Uniform (ULM) | Constant | Zero | Horizontal line | Train on a straight track at 60 km/h |
| Non-uniform (NULM) | Variable | Non-zero | Sloped or curved line | Car accelerating from rest |
Uniform Linear Motion (ULM)
An object in ULM covers equal displacements in equal time intervals. The velocity is constant and acceleration $= 0$. The position–time graph is a straight line with slope $= v$.
$$x = x_0 + vt$$
Non-Uniform Linear Motion (NULM)
Here, the velocity changes with time. If acceleration is constant (uniformly accelerated motion), we apply the equations of motion. If acceleration varies, we use calculus.
All Formulas — Motion in a Straight Line
Average vs Instantaneous Quantities
This distinction is heavily tested in JEE. Know when to use which.
| Quantity | Average (over interval) | Instantaneous (at a point) |
|---|---|---|
| Velocity | $\bar{v} = \dfrac{\Delta x}{\Delta t} = \dfrac{x_2 - x_1}{t_2 - t_1}$ | $v = \dfrac{dx}{dt}$ |
| Speed | $\bar{s} = \dfrac{\text{Total distance}}{\Delta t}$ | $|\,v\,|$ (magnitude of instantaneous velocity) |
| Acceleration | $\bar{a} = \dfrac{\Delta v}{\Delta t} = \dfrac{v_2 - v_1}{t_2 - t_1}$ | $a = \dfrac{dv}{dt} = v\dfrac{dv}{dx}$ |
The Three Equations of Motion (Constant Acceleration)
These hold only when the acceleration $a$ is constant. In JEE, whenever you see "uniformly accelerated," reach for these.
Where: $u$ = initial velocity, $v$ = final velocity, $a$ = acceleration (const.), $t$ = time, $s$ = displacement.
The $s_n$ formula (displacement in the $n$-th second) appears in JEE at least once every few years. Don't confuse $s_n$ with total displacement — it's the displacement during the $n$-th second only. Derive it by computing $s_n - s_{n-1}$ from Eq. 2 to understand where it comes from.
Position–Time and Velocity–Time Graphs
Graph-based questions account for roughly 1–2 questions per JEE Main paper. Master this section, and you claim those marks instantly.
What does a position–time (x–t) graph tell you?
The slope of an x–t graph at any point equals the instantaneous velocity at that point. A steeper slope = faster motion. Negative slope = motion in the negative direction.
Figure 1 — Three common x–t graph shapes in JEE: uniform motion (straight line), uniformly accelerated (parabola), and a stationary object (horizontal line).
What does a velocity–time (v–t) graph tell you?
From a v–t graph, you can extract two critical pieces of information:
- Slope at any point = instantaneous acceleration at that time.
- Area under the graph (between the curve and the time-axis) = displacement in that time interval. Area above the axis = positive displacement; below the axis = negative displacement.
Figure 2 — v–t graphs: (a) constant velocity, (b) uniform acceleration — area gives displacement, (c) deceleration with reversal — areas above and below the axis give +ve and −ve displacements respectively.
| Graph Type | What Slope Gives | What Area Gives | Shape for const. a |
|---|---|---|---|
| x–t (position–time) | Instantaneous velocity | — | Parabola (if $a \neq 0$) |
| v–t (velocity–time) | Instantaneous acceleration | Displacement | Straight line |
| a–t (acceleration–time) | Jerk (rate of change of a) | Change in velocity ($\Delta v$) | Horizontal line |
How to Use Calculus in Kinematics (JEE Advanced)
For JEE Advanced and problems where acceleration is not constant, the calculus approach is essential. These three relations are the backbone:
When Acceleration is a Function of Time: $a = f(t)$
Integrate to find velocity, then integrate again to find displacement:
$$v(t) = u + \int_0^t a(t')\,dt', \qquad x(t) = x_0 + \int_0^t v(t')\,dt'$$When Acceleration is a Function of Velocity: $a = f(v)$
Use the separation-of-variables form:
$$\frac{dv}{dt} = f(v) \implies \int_{u}^{v}\frac{dv'}{f(v')} = t$$When Acceleration is a Function of Position: $a = f(x)$
Use $a = v\,dv/dx$:
$$\int_{u}^{v} v'\,dv' = \int_{x_0}^{x} f(x')\,dx'$$Relative Motion in a Straight Line
Relative motion is a frequent JEE source of tricky questions, especially involving trains, boats, and chasing problems.
Chasing / Overtaking Problem
Object A (behind) chases object B (ahead). They meet when $x_{AB} = 0$, i.e., when $x_A = x_B$. The minimum velocity to catch B requires $v_{AB}$ at the moment of closest approach to be zero.
Free Fall and Motion Under Gravity
Free fall is uniformly accelerated motion with $a = -g$ (taking upward as positive). This is the most common application of the kinematic equations in JEE.
Figure 3 — Free fall: ball projected upward with initial velocity $u$. At maximum height $H_{max}$, velocity $= 0$. Time of ascent = time of descent.
Solved Examples — JEE Main Level
Work through these carefully. Each one is modelled on a real JEE Main question pattern. Physics ki taiyari mein examples skip karna sabse badi galti hai.
A car starts from rest and accelerates uniformly at 4 m/s². Find (a) velocity after 5 s, (b) distance covered in 5 s, (c) distance in the 5th second.
$v = 0 + 4 \times 5 = 20$ m/s
$s = 0 + \frac{1}{2}(4)(25) = 50$ m
$s_5 = 0 + \frac{4}{2}(2 \times 5 - 1) = 2 \times 9 = 18$ m
A ball is thrown vertically upward with a speed of 20 m/s. Find the maximum height reached and the time to return to the starting point. (Take g = 10 m/s²)
$0 = 400 - 2(10)H \implies H = 20$ m
$0 = 20 - 10t \implies t_{up} = 2$ s
The acceleration of a particle is given by $a = (3t^2 - 2t)$ m/s². If the initial velocity is 2 m/s and the initial position is 1 m, find the velocity and position at $t = 2$ s.
$$v(t) = u + \int_0^t (3t'^2 - 2t')\,dt' = 2 + \left[t'^3 - t'^2\right]_0^t = 2 + t^3 - t^2$$
$$x(t) = 1 + \int_0^t (2 + t'^3 - t'^2)\,dt' = 1 + 2t + \frac{t^4}{4} - \frac{t^3}{3}$$
Quick Formula Sheet — Paste in Your Notes
JEE Main Weightage and Previous Year Trends
According to NTA data and eSaral's internal analysis of the last 10 years of JEE Main papers, Motion in a Straight Line contributes consistently across sessions.
| Topic | Avg. Questions/Year | Difficulty | JEE Adv. Relevance |
|---|---|---|---|
| Equations of motion (const. a) | 1–2 | Easy–Medium | ✅ High |
| Graphs (x–t and v–t) | 1 | Medium | ✅ High |
| Free fall / projectile base | 1 | Easy | ✅ High |
| Variable acceleration (calculus) | 0–1 | Hard | ✅✅ Very High |
| Relative motion | 0–1 | Medium | ✅ High |
| s_n formula (n-th second) | 0–1 | Medium | Medium |
Source: NTA JEE Main official website & eSaral internal paper analysis (2015–2025).
If you want to build on this chapter with structured problem sets, topic-wise PYQ practice, and live doubt resolution from IIT Bombay faculty, eSaral's JEE Main + Advanced course is designed to take you from notes like these to exam-ready in the most efficient path possible.
Bookmark this page for quick formula revision before your exam, and explore eSaral's chapter-wise JEE PYQs to test what you have learned here.JEE Main 2026
Frequently Asked Questions
Find answers to common questions.
What is the difference between distance and displacement in motion in a straight line?
Distance is the total path length covered — a scalar quantity that is always positive. Displacement is the shortest straight-line change in position from start to finish — a vector that can be positive, negative, or zero. An object that moves 5 m forward and 5 m back has a distance of 10 m but a displacement of 0 m.
How many questions come from motion in a straight line in JEE Main?
Motion in a Straight Line typically contributes 1–3 questions per JEE Main paper, accounting for roughly 4–12 marks. This includes direct numerical problems using kinematic equations, graph-based questions, and occasionally relative motion problems. The chapter also forms the conceptual base for projectile motion, which adds more questions indirectly.
What happens to displacement when an object reverses direction in free fall?
When an object is thrown upward and returns, displacement and distance differ. Displacement is the net change in position — if it returns to the starting point, displacement is zero. Distance, however, is the total path covered: twice the maximum height. On the v–t graph, the area above the time-axis gives positive displacement and the area below gives negative displacement.
What does negative velocity indicate?
Negative velocity indicates motion in the opposite direction of the chosen positive axis. It does not mean the object is slowing down.
What is retardation?
Retardation is simply negative acceleration — when acceleration acts opposite to velocity, reducing the speed of the object.