Motion in a Straight Line JEE Notes 2026 | Formulas, PYQs

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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.

💡Expert Tip — Saransh Gupta, IIT Bombay AIR-41

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.

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.

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.

🔢 Kinematic Equations — Constant Acceleration

Eq. 1

$$v = u + at$$

velocity–time relation

Eq. 2

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

displacement–time relation

Eq. 3

$$v^2 = u^2 + 2as$$

velocity–displacement relation

Bonus

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

displacement in the n^th second

Where: $u$ = initial velocity, $v$ = final velocity, $a$ = acceleration (const.), $t$ = time, $s$ = displacement.

💡Expert Tip — eSaral Academic Team

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.

Uniform Motion (a = 0) x t slope = v Constant Acceleration x t parabola Object at Rest (v = 0) x t slope = 0

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.

Uniform Motion (a = 0) v t Area = s slope = 0 Uniform Acceleration v t Area = s slope = a Deceleration (reversal) v t +s −s v=0 here

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.

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:

📐 Calculus Relations

From the definition of velocity

$$v = \frac{dx}{dt} \implies dx = v\,dt$$

From the definition of acceleration

$$a = \frac{dv}{dt} \implies dv = a\,dt$$

Chain rule form

$$a = v\frac{dv}{dx} \implies v\,dv = a\,dx$$

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.

📐 Relative Motion Formulas

Relative velocity of A w.r.t. B

$$v_{AB} = v_A - v_B$$

Relative acceleration

$$a_{AB} = a_A - a_B$$

Relative displacement

$$x_{AB} = x_A - x_B$$

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.

⬇️ Free Fall Equations (Taking upward as +ve, $g = 9.8$ m/s² or $10$ m/s²)

E1

$$v = u - gt$$

velocity at time t

E2

$$h = ut - \frac{1}{2}gt^2$$

height gained in time t

E3

$$v^2 = u^2 - 2gh$$

velocity at height h

E4

$$H_{max} = \frac{u^2}{2g}$$

maximum height (when u is upward)

E5

$$t_{up} = t_{down} = \frac{u}{g}$$

time of ascent = time of descent

Ground (reference) u↑ H_max g v decreases on way up v increases on way down H

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.

✅ Solved Example 1 — Equations of Motion

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.

1

Given: $u = 0$, $a = 4$ m/s², $t = 5$ s.

2

(a) Final velocity using $v = u + at$:
$v = 0 + 4 \times 5 = 20$ m/s

3

(b) Distance in 5 s using $s = ut + \frac{1}{2}at^2$:
$s = 0 + \frac{1}{2}(4)(25) = 50$ m

4

(c) Distance in 5th second using $s_n = u + \frac{a}{2}(2n-1)$:
$s_5 = 0 + \frac{4}{2}(2 \times 5 - 1) = 2 \times 9 = 18$ m

Answers: (a) 20 m/s  |  (b) 50 m  |  (c) 18 m

✅ Solved Example 2 — Free Fall

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)

1

Given: $u = 20$ m/s (upward), $g = 10$ m/s², at max height $v = 0$.

2

Maximum height using $v^2 = u^2 - 2gH$:
$0 = 400 - 2(10)H \implies H = 20$ m

3

Time of ascent using $v = u - gt$:
$0 = 20 - 10t \implies t_{up} = 2$ s

4

Total time: Since $t_{up} = t_{down}$, total time $= 2 \times 2 = 4$ s.

H_max = 20 m  |  Total time = 4 s

✅ Solved Example 3 — Variable Acceleration (JEE Advanced Level)

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.

1

Find velocity: Integrate acceleration.
$$v(t) = u + \int_0^t (3t'^2 - 2t')\,dt' = 2 + \left[t'^3 - t'^2\right]_0^t = 2 + t^3 - t^2$$

2

At $t = 2$: $v = 2 + 8 - 4 = 6$ m/s

3

Find position: Integrate velocity.
$$x(t) = 1 + \int_0^t (2 + t'^3 - t'^2)\,dt' = 1 + 2t + \frac{t^4}{4} - \frac{t^3}{3}$$

4

At $t = 2$: $x = 1 + 4 + 4 - \frac{8}{3} = 9 - 2.67 \approx 6.33$ m

v(2) = 6 m/s  |  x(2) ≈ 6.33 m

Quick Formula Sheet — Paste in Your Notes

 Complete Formula Reference — Motion in a Straight Line

Displacement

$\Delta x = x_f - x_i$

Avg. Velocity

$\bar{v} = \dfrac{\Delta x}{\Delta t}$

Avg. Speed

$\bar{s} = \dfrac{d}{\Delta t}$

Inst. Velocity

$v = \dfrac{dx}{dt}$

Inst. Acceleration

$a = \dfrac{dv}{dt} = v\dfrac{dv}{dx}$

1st Equation

$v = u + at$

2nd Equation

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

3rd Equation

$v^2 = u^2 + 2as$

n-th second

$s_n = u + \tfrac{a}{2}(2n-1)$

Max Height

$H = \dfrac{u^2}{2g}$

Time of flight

$T = \dfrac{2u}{g}$

Relative velocity

$v_{AB} = v_A - v_B$

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.

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.

Once one-dimensional kinematics is clear, the natural next step is extending motion to two dimensions with vectors, where direction and component resolution become essential.

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

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