Motion in a Straight Line JEE Notes 2026 | Formulas, PYQs
eSaral Academic and Editorial Team
Summary
Motion in a Straight Line JEE Notes cover position, distance, displacement, speed, velocity and acceleration, average and instantaneous values, the equations of motion for constant acceleration, position–time and velocity–time graphs, relative motion and free fall, with a calculus method for variable acceleration.
Motion in a Straight Line is Chapter 2 of NCERT Class 11 Physics, and it is the base of kinematics. The same ideas return in projectile motion, circular motion, Laws of Motion and Work, Energy and Power. These Motion in a Straight Line JEE Notes follow the topics of Unit 2 (Kinematics) of the JEE Main 2026 syllabus, and they add the calculus approach for JEE Advanced.
Motion in a Straight Line JEE Notes: Chapter Overview and Syllabus
The JEE Main 2026 syllabus (Unit 2, Kinematics) lists these topics for motion in a straight line:
- The frame of reference
- Speed and velocity, uniform and non-uniform motion
- Average speed and instantaneous velocity
- Uniformly accelerated motion, velocity–time and position–time graphs, and the relations for uniformly accelerated motion
- Relative velocity
The calculus method for non-uniform acceleration is not listed separately, but it is the standard tool for non-uniform motion problems, especially in JEE Advanced. Motion in a plane and projectile motion are the next topics in the same unit.
Tip from the eSaral Physics team: Before you use a formula, ask one question: is the acceleration constant? The three equations of motion work only if it is. If the acceleration changes, use graphs or calculus.
Key Concepts in Motion in a Straight Line JEE Notes
Quantity | Meaning | Type | Formula |
|---|---|---|---|
Position | Location with respect to a reference origin | Vector (1D: a signed number) | $x$ |
Distance | Total length of the path | Scalar | $d \geq \lvert \Delta x \rvert$ |
Displacement | Change in position | Vector | $\Delta x = x_f - x_i$ |
Speed | Rate of change of distance | Scalar | $\text{speed} = \dfrac{d}{t}$ |
Velocity | Rate of change of displacement | Vector | $v = \dfrac{dx}{dt}$ |
Acceleration | Rate of change of velocity | Vector | $a = \dfrac{dv}{dt}$ |
The displacement can be zero even when the distance is not, and the speed is never less than the magnitude of the velocity. Retardation is acceleration in the direction opposite to the velocity, and a negative velocity means motion in the negative direction.
Example of the trap: a ball thrown upward and caught back has zero displacement, so its average velocity is zero, but its average speed is not zero. Ask yourself whether the path is in one direction only.
Types of Straight-Line Motion in Motion in a Straight Line JEE Notes
Type | Speed | Acceleration | v–t graph | Example |
|---|---|---|---|---|
Uniform motion | Constant | Zero | Horizontal line | A train moving at constant speed on a straight track |
Non-uniform motion | Changing | Not zero | Sloped or curved line | A car speeding up from rest |
For uniform motion, the position is $x = x_0 + vt$, and the x–t graph is a straight line with slope $v$.
Average and Instantaneous Quantities in Motion in a Straight Line JEE Notes
Quantity | Average (over an 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 | $\bar{s} = \dfrac{\text{total distance}}{\Delta t}$ | $\lvert v \rvert$ |
Acceleration | $\bar{a} = \dfrac{\Delta v}{\Delta t} = \dfrac{v_2 - v_1}{t_2 - t_1}$ | $a = \dfrac{dv}{dt} = v\dfrac{dv}{dx}$ |
Average speed for two equal parts:
- Equal times at speeds $v_1$ and $v_2$: $\bar{s} = \dfrac{v_1 + v_2}{2}$
- Equal distances at speeds $v_1$ and $v_2$: $\bar{s} = \dfrac{2v_1v_2}{v_1 + v_2}$
Equations of Motion for Constant Acceleration in Motion in a Straight Line JEE Notes
These hold only when the acceleration $a$ is constant.
$$v = u + at$$ $$s = ut + \frac{1}{2}at^2$$ $$v^2 = u^2 + 2as$$ $$s_n = u + \frac{a}{2}(2n - 1)$$
Here $u$ is the initial velocity, $v$ is the final velocity, $s$ is the displacement, $t$ is the time, and $s_n$ is the displacement during the $n$-th second only. To see where $s_n$ comes from, subtract the displacement in $(n-1)$ seconds from the displacement in $n$ seconds: $s_n = s(n) - s(n-1)$.
Position–Time and Velocity–Time Graphs in Motion in a Straight Line JEE Notes
Graph | Slope gives | Area gives | Shape for constant $a$ |
|---|---|---|---|
x–t (position–time) | Instantaneous velocity | Not used | Parabola, if $a \neq 0$ |
v–t (velocity–time) | Instantaneous acceleration | Displacement | Straight line |
a–t (acceleration–time) | Rate of change of acceleration | Change in velocity $\Delta v$ | Horizontal line |
- x–t graph: a straight line means uniform motion, a horizontal line means the object is at rest, and a parabola means uniform acceleration. A steeper slope means a larger speed, and a negative slope means motion in the negative direction.
- v–t graph: the area between the curve and the time axis is the displacement. The area above the axis is positive displacement, and the area below the axis is negative displacement. The total distance is the sum of the magnitudes of all the areas.
- a–t graph: the area under the graph is the change in velocity.
Calculus Method for Variable Acceleration in Motion in a Straight Line JEE Notes
The three basic relations are:
$$v = \frac{dx}{dt}, \qquad a = \frac{dv}{dt}, \qquad a = v\frac{dv}{dx}$$
Given | Method |
|---|---|
$a = f(t)$ | $v(t) = u + \displaystyle\int_0^t a\,dt$ and $x(t) = x_0 + \displaystyle\int_0^t v\,dt$ |
$a = f(v)$ | $\displaystyle\int_u^v \frac{dv}{f(v)} = t$ |
$a = f(x)$ | $\displaystyle\int_u^v v\,dv = \int_{x_0}^{x} f(x)\,dx$ |
Relative Motion in a Straight Line in Motion in a Straight Line JEE Notes
$$v_{AB} = v_A - v_B, \qquad a_{AB} = a_A - a_B, \qquad x_{AB} = x_A - x_B$$
- Two objects moving in the same direction: the relative speed is the difference of the speeds.
- Two objects moving in opposite directions: the relative speed is the sum of the speeds.
- Chase problem: object A chases object B. They meet when $x_A = x_B$, which means that the relative displacement $x_{AB}$ becomes zero. If A is slowing down or B is speeding up, A just manages to catch B only when the relative displacement and the relative velocity both become zero at the same moment.
Free Fall and Motion Under Gravity in Motion in a Straight Line JEE Notes
Free fall is uniformly accelerated motion with $a = -g$, taking upward as positive ($g = 9.8$ m/s² or 10 m/s²).
$$v = u - gt, \qquad h = ut - \frac{1}{2}gt^2, \qquad v^2 = u^2 - 2gh$$
For a body projected upward with speed $u$ and returning to the starting level:
$$H_{max} = \frac{u^2}{2g}, \qquad t_{up} = t_{down} = \frac{u}{g}, \qquad T = \frac{2u}{g}$$
At the maximum height the velocity is zero, but the acceleration is still $g$ downward. The body returns to the starting point with the same speed $u$, in the downward direction.
Quick Formula Sheet for Motion in a Straight Line JEE Notes
Concept | Formula |
|---|---|
Displacement | $\Delta x = x_f - x_i$ |
Average velocity | $\bar{v} = \dfrac{\Delta x}{\Delta t}$ |
Average speed | $\bar{s} = \dfrac{d}{\Delta t}$ |
Instantaneous velocity | $v = \dfrac{dx}{dt}$ |
Instantaneous acceleration | $a = \dfrac{dv}{dt} = v\dfrac{dv}{dx}$ |
First equation | $v = u + at$ |
Second equation | $s = ut + \tfrac{1}{2}at^2$ |
Third equation | $v^2 = u^2 + 2as$ |
$n$-th second | $s_n = u + \tfrac{a}{2}(2n - 1)$ |
Maximum height | $H = \dfrac{u^2}{2g}$ |
Time of flight | $T = \dfrac{2u}{g}$ |
Relative velocity | $v_{AB} = v_A - v_B$ |
Solved Examples from Motion in a Straight Line JEE Notes
Example 1 (equations of motion): A car starts from rest and accelerates uniformly at 4 m/s². Find (a) the velocity after 5 s, (b) the distance in 5 s, and (c) the distance in the 5th second.
(a) $v = u + at = 0 + 4 \times 5 = 20$ m/s. (b) $s = ut + \tfrac{1}{2}at^2 = \tfrac{1}{2}(4)(25) = 50$ m. (c) $s_5 = u + \tfrac{a}{2}(2n - 1) = 0 + 2 \times 9 = 18$ m.
Example 2 (free fall): A ball is thrown upward at 20 m/s. Find the maximum height and the total time to return (take $g = 10$ m/s²).
$H = \dfrac{u^2}{2g} = \dfrac{400}{20} = 20$ m. $t_{up} = \dfrac{u}{g} = 2$ s, so the total time is $T = 4$ s.
Example 3 (variable acceleration): The acceleration of a particle is $a = (3t^2 - 2t)$ m/s², with $u = 2$ m/s and $x_0 = 1$ m. Find $v$ and $x$ at $t = 2$ s.
$v(t) = 2 + \displaystyle\int_0^t (3t'^2 - 2t')\,dt' = 2 + t^3 - t^2$, so $v(2) = 2 + 8 - 4 = 6$ m/s. $x(t) = 1 + 2t + \dfrac{t^4}{4} - \dfrac{t^3}{3}$, so $x(2) = 1 + 4 + 4 - \dfrac{8}{3} \approx 6.33$ m.
Example 4 (average speed): A car covers the first half of a distance at 40 km/h and the second half at 60 km/h. Find the average speed.
For equal distances, $\bar{s} = \dfrac{2v_1v_2}{v_1 + v_2} = \dfrac{2 \times 40 \times 60}{100} = 48$ km/h.
Example 5 (relative motion): Two trains of lengths 100 m and 150 m move towards each other on parallel tracks at 36 km/h and 54 km/h. How long do they take to cross each other completely?
The relative speed is $36 + 54 = 90$ km/h $= 25$ m/s, and the distance to cover is $100 + 150 = 250$ m. So $t = \dfrac{250}{25} = 10$ s.
Example 6 (chase): A police car moving at 20 m/s is 100 m behind a thief moving at 15 m/s. When does the police car catch the thief?
The relative speed is $20 - 15 = 5$ m/s, so $t = \dfrac{100}{5} = 20$ s.
Example 7 (area of a v–t graph): The velocity of a body rises uniformly from 0 to 10 m/s in 5 s and then falls uniformly to 0 in another 5 s. Find the displacement.
The v–t graph is a triangle with base 10 s and height 10 m/s, so $s = \tfrac{1}{2} \times 10 \times 10 = 50$ m.
Common Mistakes in Motion in a Straight Line JEE Notes Problems
Mistake | Correct approach |
|---|---|
Using the equations of motion when $a$ is not constant | Use graphs or calculus |
Mixing up distance and displacement | Displacement is the net change in position. Distance is the total path. |
Taking the area of a v–t graph as the distance when it crosses the axis | Add the magnitudes of the areas above and below the axis for the distance |
Using the arithmetic mean for the average speed over equal distances | Use the harmonic mean $\dfrac{2v_1v_2}{v_1 + v_2}$ |
Using $s_n$ as the total displacement | $s_n$ is the displacement in the $n$-th second only |
Forgetting the sign of $g$ | Take upward as positive and use $a = -g$ |
How to Study Motion in a Straight Line JEE Notes
- Read the NCERT chapter first, and solve its examples and exercises. eSaral’s NCERT Solutions for Class 11 Physics Chapter 2 can help you check your answers.
- Practise graph questions daily: draw the v–t graph first, and read the displacement and acceleration from it.
- Once one-dimensional kinematics is clear, extend it to two dimensions with eSaral’s vectors and two-dimensional motion notes.
- Revise the unit list in eSaral’s JEE Main Physics revision notes, and test yourself with eSaral’s chapter-wise JEE Main PYQs and important questions.
Frequently Asked Questions
What is the difference between distance and displacement in motion in a straight line?
Distance is the total length of the path, and it is a scalar that is never negative. Displacement is the change in position from the start to the end, and it is 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 and a displacement of 0.
What are the equations of motion for constant acceleration?
They are $v = u + at$, $s = ut + \tfrac{1}{2}at^2$ and $v^2 = u^2 + 2as$, with $s_n = u + \tfrac{a}{2}(2n - 1)$ for the $n$-th second. They hold only when the acceleration is constant. These Motion in a Straight Line JEE Notes on eSaral include examples of each.
How many questions come from Motion in a Straight Line in JEE Main?
Kinematics is Unit 2 of the JEE Main Physics syllabus, and motion in a straight line is its first part. The number of questions changes from year to year, and the chapter also supports projectile motion and other topics. Check previous years’ papers for the trend.
What do the slope and the area of a velocity–time graph give?
The slope of a v–t graph gives the instantaneous acceleration, and the area under the graph gives the displacement. Areas above the time axis are positive, and areas below it are negative. For the total distance, add the magnitudes of the areas. This is a core part of the Motion in a Straight Line JEE Notes.
What is the average speed if a body covers equal distances at two different speeds?
The average speed is the harmonic mean, $\dfrac{2v_1v_2}{v_1 + v_2}$. For equal times at the two speeds, it is the arithmetic mean, $\dfrac{v_1 + v_2}{2}$.
How do you solve motion with variable acceleration?
If $a = f(t)$, integrate $a$ to get $v$ and integrate $v$ to get $x$. If $a = f(x)$, use $v\,dv = a\,dx$. If $a = f(v)$, use $dt = dv/f(v)$. In Motion in a Straight Line JEE Notes, this method is mainly useful for JEE Advanced.
What does negative velocity indicate?
Negative velocity means motion in the direction opposite to the chosen positive axis. It does not mean that the body is slowing down. A body slows down when its acceleration is opposite to its velocity.
Which chapter number is Motion in a Straight Line in Class 11 Physics?
In the current NCERT Class 11 Physics book, it is Chapter 2. These Motion in a Straight Line JEE Notes follow the chapter name.
Where can I find more Class 11 Physics notes on eSaral?
eSaral has chapter-wise Class 11 Physics notes, NCERT solutions, JEE Main chapter-wise PYQs and the Motion in a Straight Line JEE Notes on this page, so you can study the chapter and practise it in one place.
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eSaral Academic and Editorial Team
Team eSaral is the collective author profile for educational content created by eSaral’s teachers and academic contributors. The team draws on expertise from IIT graduates, doctors, experienced educators and subject specialists to develop resources for JEE, NEET and school students. Our articles aim to explain concepts clearly and help students study with confidence.
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