Learn at My Place/Competitive Exams/JEE Main & Advanced/Physics/Kinematics 1D & Calculus

JEE Main & Advanced / Physics / Chapter 2

Kinematics 1D & Calculus

Master one-dimensional motion with calculus — distance vs displacement, velocity, acceleration, equations of motion, free fall, and graphical analysis with full derivations and solved examples.

10 Solved Examples4 Session Tests60-Question Mock90s Per Question

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JEE Intro

Why Kinematics 1D Is the Engine of JEE Mechanics

Kinematics in one dimension is where the mathematical language of motion is first introduced. Every subsequent topic in mechanics — Newton's laws, work-energy theorem, collisions, circular motion — relies on the student being completely fluent in the ideas developed here.

The suvat equations (v = u + at, s = ut + ½at², v² = u² + 2as) are used in every single mechanics chapter. A student who derives these equations rather than memorising them will never misapply them. The derivations also introduce the calculus techniques (integration and differentiation) that become essential in the chapters on variable acceleration.

Graph-based questions are a recurring JEE Main pattern. Slope of x-t = velocity; area under v-t = displacement; slope of v-t = acceleration; area under a-t = Δv. Mastering these four relationships means you can decode any kinematics graph question in under 30 seconds.

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Section A

Notes: Kinematics 1D & Calculus

Original teaching copy for Learn at My Place

Motion, rest, distance vs displacement Average and instantaneous velocity Acceleration — uniform and variable Equations of motion and nth second formula Free fall under gravity Graphical analysis — s-t, v-t, a-t graphs

1. Kinematics Basics — Motion, Rest, and Reference Frame

Kinematics is the branch of mechanics that describes how objects move without asking why they move. Understanding kinematics begins with precise definitions of the simplest concepts — motion, rest, and frame of reference.

Rest and Motion are relative. A passenger sitting in a moving bus is at rest relative to the bus but in motion relative to the road. There is no absolute rest in the universe — every description of motion requires a specified observer or reference frame.

Reference Frame: A coordinate system attached to an observer. For 1D motion, a single number line is sufficient. We usually take a fixed point on the ground as origin.

Distance — total path length covered, always ≥ 0, scalar quantity.

Displacement — straight-line vector from initial to final position: $\Delta x = x_f - x_i$ . Can be zero or negative. SI unit: metre (m).

Key rule: distance ≥ |displacement|, with equality only when motion is along a straight line without reversal.

JEE tip: Questions often give a curved path and ask separately for distance and displacement. The path length is distance; the magnitude of the position change is displacement. For a semicircle of radius r: distance = πr, displacement = 2r.

2. Speed and Velocity — Average and Instantaneous

Speed and velocity are related but fundamentally different quantities. Speed is scalar (magnitude only); velocity is a vector (magnitude + direction).

\text{Average speed} = \frac{\text{total distance}}{\text{total time}} = \frac{d}{t}

\text{Average velocity} = \frac{\text{displacement}}{\text{time}} = \frac{\Delta x}{\Delta t}

Instantaneous velocity is defined using calculus as the limiting case of average velocity when the time interval shrinks to zero:

v = \lim_{\Delta t \to 0} \frac{\Delta x}{\Delta t} = \frac{dx}{dt}

This is the derivative of position with respect to time. Geometrically, instantaneous velocity is the slope of the tangent to the x-t graph at that instant.

Common JEE trap: Average speed ≠ magnitude of average velocity in general. For a round trip of equal distances: average speed =

\frac{2v_1 v_2}{v_1+v_2}

(harmonic mean). The arithmetic mean

\frac{v_1+v_2}{2}

only applies to equal time intervals.

When a particle returns to its starting point, displacement = 0, so average velocity = 0. But average speed is non-zero (total distance / total time).

3. Acceleration — Uniform and Variable

Acceleration is the rate of change of velocity. Like velocity, it is a vector — it has direction. A particle can be decelerating (slowing down) even if acceleration is in the positive direction, if velocity is negative.

\text{Average acceleration} = \frac{\Delta v}{\Delta t} = \frac{v_f - v_i}{t_f - t_i}

\text{Instantaneous acceleration} = \frac{dv}{dt} = \frac{d^2x}{dt^2}

Deceleration vs. negative acceleration: These are not the same. Deceleration means the particle is slowing down — the acceleration is opposite to the velocity. If a particle moves in the negative x direction and also has negative acceleration, it is speeding up (not decelerating).

Sign convention: Choose a positive direction at the start. Be consistent. If up = positive, then g = −10 m/s² for a freely falling body. If down = positive, g = +10 m/s². JEE problems use both conventions; choose the one that makes the algebra cleaner.

Acceleration can also be expressed in terms of position using the chain rule:

a = \frac{dv}{dt} = \frac{dv}{dx}\cdot\frac{dx}{dt} = v\frac{dv}{dx}

This form ( $a = v\,dv/dx$ ) is extremely useful when velocity is given as a function of position.

4. Equations of Motion — Uniform Acceleration

When acceleration is constant, three kinematic equations connect u (initial velocity), v (final velocity), a (acceleration), s (displacement), and t (time). These are derived from the definitions of velocity and acceleration.

Derivation of the first equation:

a = \frac{dv}{dt} \Rightarrow \int_u^v dv = a\int_0^t dt \Rightarrow v - u = at

\boxed{v = u + at} \tag{1}

Derivation of the second equation:

v = \frac{dx}{dt} = u + at \Rightarrow s = \int_0^t (u+at)\,dt = ut + \frac{1}{2}at^2

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

Derivation of the third equation:

a = v\frac{dv}{dx} \Rightarrow \int_u^v v\,dv = a\int_0^s dx \Rightarrow \frac{v^2 - u^2}{2} = as

\boxed{v^2 = u^2 + 2as} \tag{3}

nth second formula — distance covered in the nth second alone:

s_n = s(t=n) - s(t=n-1) = u + a\left(n - \frac{1}{2}\right) = u + \frac{a(2n-1)}{2}

\boxed{s_n = u + a\left(n - \frac{1}{2}\right)} \tag{4}

For a body starting from rest: $s_1 : s_2 : s_3 : \ldots = 1 : 3 : 5 : \ldots$ (odd numbers). This is one of the most-tested JEE results in kinematics.

JEE tip: Equation (3) is time-independent — it is ideal when time is neither given nor required. Equation (1) is velocity-time. Equation (2) is position-time. Equation (4) gives the distance in a specific second (not cumulative distance).

5. Free Fall Under Gravity

Free fall is the special case of uniformly accelerated motion where the only force is gravity. Near the Earth's surface, g ≈ 9.8 m/s² ≈ 10 m/s² (used in JEE problems). All objects fall with the same acceleration regardless of mass (neglecting air resistance).

Body dropped from rest (taking downward as positive):

v = gt, \ h = \frac{1}{2}gt^2, \ v^2 = 2gh

Time to fall height h: $t = \sqrt{2h/g}$ . Velocity on hitting ground: $v = \sqrt{2gh}$ .

Body thrown vertically upward with speed u (taking upward as positive):

v = u - gt \ \text{(v = 0 at top)}

H_{max} = \frac{u^2}{2g} \ \text{(time of ascent = } t_{up} = u/g\text{)}

T_{total} = \frac{2u}{g} \ \text{(symmetric: time up = time down)}

When the body returns to the same height, its speed equals u (by energy conservation or $v^2 = u^2 + 2as$ with s = 0). The velocity on return is −u (downward).

JEE trap: For a ball thrown upward from a building roof, the "displacement" when it hits the ground is negative (below the roof). Using

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

with s negative gives the height of the building. Many students get confused by sign — always define direction clearly first.

For multiple objects dropped from different heights or at different times, find a common reference point and write equations for each, then equate positions or times.

6. Graphical Analysis — s-t, v-t, and a-t Graphs

Graphs are one of the most powerful tools in kinematics. The relationships between x-t, v-t, and a-t graphs are governed by differentiation and integration.

Position-time (x-t) graph:

Slope at any point = instantaneous velocity
Straight line → constant velocity (zero acceleration)
Parabola opening up → positive constant acceleration
A horizontal line → object at rest
Slope of chord between two points = average velocity over that interval

Velocity-time (v-t) graph:

Slope at any point = instantaneous acceleration
Area under the curve = displacement (signed area)
Area above x-axis = forward displacement; below = backward displacement
Straight line → constant acceleration (uniform acceleration)
Area above v-t graph for distance: take absolute values of all areas

Acceleration-time (a-t) graph:

Area under the curve = change in velocity ( $\Delta v$ )
Horizontal line → constant acceleration (suvat equations apply)

Chain of graphs: If you have the a-t graph, integrate to get v-t; integrate again to get x-t. If you have the x-t graph, differentiate to get v-t; differentiate again to get a-t. JEE frequently tests your ability to go in both directions.

A parabolic v-t graph means linearly varying acceleration. A cubic x-t graph gives a parabolic v-t graph and a linear a-t graph.

Solved Practice

10 Solved Examples

Open only after you try the question yourself

JEE Exam Trap: When a question asks for "distance" and "displacement" separately, never use the suvat equations for distance if the particle reverses direction. Split the motion at the turning point (where v = 0) and calculate each segment separately.

Example 1. A car travels from A to B (100 km) in 2 hours and returns from B to A in 3 hours. Find the average speed and average velocity for the round trip.

Average speed = total distance / total time = 200 km / 5 h = 40 km/h.

Average velocity = total displacement / total time = 0 km / 5 h = 0 km/h. (The car returns to the starting point, so displacement = 0.)

Example 2. Position of a particle is given by $x = 3t^2 - 12t + 5$ metres. Find (a) velocity at t = 3 s and (b) when velocity is zero.

(a) $v = dx/dt = 6t - 12$ . At t = 3: $v = 18 - 12 = 6$ m/s.

(b) $v = 0$ : $6t - 12 = 0 \Rightarrow t = 2$ s.

Example 3. A train starts from rest and attains 54 km/h in 3 minutes with uniform acceleration. Find (a) acceleration and (b) distance covered.

u = 0, v = 54 km/h = 15 m/s, t = 180 s.

(a) $a = (v-u)/t = 15/180 = 1/12$ m/s².

(b) $s = ut + \frac{1}{2}at^2 = 0 + \frac{1}{2}\times\frac{1}{12}\times32400 = 1350$ m.

Example 4. A body starts from rest with acceleration 2 m/s². Find the distance covered in the 7th second.

$s_7 = u + a(7 - 0.5) = 0 + 2\times6.5 = 13$ m.

Example 5. A stone is dropped from a height of 125 m. Taking g = 10 m/s², find (a) time to reach the ground, (b) velocity on hitting ground, and (c) distance covered in the last second.

(a) $h = \frac{1}{2}gt^2 \Rightarrow 125 = 5t^2 \Rightarrow t = 5$ s.

(b) $v = gt = 10\times5 = 50$ m/s.

Example 6. The acceleration of a particle is $a = (3t^2 - 2)$ m/s². If v = 4 m/s at t = 0, find velocity at t = 2 s.

$v = u + \int_0^2 a\,dt = 4 + \int_0^2(3t^2-2)dt = 4 + [t^3-2t]_0^2 = 4 + (8-4) = 4 + 4 = 8$ m/s.

Example 7. Velocity of a particle is $v = \sqrt{6x+4}$ m/s. Find the acceleration at x = 2 m.

Use $a = v\,dv/dx$ . $v = (6x+4)^{1/2}$ , $\frac{dv}{dx} = \frac{6}{2\sqrt{6x+4}} = \frac{3}{\sqrt{6x+4}}$ .

$a = v\cdot\frac{dv}{dx} = \sqrt{6x+4}\times\frac{3}{\sqrt{6x+4}} = 3$ m/s² (constant, independent of x!).

Example 8. A ball is thrown vertically upward from a building roof at 20 m/s. It hits the ground 6 s later (g = 10 m/s²). Find the height of the building.

Taking upward as positive: $s = ut + \frac{1}{2}at^2 = 20(6) + \frac{1}{2}(-10)(36) = 120 - 180 = -60$ m.

Negative means 60 m below launch point (roof). Building height = 60 m.

Example 9. A v-t graph is a straight line from (0, 20 m/s) to (10 s, 0). Find (a) acceleration, (b) displacement in 10 s, and (c) distance from t = 0 to t = 12 s.

(a) Slope = $(0-20)/(10-0) = -2$ m/s². (b) Area of triangle = $\frac{1}{2}\times10\times20 = 100$ m. (c) At t = 10 s, v = 0. From t = 10 to 12: v = 0 + (−2)(t−10) = −2(t−10). Distance 10 to 12 s = $\frac{1}{2}\times2\times4 = 4$ m below. Total distance = 100 + 4 = 104 m. Displacement = 100 − 4 = 96 m.

Example 10. Two particles start at the same point. Particle A has u = 12 m/s, a = 0. Particle B starts from rest with a = 2 m/s². When does B overtake A and at what displacement?

Position of A: $x_A = 12t$ . Position of B: $x_B = t^2$ .

B overtakes when $x_B = x_A$ : $t^2 = 12t \Rightarrow t = 12$ s (t = 0 is start).

Displacement = $12\times12 = 144$ m.

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Section B

The Practice Zone

This topic has a dedicated practice route with 4 session-wise tests of 15 questions each and a full-length 60-question mock. Each question runs on a 90-second countdown — matching JEE exam pacing.

Session Tests

4 focused sessions: distance/velocity basics, suvat equations, variable acceleration, and graph analysis.

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