Kinematics — Motion in One & Two Dimensions

Kinematics is the starting point of mechanics — it describes how objects move without worrying about why they move. Get this chapter right and the rest of mechanics becomes much easier because you’ll be fluent with the language of motion.

This is a high-weightage topic in CBSE, JEE Main, and JEE Advanced. Projectile motion alone can fetch you 2-3 marks in boards and up to 8 marks in JEE if you know the patterns.

Basic Definitions — Get These Right First

Before equations, we need the vocabulary. Students who mix up displacement and distance, or speed and velocity, lose marks in conceptual questions.

Distance — Total path length traveled. Scalar. Always positive.

Displacement — Change in position from start to end. Vector. Can be zero even if the body has moved (e.g., one complete circular trip).

Speed — Distance per unit time. Scalar.

Velocity — Displacement per unit time. Vector. $v = \Delta x / \Delta t$

Acceleration — Rate of change of velocity. Vector. $a = \Delta v / \Delta t$

CBSE and JEE both ask: “A body travels a semicircle of radius R. Find displacement and distance.” Distance $= \pi R$ (half circumference). Displacement $= 2R$ (diameter, straight line). Don’t confuse the two.

Equations of Motion — The Core Tools

These three equations apply when acceleration is constant. Memorize them, understand their derivation, and recognize which one to use in each situation.

Equation 1: $v = u + at$

Equation 2: $s = ut + \dfrac{1}{2}at^2$

Equation 3: $v^2 = u^2 + 2as$

Bonus (average velocity): $s = \dfrac{u+v}{2} \cdot t$

Where:

$u$ = initial velocity (m/s)
$v$ = final velocity (m/s)
$a$ = acceleration (m/s²)
$s$ = displacement (m)
$t$ = time (s)

Derivation of the Three Equations

You should know these derivations — CBSE boards ask for them regularly.

Equation 1 — from definition of acceleration:

a = \frac{v - u}{t}

Rearranging: $v = u + at$

Equation 2 — from area under v-t graph:

Area under a v-t graph = displacement

\text{Area of trapezoid} = \frac{1}{2}(u + v) \times t = \frac{1}{2}(u + u + at) \times t = ut + \frac{1}{2}at^2

Equation 3 — eliminating time:

From eq 1: $t = \dfrac{v - u}{a}$ . Substitute into eq 2 and simplify:

v^2 = u^2 + 2as

Which equation to use? Count your knowns and unknowns. Each equation has 4 variables. You know 3, find the 4th. If time $t$ is not involved at all, use Equation 3.

Motion Graphs — A Separate Skill

Graph-based questions are very common in JEE Main and CBSE. The key relationships:

Position-Time (x-t) graph:

Slope = velocity
Straight line → constant velocity
Curve → changing velocity (acceleration present)
Parabola for uniform acceleration

Velocity-Time (v-t) graph:

Slope = acceleration
Area under curve = displacement (careful with sign — area below time axis = negative displacement)
Straight line → constant acceleration
Horizontal line → zero acceleration (constant velocity)

Acceleration-Time (a-t) graph:

Area under curve = change in velocity

JEE Main frequently gives a v-t graph and asks for displacement in a time interval. Calculate the area geometrically (triangles + rectangles + trapezoids). Watch out for sections where velocity is negative — subtract that area.

Free Fall and Vertical Motion

Free fall is just uniform acceleration with $a = g = 10 \text{ m/s}^2$ (downward). Use the same three equations — just take downward as positive (or negative, be consistent).

v = u + gt

h = ut + \frac{1}{2}gt^2

v^2 = u^2 + 2gh

For a body dropped from rest: $u = 0$ , so $h = \dfrac{1}{2}gt^2$ and $v = gt$

Key result for free fall: Time to fall from height $h$ when dropped from rest:

t = \sqrt{\frac{2h}{g}}

For a body thrown upward, many students forget that at maximum height, velocity $= 0$ but acceleration $= g$ downward (not zero). Acceleration is always $g$ during the entire flight, even at the topmost point.

Projectile Motion

A projectile is any object launched at an angle with only gravity acting on it (air resistance neglected). The key insight: horizontal and vertical motions are independent.

Horizontal: no force, so constant velocity $u\cos\theta$
Vertical: gravity acts, so uniformly accelerated motion with $a = -g$

Initial horizontal velocity: $u_x = u\cos\theta$

Initial vertical velocity: $u_y = u\sin\theta$

Time of flight: $T = \dfrac{2u\sin\theta}{g}$

Maximum height: $H = \dfrac{u^2\sin^2\theta}{2g}$

Range: $R = \dfrac{u^2\sin 2\theta}{g}$

Maximum range (at $\theta = 45°$ ): $R_{\max} = \dfrac{u^2}{g}$

Important Results to Remember

Equal ranges at complementary angles: A projectile at $\theta$ and $(90° - \theta)$ gives the same range. So 30° and 60° give the same range; 20° and 70° give the same range.

Range at 45°: Maximum range. Remember: $R_{\max} = \dfrac{u^2}{g}$ , and $H$ at 45° $= \dfrac{R_{\max}}{4}$ .

Velocity at any point: At time $t$ ,

$v_x = u\cos\theta$ (constant)
$v_y = u\sin\theta - gt$
Speed $= \sqrt{v_x^2 + v_y^2}$

JEE Advanced has asked: at what angle does the velocity vector make 45° with horizontal? Set $v_y = v_x$ , i.e., $u\sin\theta - gt = u\cos\theta$ , solve for $t$ . This type of question requires you to think about the velocity components, not just plug into formulas.

Projectile on Inclined Plane

For JEE Advanced, projectile launched on/from an incline requires resolving along and perpendicular to the incline. Take the incline surface as x-axis, perpendicular as y-axis. Then:

Acceleration along incline: $-g\sin\alpha$ (deceleration along surface)
Acceleration perpendicular to incline: $-g\cos\alpha$

The time of flight is when displacement perpendicular to incline $= 0$ again.

Relative Velocity

Relative velocity tells us how one object appears to move from the reference frame of another. Very useful for rain-umbrella problems, river-boat problems, and two-train problems.

Velocity of A relative to B:

\vec{v}_{AB} = \vec{v}_A - \vec{v}_B

For 1D motion:

Same direction: $v_{rel} = v_A - v_B$ (magnitude)
Opposite direction: $v_{rel} = v_A + v_B$ (magnitude)

River-Boat Problem

A classic JEE topic. A boat of speed $v_b$ in still water crosses a river of width $d$ with current speed $v_r$ .

To cross in minimum time: Point the boat straight across (perpendicular to river). Time $= \dfrac{d}{v_b}$ . Drift $= v_r \times \dfrac{d}{v_b}$

To cross with zero drift: Aim the boat upstream at angle $\theta$ where $\sin\theta = \dfrac{v_r}{v_b}$ (only possible if $v_b > v_r$ ).

For the rain-umbrella problem: velocity of rain relative to person $= \vec{v}_{rain} - \vec{v}_{person}$ . The umbrella should be tilted in the direction of this relative velocity vector.

Solved Numericals

Numerical 1 — Basic Kinematics (CBSE Level)

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

$u = 0$ , $a = 2$ m/s², $t = 10$ s

v = u + at = 0 + 2 \times 10 = 20 \text{ m/s}

s = ut + \frac{1}{2}at^2 = 0 + \frac{1}{2} \times 2 \times 100 = 100 \text{ m}

Numerical 2 — Projectile Motion (JEE Main Level)

A ball thrown at 30° with speed 20 m/s. Find max height, range, and time of flight. ( $g = 10$ m/s²)

$u = 20$ m/s, $\theta = 30°$

H = \frac{u^2\sin^2\theta}{2g} = \frac{400 \times 0.25}{20} = 5 \text{ m}

T = \frac{2u\sin\theta}{g} = \frac{2 \times 20 \times 0.5}{10} = 2 \text{ s}

R = \frac{u^2\sin 2\theta}{g} = \frac{400 \times \sin 60°}{10} = \frac{400 \times 0.866}{10} \approx 34.64 \text{ m}

Numerical 3 — Relative Velocity (CBSE/JEE Main)

Train A moves at 60 m/s east. Train B moves at 40 m/s west. Find velocity of A relative to B.

Taking east as positive: $v_A = +60$ m/s, $v_B = -40$ m/s

v_{AB} = v_A - v_B = 60 - (-40) = 100 \text{ m/s (eastward)}

From B’s perspective, A appears to move east at 100 m/s. This is why head-on collisions are so dangerous — the relative speed is the sum of both speeds.

Numerical 4 — Free Fall (NCERT Level)

Stone dropped from 80 m height. Find time to reach ground. ( $g = 10$ m/s²)

$u = 0$ , $s = 80$ m, $a = 10$ m/s²

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

80 = 0 + \frac{1}{2} \times 10 \times t^2

t^2 = 16 \implies t = 4 \text{ seconds}

Numerical 5 — JEE Advanced Level

A particle is projected horizontally at 10 m/s from a 20 m high cliff. Find the speed and direction of velocity just before hitting the ground. ( $g = 10$ m/s²)

Horizontal: $v_x = 10$ m/s (constant)

Vertical: $v_y^2 = 0 + 2 \times 10 \times 20 = 400$ , so $v_y = 20$ m/s

\text{Speed} = \sqrt{v_x^2 + v_y^2} = \sqrt{10^2 + 20^2} = \sqrt{500} = 10\sqrt{5} \approx 22.4 \text{ m/s}

Angle with horizontal: $\tan\theta = \dfrac{v_y}{v_x} = \dfrac{20}{10} = 2$ , so $\theta = \tan^{-1}(2) \approx 63.4°$

5 Common Mistakes

Mistake 1 — Using equations of motion when acceleration is not constant. The three equations only work for uniform (constant) acceleration. For variable acceleration, you need calculus — integrate $a(t)$ to get $v(t)$ , then integrate again for $x(t)$ .

Mistake 2 — Taking $g = 9.8$ in JEE when the problem says $g = 10$ . Always use the value given in the problem. Most JEE problems say “take $g = 10$ m/s²” — use 10.

Mistake 3 — Sign errors in projectile motion. Choose a sign convention (up positive or down positive) and stick to it throughout. If you take up as positive, then $g = -10$ m/s², and downward displacement is negative.

Mistake 4 — Forgetting that range formula $R = \dfrac{u^2\sin 2\theta}{g}$ assumes projection and landing at the same horizontal level. If the landing point is at a different height, you must use the kinematic equations directly.

Mistake 5 — Confusing velocity of A relative to B with velocity of B relative to A. $v_{AB} = v_A - v_B$ and $v_{BA} = v_B - v_A$ . They are equal in magnitude but opposite in direction.

Exam Tips

JEE Main: 2-3 questions per paper on average. Most common types: projectile motion (finding time, range, height), graph-based problems (reading v-t or x-t graphs), relative velocity (rain/river/train). Memorize the formula for range at complementary angles — saves time.

JEE Advanced: Expects you to apply calculus. Questions on variable acceleration like $a = kv$ or $a = kx$ require separation of variables. Projectile on inclined plane, and multi-body relative motion problems are frequently tested.

CBSE Board: Derivations of all three equations of motion are asked almost every year. Definitions, units, dimensional analysis of kinematic quantities. Projectile motion: one 3-5 mark numerical is standard.

Real-World Examples

Example 1: MS Dhoni’s Helicopter Shot

When Dhoni launches one of his trademark helicopter shots at Chepauk, the ball leaves the bat at roughly 30–35 m/s at an angle of about 40–45° above the horizontal. From that moment, only gravity acts on it — horizontal velocity stays constant while the vertical component decreases, reverses, and the ball traces a perfect parabolic arc over the boundary rope.

This is projectile motion in its purest form. The range formula $R = \frac{u^2 \sin 2\theta}{g}$ tells us exactly why a 45° launch angle maximises the distance — something Dhoni’s muscle memory has independently verified over thousands of innings.

Connect to the syllabus: The independence of horizontal and vertical motion — $x = u\cos\theta \cdot t$ and $y = u\sin\theta \cdot t - \frac{1}{2}gt^2$ — is the core of two-dimensional kinematics and a direct JEE Main favourite.

Example 2: Mumbai Local Train Braking Distance

A Virar-Churchgate local enters Dadar station at 60 km/h and the motorman applies brakes uniformly, bringing it to rest in about 400 metres. Passengers feel that constant backward push — that’s uniform deceleration, typically around $-0.7 \text{ m/s}^2$ for these rakes.

The motorman doesn’t calculate anything consciously, but the physics is exact: $v^2 = u^2 + 2as$ predicts the stopping distance to the metre. Overestimate the braking and you overshoot the platform; underestimate and you stop short — both are daily annoyances Mumbai commuters know too well.

Connect to the syllabus: This is the third equation of motion $v^2 = u^2 + 2as$ in action — one of the three kinematic equations under uniform acceleration that appear in literally every CBSE Class 11 and JEE Main paper.

Example 3: ISRO Rocket Relative Velocity at Stage Separation

When ISRO’s PSLV drops its second stage at around 120 km altitude, both the rocket body and the separated stage are moving eastward at roughly 5 km/s relative to the ground. From the perspective of the main vehicle, the spent stage appears to drift slowly backward — its relative velocity is only a few metres per second, even though both objects are screaming across the sky.

This is exactly the concept of relative velocity: $\vec{v}_{AB} = \vec{v}_A - \vec{v}_B$ . Two trains passing each other on parallel tracks at 80 km/h and 60 km/h give you the same arithmetic — the faster train sees the slower one crawling at just 20 km/h.

Connect to the syllabus: Relative velocity in one dimension, $v_{rel} = v_1 - v_2$ for same direction and $v_{rel} = v_1 + v_2$ for opposite directions, is a high-weightage short-answer type in JEE Main and a guaranteed 2-mark question in CBSE board practicals and theory both.

Practice Questions

A train starts from rest and attains speed of 72 km/h in 5 minutes. Find acceleration and distance covered.
A ball is thrown vertically upward with speed 30 m/s. Find time to reach maximum height, maximum height, and total time of flight. ( $g = 10$ m/s²)
A stone is thrown at 45° with speed 20 m/s. Find the range and maximum height. ( $g = 10$ m/s²)
Two cars approach each other on a straight road. Car A moves at 60 km/h and car B at 80 km/h. If they are 700 m apart, when do they meet?
A ball is thrown horizontally from a 45 m high building at 10 m/s. Find where it lands (horizontal distance from base of building). ( $g = 10$ m/s²)
A particle moves with velocity $v = 3t^2 - 6t$ . Find displacement from $t = 0$ to $t = 4$ s, and when the particle momentarily stops.
A river 60 m wide has current of 4 m/s. A boat can do 5 m/s in still water. Find minimum time to cross and drift in that case. Also find the angle to cross with zero drift.
Two objects are dropped from a 100 m tall building at 1 second apart. Find their separation when the first object has been in air for 3 seconds. ( $g = 10$ m/s²)

Frequently Asked Questions

Q1: What is the difference between uniform velocity and uniform acceleration?

Uniform velocity means constant speed in a constant direction — no acceleration at all. Uniform acceleration means acceleration is constant, but velocity is changing at a steady rate. A car moving in a straight line at 60 km/h has uniform velocity. The same car speeding up by 10 km/h every second has uniform acceleration. Free fall is the most common example of uniform acceleration ( $g =$ constant).

Q2: Can a body have zero velocity but non-zero acceleration?

Yes. When a ball thrown upward reaches its highest point, its instantaneous velocity is zero but acceleration is still $g$ downward. Similarly, at the turning point of any oscillating motion, velocity is zero but acceleration is non-zero.

Q3: Why do two complementary angles give the same range in projectile motion?

$R = \dfrac{u^2\sin 2\theta}{g}$ . For angle $\theta$ and $(90° - \theta)$ :

At $\theta$ : $R_1 = \dfrac{u^2\sin 2\theta}{g}$

At $(90° - \theta)$ : $R_2 = \dfrac{u^2\sin(180° - 2\theta)}{g} = \dfrac{u^2\sin 2\theta}{g} = R_1$

Because $\sin(180° - x) = \sin x$ .

The difference: the $45° +$ angle gives a higher trajectory with more time of flight; the $45° -$ angle gives a flatter path with less time.

Q4: For JEE, what is the significance of the slope of a v-t graph being negative?

Negative slope on a v-t graph means negative acceleration (deceleration if the body is moving in positive direction). But be careful — if the body is already moving in the negative direction, a negative slope means it’s speeding up (acceleration and velocity are in the same direction). Always interpret sign with physical context.

Q5: How do I approach variable acceleration problems for JEE Advanced?

If acceleration is given as a function of time $a(t)$ : integrate to get $v(t)$ , then integrate again to get $x(t)$ . Add integration constants using initial conditions.

If $a = f(v)$ (function of velocity): Use $a = v\,\dfrac{dv}{dx}$ or $a = \dfrac{dv}{dt}$ and separate variables.

If $a = f(x)$ (function of position): Use $a = v\,\dfrac{dv}{dx}$ and integrate.

Practice recognizing which form the acceleration is given in — that tells you which approach to use.

Q6: In the river-boat problem, what if the river speed is greater than the boat speed?

If $v_r > v_b$ , the boat cannot reach a point directly across. It will always drift downstream. In this case, to minimize the angle of drift (not zero drift), point the boat at 90° to the river bank. You cannot achieve zero drift — there’s no upstream angle that works when the current is faster than the boat.