Unit 2: Kinematics

Kinematics is the description of motion without asking about its causes. NEET expects 2–3 MCQs per year from this unit, focused on equations of motion, projectile, graphs, and relative velocity. Numericals are short (one or two steps) and almost always reduce to direct substitution into a formula. Graph-reading and assertion-reason questions appear frequently.

The key skill: recognise the type of motion (uniform, uniformly accelerated, projectile, circular) and pick the right formula immediately.

Concept Map

• Motion in 1D
  • Position, displacement, distance
  • Velocity (avg, instantaneous)
  • Acceleration (avg, instantaneous)
  • Equations of motion (uniform acceleration)
  • Free fall under gravity
• Graphical analysis
  • x-t, v-t, a-t graphs
  • Slope and area interpretations
• Motion in 2D
  • Vector addition / resolution
  • Projectile motion
  • Uniform circular motion (kinematic part)
• Relative motion
  • 1D relative velocity
  • 2D relative velocity (boat-river, rain-man)

Topic 1: Motion in One Dimension

Sub-topic A: Distance vs Displacement

• Distance is the actual path length (scalar, always positive).
• Displacement is the vector from initial to final position. Magnitude $\le$ distance.

For an object that moves 30 m east then 40 m north: distance = 70 m, displacement magnitude = 50 m.

Sub-topic B: Speed and Velocity

• Average speed = total distance / total time.
• Average velocity $\vec v_\text{avg} = \Delta \vec r / \Delta t$.
• Instantaneous velocity $\vec v = d\vec r/dt$.

Trap: average speed $\ne$ |average velocity| in general. For a round trip, average velocity is zero but average speed is not.

For two equal distances at speeds $v_1$ and $v_2$ the average speed is the harmonic mean:

For two equal time intervals the average speed is the arithmetic mean $(v_1 + v_2)/2$. NEET loves this distinction.

Sub-topic C: Acceleration

• $\vec a_\text{avg} = \Delta \vec v / \Delta t$.
• $\vec a = d\vec v/dt = d^2\vec r/dt^2$.

Acceleration can change speed, direction, or both. In uniform circular motion the speed is constant but $\vert \vec a\vert = v^2/r$ is non-zero (centripetal).

Sub-topic D: Equations of Uniformly Accelerated Motion

For constant acceleration $a$ along a line:

The last formula gives the displacement during the $n$-th second of motion (not the total in $n$ seconds). It's a common NEET shortcut question.

Sub-topic E: Free Fall

Taking downward as positive with $a = g \approx 9.8\ \text{m/s}^2$:

For a body dropped from rest at height $h$:

For a body projected upward with initial speed $u$:

The distances covered in successive seconds during free fall from rest are in the ratio $1 : 3 : 5 : 7 \dots$ (Galileo's ratio).

Topic 2: Motion Graphs

Sub-topic A: Position-Time Graph

• Slope $=$ velocity.
• Straight line $\Rightarrow$ uniform velocity.
• Curve $\Rightarrow$ accelerated motion (concave up for $+a$, concave down for $-a$).
• Horizontal line $\Rightarrow$ object at rest.

Sub-topic B: Velocity-Time Graph

• Slope $=$ acceleration.
• Area under v-t graph = displacement (algebraic — areas below the axis subtract).
• Straight line $\Rightarrow$ uniform acceleration.
• Horizontal line $\Rightarrow$ uniform velocity.

Sub-topic C: Acceleration-Time Graph

• Area under a-t graph = change in velocity.

Sub-topic D: Reading Graphs (NEET style)

NEET often shows a v-t graph and asks one of:

• What is the displacement? (sum of areas)
• What is the average acceleration? (Δv/Δt)
• Identify when the body is at rest (v = 0).
• Identify when acceleration is maximum (steepest slope).

Topic 3: Motion in Two Dimensions

Sub-topic A: Vector Decomposition

A 2D motion with position $\vec r(t) = x(t)\hat i + y(t)\hat j$ has

The horizontal and vertical motions are independent — a key principle for projectile motion.

Sub-topic B: Projectile Motion (level ground)

A particle is projected with speed $u$ at angle $\theta$ from horizontal. Take origin at launch point, $x$ horizontal, $y$ vertical (up positive).

• Horizontal velocity: $u_x = u \cos\theta$ (constant).
• Vertical velocity: $u_y = u \sin\theta$ (decreases by $gt$).
• Position: $x = u \cos\theta \cdot t$, $y = u \sin\theta \cdot t - \tfrac{1}{2} g t^2$.

Equation of trajectory (eliminate $t$):

This is a parabola.

Key quantities:

Useful facts:

• Maximum range when $\theta = 45°$: $R_\text{max} = u^2/g$.
• For complementary angles $\theta$ and $90° - \theta$, the ranges are equal but $H$ and $T$ differ.
• Relation: $R = 4H \cot\theta$, so when $\theta = 45°$, $R = 4H$.
• Speed at highest point: $u \cos\theta$ (horizontal only).
• At highest point the velocity and acceleration are perpendicular.

Sub-topic C: Projectile from a Height

Object launched horizontally with speed $u$ from height $h$:

Sub-topic D: Projectile on an Inclined Plane

If thrown with speed $u$ at angle $\alpha$ from an incline of angle $\beta$ (taking incline as reference):

Maximum range up the incline at $\alpha = (90° - \beta)/2 = 45° - \beta/2$.

Topic 4: Relative Velocity

Sub-topic A: In One Dimension

The velocity of A relative to B is

If two trains move in the same direction at $u$ and $v$ with $u > v$, the relative speed is $u - v$. If opposite, $u + v$.

Sub-topic B: In Two Dimensions

Use vector subtraction. The magnitude is

where $\theta$ is the angle between $\vec v_A$ and $\vec v_B$.

Sub-topic C: Boat in River

Let river velocity be $v_r$ along the bank, boat velocity $v_b$ relative to water.

Shortest path (crossing perpendicular to bank): boat must be aimed upstream at angle $\theta$ such that

Time to cross width $d$:

Shortest time (boat perpendicular to bank): $t = d/v_b$, but boat drifts by $v_r t$ downstream.

Sub-topic D: Rain-Man Problem

If rain falls vertically with speed $v_r$ and the man walks horizontally with speed $v_m$, the rain appears to fall at angle $\theta$ from vertical given by

To stay dry, the man should tilt the umbrella forward by this angle.

Topic 5: Variable Acceleration (Calculus)

When $a$ is a function of $t$, $v$ or $x$, use:

For $a$ as function of $x$: $a = v\,dv/dx$, so $\int v\,dv = \int a\,dx$.

NEET asks one such calculus question every 2–3 years — usually integrating $a(t) = kt$ to get $v$ and $x$.

NEET Pattern MCQ Tips

• Direct formula plug: H, R, T of projectile given $u$ and $\theta$ → one-line answer.
• Graph interpretation: identify motion type from a v-t curve; compute displacement as area.
• Average velocity vs speed: harmonic mean trap for equal-distance segments.
• Assertion-reason: "At the highest point of a projectile, velocity and acceleration are perpendicular." (True)
• Relative motion: river-boat, rain-man, two-trains.
• Galileo ratio: distances in successive seconds in free fall.
• n-th second: $s_n = u + (a/2)(2n - 1)$.

Common Confusions and Traps

• Distance can never decrease; displacement can.
• For uniform circular motion, speed is constant but velocity is not (direction changes).
• A body thrown upward has $a = -g$ throughout (even at the top where $v = 0$).
• $\vert \vec v_\text{avg}\vert$ and $v_\text{avg-speed}$ are equal only for straight-line motion in one direction.
• In projectile motion, horizontal velocity component is always $u\cos\theta$; vertical is $u\sin\theta - gt$.
• $R = 4H/\tan\theta$ — useful for "find $\theta$" given $R$ and $H$.
• Two projectiles with complementary angles ($\theta, 90° - \theta$) at the same $u$ have the same $R$, but the higher angle gives more $H$ and more $T$.

Quick Revision Card

• $v = u + at$, $s = ut + \tfrac{1}{2} at^2$, $v^2 = u^2 + 2as$.
• $s_n = u + (a/2)(2n - 1)$.
• Galileo: $1:3:5:7\dots$ ratio for free-fall distances.
• $R = u^2 \sin 2\theta / g$, $H = u^2 \sin^2\theta / (2g)$, $T = 2u\sin\theta/g$.
• Range maximum at 45°: $R_\text{max} = u^2/g$.
• At highest point: $\vec v \perp \vec a$.
• Boat shortest path: $\sin\theta_\text{upstream} = v_r/v_b$.
• Rain angle: $\tan\theta = v_m/v_r$.
• Slope of x-t = v; slope of v-t = a; area of v-t = displacement.

Worked NEET Examples

Example 1: Drop and projection at the same time

A body $A$ is dropped from height $H$ at the same instant another body $B$ is thrown vertically upward with speed $u$ from the ground. They meet at time

(from relative-velocity argument: $A$ and $B$ approach each other at relative speed $u$, while gravity affects both equally). The meeting height above ground is $H - \tfrac{1}{2}g t^2$.

Example 2: Two cars, head-on approach

Two cars on a straight road, 200 m apart, approach each other at 30 m/s and 20 m/s. Relative speed = 50 m/s. Time to meet = 200/50 = 4 s. The faster car covers 120 m in this time; they meet 120 m from the starting point of the faster car.

Example 3: Vertical projectile, second projectile launched later

A ball thrown vertically up with $u = 20$ m/s. After 1 s, a second ball is thrown with same $u$. Where and when do they meet?

Position of first ball at time $t$ (from launch of first): $y_1 = 20t - 5t^2$.

Position of second ball: $y_2 = 20(t - 1) - 5(t-1)^2$, valid for $t \ge 1$.

Setting $y_1 = y_2$ gives $t = 2.5$ s, at $y = 18.75$ m.

Example 4: Projectile striking an incline

Particle projected at angle $\alpha$ from a horizontal ground reaches an incline of angle $\beta$ to the horizontal at distance $L$ along the incline. Use formula

For $\alpha = 60°$, $\beta = 30°$, $u = 20$ m/s, $g = 10$ m/s²:

So $\alpha + \beta = 90°$ means the projectile lands exactly at the foot of the launch — the projectile rises perpendicular to the incline.

Example 5: Time of flight from a height

A ball thrown horizontally at 10 m/s from a 20 m cliff. Time to hit ground: $t = \sqrt{2(20)/10} = 2$ s. Range: $R = 10 \times 2 = 20$ m. Final velocity: $v = \sqrt{10^2 + (10 \times 2)^2} = \sqrt{500} \approx 22.4$ m/s.

Derivations Summary

Range Formula

Range $R$ is horizontal distance traveled during time of flight $T$. Horizontal velocity is $u\cos\theta$ (constant).

Maximum at $\theta = 45°$.

Maximum Height

At max height, $v_y = 0$. Using $v_y^2 = u_y^2 - 2gH$:

Equation of Trajectory

From $x = u\cos\theta \cdot t$, get $t = x/(u\cos\theta)$. Substitute in $y = u\sin\theta \cdot t - \tfrac{1}{2}g t^2$:

This is a parabola opening downward.

n-th Second Displacement

Total displacement in $n$ seconds: $s_n^\text{total} = un + \tfrac{1}{2}a n^2$.

Total in $(n-1)$ seconds: $s_{n-1}^\text{total} = u(n-1) + \tfrac{1}{2}a(n-1)^2$.

Difference (displacement during $n$-th second):

Graphical Problem-Solving

Reading a v-t Graph

Given a v-t graph:

1. Slope at any instant = instantaneous acceleration.
2. Area under graph from $t_1$ to $t_2$ = displacement during that interval.
3. Sign of area matters: areas below the time axis represent displacement in the opposite direction.

Conversion Between Graphs

If x(t) is parabolic up ⇒ v(t) is linear up ⇒ a is constant positive.

If x(t) is parabolic down ⇒ v(t) is linear down ⇒ a is constant negative.

If v(t) has a peak (max) ⇒ at peak $a = 0$, before peak $a > 0$, after peak $a < 0$.

Calculus-Based Kinematics Problems

Problem 1

A particle has velocity $v(t) = 6t^2 - 2t$. Find acceleration at $t = 2$ and total displacement in first 3 s.

Solution: $a = dv/dt = 12t - 2$, so $a(2) = 22$ m/s².

Displacement: $\int_0^3 (6t^2 - 2t)\,dt = [2t^3 - t^2]_0^3 = 54 - 9 = 45$ m.

Problem 2

A particle starts from rest with $a = kt$ (variable). Then $v = \tfrac{1}{2}k t^2$ and $x = \tfrac{1}{6}k t^3$. Velocity is quadratic in time, displacement cubic.

Problem 3

Acceleration depends on position: $a = -kx$. Then $v\,dv/dx = -kx$, so $\int v\,dv = -k\int x\,dx$, giving $v^2 = v_0^2 - kx^2$ — this is SHM!

Formula Sheet