⚡Physics • Chapter Revision
Kinematics Revision Notes
Quick facts, formula checklists, and concept reminders for NEET UG preparation.
3–5%
Weightage
~2 Q
Avg Qs
Medium
Difficulty
Chapter Overview
Kinematics is the branch of mechanics that describes the motion of points, bodies, and systems of bodies without consideration of the forces that cause the motion. It forms the core foundation of Newtonian Mechanics and is heavily tested in JEE, NEET, and EAPCET.
Theory & Concepts
1Equations of Motion (Constant Acceleration)
For motion in a straight line with uniform acceleration $a$:
1. **First Equation**: $v = u + at$ (relates velocity and time).
2. **Second Equation**: $s = ut + \frac{1}{2}at^2$ (relates displacement and time).
3. **Third Equation**: $v^2 = u^2 + 2as$ (relates velocity and displacement).
4. **Displacement in $n$-th second**: $s_n = u + \frac{a}{2}(2n - 1)$.
*Note: These equations are vector equations and must be applied with proper sign conventions relative to a chosen origin.*
2Projectile Motion (2D Motion)
When an object is thrown near the Earth's surface at an angle $\theta$ to the horizontal with velocity $u$:
* **Horizontal Component**: $u_x = u\cos\theta$, acceleration $a_x = 0$.
* **Vertical Component**: $u_y = u\sin\theta$, acceleration $a_y = -g$.
* **Time of Flight ($T$)**: The total time the projectile remains in air: $T = \frac{2u\sin\theta}{g}$.
* **Maximum Height ($H$)**: The maximum vertical displacement: $H = \frac{u^2\sin^2\theta}{2g}$.
* **Horizontal Range ($R$)**: The total horizontal displacement: $R = \frac{u^2\sin 2\theta}{g}$. Range is maximum when $\theta = 45^\circ$ ($R_{max} = \frac{u^2}{g}$).
3Relative Motion in 1D and 2D
Relative velocity describes the velocity of one body with respect to another moving frame:
* **Velocity of A relative to B**: $\vec{v}_{AB} = \vec{v}_A - \vec{v}_B$.
* **Acceleration of A relative to B**: $\vec{a}_{AB} = \vec{a}_A - \vec{a}_B$.
* **River-Boat / Swimmer Problems**:
- *Shortest Time to Cross*: Swimmer must head perpendicular to flow. $t_{min} = \frac{d}{v_{sr}}$, Drift $x = v_r \cdot t_{min}$.
- *Shortest Path (Zero Drift)*: Swimmer must head upstream at angle $\theta$ where $\sin\theta = \frac{v_r}{v_{sr}}$ (valid if $v_{sr} > v_r$).
Core Terminology
Displacement
The shortest straight-line vector distance pointing from the initial position to the final position of a moving body.
Instantaneous Acceleration
The rate of change of velocity at a specific point in time, mathematically defined as the limit of average acceleration as time interval approaches zero: $a = \frac{dv}{dt} = v\frac{dv}{dx}$.
Trajectory
The curved path followed by a projectile under the action of gravity, mathematically modeled as a parabola: $y = x\tan\theta - \frac{gx^2}{2u^2\cos^2\theta}$.
Concept Application (Solved Examples)
Example 1: Question
A ball is projected from the ground with an initial velocity of $20\text{ m/s}$ at an angle of $30^\circ$ to the horizontal. Calculate the Time of Flight and the Horizontal Range. (Take $g = 10\text{ m/s}^2$)
Step-by-Step Solution:
1. **Identify Given Parameters**:
- Initial velocity, $u = 20\text{ m/s}$
- Angle of projection, $\theta = 30^\circ$
- Acceleration due to gravity, $g = 10\text{ m/s}^2$
2. **Time of Flight ($T$)**:
$$T = \frac{2u\sin\theta}{g} = \frac{2 \times 20 \times \sin(30^\circ)}{10} = \frac{40 \times 0.5}{10} = 2\text{ seconds}$$
3. **Horizontal Range ($R$)**:
$$R = \frac{u^2\sin(2\theta)}{g} = \frac{20^2 \times \sin(60^\circ)}{10} = \frac{400 \times \frac{\sqrt{3}}{2}}{10} = 20\sqrt{3}\text{ m} \approx 34.64\text{ m}$$
Exam Tip: Always double-check if range is maximized. For $\theta = 30^\circ$ or $60^\circ$, ranges are identical because $\sin(2\theta)$ values are equal ($\sin(60^\circ) = \sin(120^\circ)$).
Example 2: Question
A particle moves along a straight line such that its displacement $x$ at time $t$ is given by $x = t^3 - 6t^2 + 9t + 2$ meters. Find the acceleration of the particle when its velocity becomes zero.
Step-by-Step Solution:
1. **Find Velocity ($v$)**:
$$v = \frac{dx}{dt} = \frac{d}{dt}(t^3 - 6t^2 + 9t + 2) = 3t^2 - 12t + 9$$
2. **Set Velocity to Zero to find $t$**:
$$3t^2 - 12t + 9 = 0 \implies 3(t^2 - 4t + 3) = 0 \implies 3(t - 1)(t - 3) = 0$$
So velocity is zero at $t = 1\text{ s}$ and $t = 3\text{ s}$.
3. **Find Acceleration ($a$)**:
$$a = \frac{dv}{dt} = \frac{d}{dt}(3t^2 - 12t + 9) = 6t - 12$$
4. **Calculate Acceleration at $t = 1\text{ s}$ and $t = 3\text{ s}$**:
- At $t = 1\text{ s}$: $a = 6(1) - 12 = -6\text{ m/s}^2$
- At $t = 3\text{ s}$: $a = 6(3) - 12 = +6\text{ m/s}^2$
Exam Tip: Acceleration can be positive or negative when velocity is zero. A negative sign simply represents acceleration pointing opposite to the positive coordinate direction.
Common Mistakes & Pitfalls to Avoid
- Confusing Distance with Displacement: Remember that displacement is a vector and can be zero if a particle returns to its starting point, whereas distance is scalar and always increases during motion.
- Wrong sign for acceleration due to gravity: If you define the upward direction as positive, $g$ must be entered as negative ($-9.8\text{ m/s}^2$ or $-10\text{ m/s}^2$) in all equations.
- Applying constant-acceleration equations when acceleration varies: If acceleration is a function of time or displacement (e.g. $a = 2t$), do NOT use equations of motion. Instead, use integration/differentiation methods: $v = \int a\,dt$ and $x = \int v\,dt$.
Syllabus Guidelines
Frame of reference, motion in a straight line: position-time graph, speed and velocity
Elementary concepts of differentiation and integration for describing motion
Uniform and non-uniform motion, average speed and instantaneous velocity
Uniformly accelerated motion, velocity-time and position-time graphs, relations for uniformly accelerated motion (graphical treatment)
Scalar and vector quantities; position and displacement vectors, general vectors and notation, equality of vectors, multiplication of vectors by a real number
Relative velocity, motion in a plane, cases of uniform velocity and uniform acceleration (projectile motion)
Uniform circular motion