Kinematics

Rigorous Derivation of Kinematic Equations for Constant Acceleration:

In classical mechanics, motion along a straight line with constant acceleration $a$ can be elegantly formulated using differential and integral calculus:

1. Velocity-Time Relation ($v = u + at$):

By definition, instantaneous acceleration is $a = \frac{dv}{dt} \implies dv = a\, dt$. Integrating both sides from $t = 0$ (velocity $u$) to time $t$ (velocity $v$): $$\int_{u}^{v} dv = a \int_{0}^{t} dt \implies [v]_{u}^{v} = a [t]_{0}^{t} \implies v - u = at \implies \mathbf{v = u + at}$$

2. Displacement-Time Relation ($s = ut + \frac{1}{2}at^2$):

By definition, instantaneous velocity is $v = \frac{ds}{dt} \implies ds = v\, dt = (u + at)\, dt$. Integrating from $t = 0$ ($s = 0$) to time $t$ (displacement $s$): $$\int_{0}^{s} ds = \int_{0}^{t} (u + at)\, dt \implies s = u\int_{0}^{t} dt + a\int_{0}^{t} t\, dt \implies \mathbf{s = ut + \frac{1}{2}at^2}$$

3. Velocity-Displacement Relation ($v^2 = u^2 + 2as$):

Using the chain rule: $a = \frac{dv}{dt} = \frac{dv}{ds} \cdot \frac{ds}{dt} = v \frac{dv}{ds} \implies a\, ds = v\, dv$. Integrating: $$\int_{0}^{s} a\, ds = \int_{u}^{v} v\, dv \implies a[s]_{0}^{s} = \left[\frac{v^2}{2}\right]_{u}^{v} \implies as = \frac{v^2 - u^2}{2} \implies \mathbf{v^2 = u^2 + 2as}$$

4. Distance Traversed in the $n^{\text{th}}$ Second ($s_n$):

The displacement during the $n^{\text{th}}$ second is the difference between displacement at $t = n$ and $t = n-1$: $$s_n = s(n) - s(n-1) = \left[un + \frac{1}{2}an^2\right] - \left[u(n-1) + \frac{1}{2}a(n-1)^2\right]$$ $$s_n = u + \frac{1}{2}a [n^2 - (n^2 - 2n + 1)] \implies \mathbf{s_n = u + \frac{a}{2}(2n - 1)}$$

Decoding Motion Graphs:

Vector Analytical Foundations:

A vector quantity possesses both magnitude and direction and obeys vector laws of addition. In Cartesian coordinates, a 3D vector is expressed in terms of orthogonal unit vectors $\hat{i}, \hat{j}, \hat{k}$:

Parallelogram Law of Vector Addition:

If two vectors $\vec{A}$ and $\vec{B}$ are represented in magnitude and direction by two adjacent sides of a parallelogram drawn from a common point, their resultant $\vec{R} = \vec{A} + \vec{B}$ is given by the diagonal:

where $\theta$ is the angle between $\vec{A}$ and $\vec{B}$, and $\alpha$ is the inclination of $\vec{R}$ with respect to $\vec{A}$.

Comparative Matrix of Vector Multiplications:

Full Analytical Derivations for Oblique Projectile Motion:

A projectile launched from origin $(0,0)$ with initial velocity $u$ at elevation angle $\theta$ experiences zero horizontal acceleration ($a_x = 0$) and downward gravitational acceleration ($a_y = -g$):

• $u_x = u\cos\theta, \quad v_x(t) = u\cos\theta$
• $u_y = u\sin\theta, \quad v_y(t) = u\sin\theta - gt$
• $x(t) = (u\cos\theta)t \implies t = \frac{x}{u\cos\theta}$
• $y(t) = (u\sin\theta)t - \frac{1}{2}gt^2$

1. Equation of Trajectory (Parabolic Path):

Substituting $t = \frac{x}{u\cos\theta}$ into $y(t)$: $$y = (u\sin\theta)\left(\frac{x}{u\cos\theta}\right) - \frac{1}{2}g\left(\frac{x}{u\cos\theta}\right)^2 \implies \mathbf{y = x\tan\theta - \frac{g}{2u^2\cos^2\theta}x^2}$$ Since this matches the standard quadratic form $y = Ax - Bx^2$, the trajectory is strictly a parabola.

2. Time of Flight ($T$):

At landing, vertical displacement $y = 0$: $$(u\sin\theta)T - \frac{1}{2}gT^2 = 0 \implies \mathbf{T = \frac{2u\sin\theta}{g}}$$

3. Maximum Height ($H$):

At the apex, vertical velocity $v_y = 0$. Using $v_y^2 = u_y^2 - 2gH$: $$0 = (u\sin\theta)^2 - 2gH \implies \mathbf{H = \frac{u^2\sin^2\theta}{2g}}$$

4. Horizontal Range ($R$) & Two Projection Angles:

Horizontal distance covered in time $T$: $$R = u_x \cdot T = (u\cos\theta)\left(\frac{2u\sin\theta}{g}\right) = \frac{u^2(2\sin\theta\cos\theta)}{g} \implies \mathbf{R = \frac{u^2\sin 2\theta}{g}}$$

• Maximum range occurs at $\sin 2\theta = 1 \implies 2\theta = 90^\circ \implies \mathbf{\theta = 45^\circ}$, giving $R_{\max} = \frac{u^2}{g} = 4H_{\max}$.
• For complementary angles $\theta$ and $90^\circ - \theta$: $\sin[2(90^\circ - \theta)] = \sin(180^\circ - 2\theta) = \sin 2\theta$. Hence, the range is identical for complementary angles!

Kinematics of Circular Motion:

When a particle travels along a circular path of radius $r$ at constant speed $v$, its direction changes continuously, generating an inward-directed acceleration.

• Angular Velocity ($\omega$): Rate of change of angular displacement: $\omega = \frac{d\theta}{dt} = \frac{2\pi}{T} = 2\pi\nu$.
• Linear-Angular Relation: Since arc length $s = r\theta$, differentiating yields $v = \frac{ds}{dt} = r\frac{d\theta}{dt} \implies \mathbf{v = \omega r}$.

Vector Derivation of Centripetal Acceleration:

Let position vector be $\vec{r} = r\cos(\omega t)\hat{i} + r\sin(\omega t)\hat{j}$.

Velocity is the first time derivative: $$\vec{v} = \frac{d\vec{r}}{dt} = -\omega r\sin(\omega t)\hat{i} + \omega r\cos(\omega t)\hat{j}$$

Acceleration is the second time derivative: $$\vec{a} = \frac{d\vec{v}}{dt} = -\omega^2 r\cos(\omega t)\hat{i} - \omega^2 r\sin(\omega t)\hat{j} = -\omega^2 \vec{r}$$ The negative sign proves that the acceleration vector points in the direction opposite to the position vector $\vec{r}$, i.e., radially inward toward the center. Its magnitude is: $$\mathbf{a_c = \omega^2 r = \frac{v^2}{r} = \frac{4\pi^2 r}{T^2}}$$

Relative Velocity Mechanics:

The relative velocity of a body $A$ with respect to another body $B$ is the velocity with which $A$ appears to move as observed from the reference frame of $B$:

Classic Two-Dimensional Applications:

• Rain-Man Problems: If rain falls vertically with velocity $\vec{v}_r = -v_r\hat{j}$ and a man walks horizontally with $\vec{v}_m = v_m\hat{i}$, the relative velocity of rain with respect to the man is: $$\vec{v}_{rm} = \vec{v}_r - \vec{v}_m = -v_m\hat{i} - v_r\hat{j}$$ To shield against rain, the man must hold the umbrella at angle $\theta$ with the vertical: $$\tan\theta = \frac{v_m}{v_r}$$
• River-Boat Crossings: For a boat of speed $v_b$ crossing a river of speed $v_r$ and width $d$:
  • Shortest Path (Perpendicular Crossing): The boat must steer upstream at angle $\theta$ where $\sin\theta = \frac{v_r}{v_b}$; time $t = \frac{d}{\sqrt{v_b^2 - v_r^2}}$.
  • Shortest Time: The boat steers directly across ($\theta = 0$ relative to bank); minimum time $t_{\min} = \frac{d}{v_b}$; downstream drift $x = v_r \cdot t_{\min} = \frac{v_r d}{v_b}$.

Comprehensive Problem-Solving Walkthrough: