Kinematics

Motion is fundamentally relative; an object is in motion if its position changes relative to a chosen reference point over time. A Frame of Reference consists of a coordinate system (Cartesian axes $x, y, z$) coupled with a clock:

• Inertial Frame of Reference: A reference frame that is at rest or moving with constant velocity (zero acceleration) where Newton's First Law holds strictly without introducing fictitious forces.
• Non-Inertial Frame of Reference: An accelerated or rotating reference frame where pseudo-forces (e.g., centrifugal force) must be invoked to describe dynamics.

For a particle moving along a straight line ($x$-axis):

We classify the rates of change of position as follows:

• Average Speed ($v_{ ext{avg}}$): Total distance divided by total elapsed time: $$v_{ ext{avg}} = rac{ ext{Total Path Length}}{ ext{Total Elapsed Time}} = rac{\Delta s}{\Delta t}$$
• Average Velocity ($ec{v}_{ ext{avg}}$): Net displacement divided by total time: $$ec{v}_{ ext{avg}} = rac{\Delta ec{x}}{\Delta t} = rac{ec{x}_f - ec{x}_i}{t_f - t_i}$$
• Instantaneous Velocity ($ec{v}$): The limiting value of average velocity as the time interval approaches zero (the first time derivative of position): $$ec{v} = \lim_{\Delta t o 0} rac{\Delta ec{x}}{\Delta t} = rac{dec{x}}{dt}$$
• Instantaneous Acceleration ($ec{a}$): The time derivative of velocity: $$ec{a} = rac{dec{v}}{dt} = rac{d^2ec{x}}{dt^2} = vrac{dv}{dx}$$

Graphical representations provide profound physical insights through slopes (derivatives) and enclosed areas (integrals):

When acceleration $a$ is constant (uniform), the equations of motion can be derived rigorously using definite integrals:

Equation 1: First Kinematic Equation ($v = u + at$)

By definition of instantaneous acceleration: $$a = rac{dv}{dt} \implies dv = a\,dt$$ Integrating both sides between limits $t = 0$ (velocity $u$) and $t = t$ (velocity $v$): $$\int_u^v dv = a \int_0^t dt \implies [v]_u^v = a [t]_0^t \implies v - u = at \implies \mathbf{v = u + at}$$

Equation 2: Second Kinematic Equation ($s = ut + rac{1}{2}at^2$)

By definition of instantaneous velocity: $$v = rac{ds}{dt} \implies ds = v\,dt = (u + at)\,dt$$ Integrating both sides between limits $t = 0$ (displacement $s = 0$) and $t = t$ (displacement $s$): $$\int_0^s ds = \int_0^t (u + at)\,dt = u \int_0^t dt + a \int_0^t t\,dt \implies s = u[t]_0^t + a\left[rac{t^2}{2} ight]_0^t \implies \mathbf{s = ut + rac{1}{2}at^2}$$

Equation 3: Third Kinematic Equation ($v^2 = u^2 + 2as$)

Using the chain rule of differentiation: $$a = rac{dv}{dt} = rac{dv}{ds} \cdot rac{ds}{dt} = v rac{dv}{ds} \implies v\,dv = a\,ds$$ Integrating both sides from initial state ($s = 0, v = u$) to final state ($s = s, v = v$): $$\int_u^v v\,dv = a \int_0^s ds \implies \left[rac{v^2}{2} ight]_u^v = a[s]_0^s \implies rac{v^2 - u^2}{2} = as \implies \mathbf{v^2 = u^2 + 2as}$$

Equation 4: Displacement in the $n^{ ext{th}}$ Second ($s_n$)

The displacement traversed specifically during the $n^{ ext{th}}$ second is the difference between total displacement after $n$ seconds and after $(n-1)$ seconds: $$s_n = s(n) - s(n-1) = \left[un + rac{1}{2}an^2 ight] - \left[u(n-1) + rac{1}{2}a(n-1)^2 ight]$$ $$s_n = un + rac{1}{2}an^2 - un + u - rac{1}{2}a(n^2 - 2n + 1) = u + rac{1}{2}a(2n - 1) \implies \mathbf{s_n = u + rac{a}{2}(2n - 1)}$$

When air resistance is negligible, all bodies fall with the same downward acceleration due to gravity $g pprox 9.8 ext{ m/s}^2$ (or $980 ext{ cm/s}^2$ in CGS). Adopting upward as positive ($+y$): $a = -g$.

• Body Projected Vertically Upward with Speed $u$:
  • At maximum height ($H$): vertical velocity becomes momentarily zero ($v = 0$). $$0 = u^2 - 2gH \implies \mathbf{H = rac{u^2}{2g}}$$
  • Time of ascent ($t_a$): $0 = u - gt_a \implies \mathbf{t_a = rac{u}{g}}$.
  • Time of descent ($t_d$): By energy conservation and symmetry, $\mathbf{t_d = rac{u}{g}}$.
  • Total time of flight ($T$): $\mathbf{T = t_a + t_d = rac{2u}{g}}$.
  • Striking velocity upon returning to the ground: $v = -u$ (equal in magnitude, opposite in direction).
• Body Dropped from Rest ($u = 0$) from Height $h$:
  • Striking velocity at the bottom: $\mathbf{v = \sqrt{2gh}}$.
  • Time taken to reach the ground: $\mathbf{t = \sqrt{rac{2h}{g}}}$.

Galileo Galilei proved that for a body dropped from rest ($u = 0$) under uniform acceleration, the distances covered in successive equal time intervals $ au$ are proportional to consecutive odd integers:

Thus, $\Delta s_1 : \Delta s_2 : \Delta s_3 : \Delta s_4 : \dots = 1 : 3 : 5 : 7 : \dots$

Physical quantities are classified based on their transformation properties under coordinate rotations:

• Scalar Quantities: Specified completely by magnitude with proper units (e.g., mass, time, speed, distance, work, temperature). They follow ordinary scalar algebra.
• Vector Quantities: Specified by magnitude, directional orientation, and obey the laws of vector addition (e.g., displacement, velocity, acceleration, force, linear momentum).
• Unit Vector ($\hat{n}$): A vector of unit magnitude whose sole purpose is to specify direction: $\hat{A} = rac{ec{A}}{|ec{A}|} = rac{ec{A}}{A}$. In Cartesian coordinates, $\hat{i}, \hat{j}, \hat{k}$ represent unit vectors along the positive $x, y, z$ axes respectively.

1. Triangle Law of Vector Addition: If two vectors are represented in magnitude and direction by two adjacent sides of a triangle taken in order, their resultant is represented in magnitude and direction by the closing third side taken in the reverse order.

2. Parallelogram Law of Vector Addition: If two vectors $ec{P}$ and $ec{Q}$ acting simultaneously at a point are represented in magnitude and direction by two adjacent sides of a parallelogram drawn from that point, their resultant $ec{R} = ec{P} + ec{Q}$ is represented by the diagonal passing through their common point.

• Magnitude of Resultant ($R$): $$R = \sqrt{P^2 + Q^2 + 2PQ\cos heta}$$ where $ heta$ is the angle between $ec{P}$ and $ec{Q}$.
• Direction of Resultant ($lpha$ w.r.t. $ec{P}$): $$ anlpha = rac{Q\sin heta}{P + Q\cos heta}$$
• Special Cases:
  • $ heta = 0^\circ$ (same direction): $R_{\max} = P + Q$, $lpha = 0^\circ$.
  • $ heta = 180^\circ$ (opposite direction): $R_{\min} = |P - Q|$, $lpha = 0^\circ$ or $180^\circ$.
  • $ heta = 90^\circ$ (orthogonal): $R = \sqrt{P^2 + Q^2}$, $ anlpha = Q/P$.

A vector $ec{A}$ in a plane can be uniquely resolved into mutually perpendicular (rectangular) components along the $x$ and $y$ axes:

In three-dimensional space: $$ec{A} = A_x\hat{i} + A_y\hat{j} + A_z\hat{k}$$ $$A = \sqrt{A_x^2 + A_y^2 + A_z^2}$$ If $lpha, eta, \gamma$ are the angles made by $ec{A}$ with the positive $x, y, z$ axes, the direction cosines are: $$l = \coslpha = rac{A_x}{A}, \quad m = \coseta = rac{A_y}{A}, \quad n = \cos\gamma = rac{A_z}{A}$$ A fundamental identity of 3D geometry states: $$\mathbf{\cos^2lpha + \cos^2eta + \cos^2\gamma = 1} \quad ext{or} \quad \mathbf{\sin^2lpha + \sin^2eta + \sin^2\gamma = 2}$$

Motion in a plane can be treated as two independent, simultaneous one-dimensional motions along mutually perpendicular axes ($x$ and $y$):

• Position Vector: $ec{r}(t) = x(t)\hat{i} + y(t)\hat{j}$
• Velocity Vector: $ec{v}(t) = rac{dx}{dt}\hat{i} + rac{dy}{dt}\hat{j} = v_x\hat{i} + v_y\hat{j}$, with magnitude $v = \sqrt{v_x^2 + v_y^2}$ and direction $ an heta = v_y / v_x$.
• Acceleration Vector: $ec{a}(t) = rac{dv_x}{dt}\hat{i} + rac{dv_y}{dt}\hat{j} = a_x\hat{i} + a_y\hat{j}$, with magnitude $a = \sqrt{a_x^2 + a_y^2}$.

The relative velocity of an object $A$ with respect to an observer/object $B$ is the rate of change of position of $A$ as seen by $B$:

• In 1D motion along a line:
  • If moving in the same direction: $v_{AB} = v_A - v_B$.
  • If moving in opposite directions: $v_{AB} = v_A - (-v_B) = v_A + v_B$.
• In 2D planar motion: $$|ec{v}_{AB}| = \sqrt{v_A^2 + v_B^2 - 2v_A v_B \cos heta}$$ where $ heta$ is the angle between vectors $ec{v}_A$ and $ec{v}_B$.

Suppose rain is falling vertically downward with velocity $ec{v}_r = -v_r\hat{j}$, and a person is walking or bicycling horizontally eastward with velocity $ec{v}_m = v_m\hat{i}$.

• The relative velocity of rain with respect to the person is: $$ec{v}_{rm} = ec{v}_r - ec{v}_m = -v_m\hat{i} - v_r\hat{j}$$
• The apparent speed of rain observed by the person is: $$v_{rm} = \sqrt{v_m^2 + v_r^2}$$
• To protect themselves from the rain, the person must hold the umbrella in the exact opposite direction of $ec{v}_{rm}$, which is at an angle $\phi$ tilted forward toward the direction of motion w.r.t the vertical: $$\mathbf{ an\phi = rac{v_m}{v_r}}$$

Let a river of width $d$ flow with speed $u$ (along $+x$). A boat/swimmer has speed $v$ relative to still water and aims at an angle $ heta$ relative to the riverbank.

Case 1: Crossing by the Shortest Path (Direct Perpendicular Crossing, Zero Drift)

To reach the point directly opposite on the other bank, the resultant velocity $ec{v}_{ ext{res}} = ec{v}_{SR} + ec{v}_R$ must point perpendicular to the flow (along the $y$-axis):

• The swimmer must head upstream at an angle $lpha$ with the normal to the bank such that the downstream flow is exactly cancelled: $$v\sinlpha = u \implies \mathbf{\sinlpha = rac{u}{v}} \quad ( ext{requires } v > u)$$
• The effective speed across the river is: $v_{ ext{res}} = \sqrt{v^2 - u^2}$.
• The time taken to cross is: $$\mathbf{t = rac{d}{v_{ ext{res}}} = rac{d}{\sqrt{v^2 - u^2}}}$$

Case 2: Crossing in the Shortest Possible Time ($t_{\min}$)

To minimize time, the swimmer must direct their entire swimming effort purely across the river perpendicular to the bank ($ heta = 90^\circ$ w.r.t flow):

• The minimum time taken is: $$\mathbf{t_{\min} = rac{d}{v}}$$
• During this crossing, the river carries the swimmer downstream by a drift distance $x$: $$\mathbf{x = u \cdot t_{\min} = rac{u d}{v}}$$

A projectile is any object thrown into space upon which the only acting force is gravity (neglecting atmospheric air resistance and the curvature/rotation of the Earth). Projectile motion is a two-dimensional motion with constant downward acceleration:

Consider a particle launched from origin $(0,0)$ with initial speed $u$ at an angle of elevation $ heta$ to the horizontal:

• Initial velocity components: $u_x = u\cos heta, \quad u_y = u\sin heta$.
• Velocity components at any time $t$: $$v_x(t) = u_x = u\cos heta \quad ( ext{remains constant})$$ $$v_y(t) = u_y - gt = u\sin heta - gt$$
• Instantaneous speed: $v = \sqrt{v_x^2 + v_y^2} = \sqrt{(u\cos heta)^2 + (u\sin heta - gt)^2}$.
• Position coordinates at time $t$: $$x = (u\cos heta)t \implies t = rac{x}{u\cos heta}$$ $$y = (u\sin heta)t - rac{1}{2}gt^2$$

Equation of Trajectory (Parabola):

Substituting $t = rac{x}{u\cos heta}$ into the $y$-equation: $$y = u\sin heta\left(rac{x}{u\cos heta} ight) - rac{1}{2}g\left(rac{x}{u\cos heta} ight)^2$$ $$\mathbf{y = x an heta - rac{g}{2u^2\cos^2 heta}x^2}$$ This is the standard equation of a downward-opening parabola: $y = Ax - Bx^2$.

In terms of horizontal range $R$, the trajectory can also be elegantly written as: $$\mathbf{y = x an heta\left(1 - rac{x}{R} ight)}$$

Since $\sin(2(90^\circ - heta)) = \sin(180^\circ - 2 heta) = \sin 2 heta$, a projectile fired at angle $ heta$ and another fired at complementary angle $(90^\circ - heta)$ with identical launch speed $u$ produce the exact same horizontal range $R$!

• Ratio of maximum heights: $rac{H_1}{H_2} = rac{\sin^2 heta}{\sin^2(90^\circ - heta)} = rac{\sin^2 heta}{\cos^2 heta} = \mathbf{ an^2 heta}$.
• Product of times of flight: $T_1 T_2 = \left(rac{2u\sin heta}{g} ight)\left(rac{2u\cos heta}{g} ight) = rac{2}{g}\left[rac{u^2\sin 2 heta}{g} ight] = \mathbf{rac{2R}{g}}$.
• Sum of maximum heights: $H_1 + H_2 = rac{u^2}{2g}(\sin^2 heta + \cos^2 heta) = \mathbf{rac{u^2}{2g}}$.

When a body is projected horizontally with velocity $u$ from the top of a tower of height $h$ ($u_x = u, u_y = 0$):

• Time taken to reach ground: $h = rac{1}{2}gt^2 \implies \mathbf{t = \sqrt{rac{2h}{g}}}$ (independent of horizontal speed $u$!).
• Horizontal range: $\mathbf{R = u \cdot t = u\sqrt{rac{2h}{g}}}$.
• Velocity upon impact: $v_x = u, v_y = \sqrt{2gh} \implies \mathbf{v = \sqrt{u^2 + 2gh}}$ at angle $ aneta = rac{\sqrt{2gh}}{u}$ below horizontal.

When a particle travels along a circular path of radius $r$:

• Angular Displacement ($ heta$): The angle swept by the position vector at the center, measured in radians ($ ext{rad}$): $ heta = rac{s}{r}$, where $s$ is the arc length.
• Angular Velocity ($\omega$): The rate of change of angular displacement: $$\omega = \lim_{\Delta t o 0} rac{\Delta heta}{\Delta t} = rac{d heta}{dt} = rac{2\pi}{T} = 2\pi f$$ where $T$ is the period of revolution and $f = 1/T$ is frequency in hertz ($ ext{Hz}$ or $ ext{s}^{-1}$).
• Relation between Linear Velocity and Angular Velocity: $$ec{v} = ec{\omega} imes ec{r} \implies \mathbf{v = \omega r}$$ Linear velocity is always directed tangentially to the circular trajectory.
• Angular Acceleration ($lpha$): The rate of change of angular velocity: $$lpha = rac{d\omega}{dt} = rac{d^2 heta}{dt^2} \implies \mathbf{a_t = lpha r}$$ where $a_t$ is the tangential linear acceleration.

In Uniform Circular Motion (UCM), the speed $v$ is constant, but because the direction of motion changes continuously, the velocity vector $ec{v}$ is constantly changing. Hence, the particle is accelerated!

Vector Calculus Derivation:

Let the center of the circle be the origin $(0,0)$. The position vector of the particle at angle $ heta = \omega t$ is: $$ec{r}(t) = r\cos(\omega t)\hat{i} + r\sin(\omega t)\hat{j}$$ Differentiating with respect to time $t$ gives the velocity vector: $$ec{v}(t) = rac{dec{r}}{dt} = -r\omega\sin(\omega t)\hat{i} + r\omega\cos(\omega t)\hat{j}$$ Note that $|ec{v}| = \sqrt{(-r\omega\sin\omega t)^2 + (r\omega\cos\omega t)^2} = \omega r = v$.

Differentiating once more gives the acceleration vector: $$ec{a}(t) = rac{dec{v}}{dt} = -r\omega^2\cos(\omega t)\hat{i} - r\omega^2\sin(\omega t)\hat{j} = -\omega^2\left[r\cos(\omega t)\hat{i} + r\sin(\omega t)\hat{j} ight]$$ $$\mathbf{ec{a} = -\omega^2ec{r} = -rac{v^2}{r}\hat{r}}$$

The negative sign indicates that the acceleration vector points in the direction opposite to the position vector $ec{r}$, which means it is directed radially inward towards the center of the circle. This is called the Centripetal Acceleration ($a_c$).

The magnitude of centripetal acceleration can be expressed in three equivalent forms: $$\mathbf{a_c = rac{v^2}{r} = \omega^2 r = rac{4\pi^2 r}{T^2} = 4\pi^2 f^2 r}$$

If the speed of the particle also changes with time ($rac{dv}{dt} eq 0$), the particle possesses two perpendicular acceleration components simultaneously:

The total net acceleration vector is the vector sum: $$ec{a} = ec{a}_c + ec{a}_t$$ Since $ec{a}_c \perp ec{a}_t$, the total magnitude of acceleration is: $$\mathbf{a = \sqrt{a_c^2 + a_t^2} = \sqrt{\left(rac{v^2}{r} ight)^2 + \left(rac{dv}{dt} ight)^2}}$$ The angle $\phi$ that the net acceleration makes with the radial direction is: $$\mathbf{ an\phi = rac{a_t}{a_c} = rac{lpha r}{v^2 / r}}$$