Kinematics

Position $x(t)$ specifies the location of a particle relative to an origin at time $t$. Displacement is the change in position: $\Delta x = x_2 - x_1$ — a vector with magnitude and sign. Distance is the total path length traversed — a scalar that is always non-negative, with $|\Delta x| \leq \text{distance}$.

Average velocity over a time interval $[t_1, t_2]$: $$\bar{v} = \frac{\Delta x}{\Delta t} = \frac{x(t_2) - x(t_1)}{t_2 - t_1}$$

Instantaneous velocity is the limit as the interval shrinks: $$v(t) = \lim_{\Delta t \to 0} \frac{\Delta x}{\Delta t} = \frac{dx}{dt}$$

Speed is the magnitude of velocity, $|v|$, and the average speed is (total distance)/(total time), generally $\geq |\bar{v}|$.

Average and instantaneous acceleration are defined analogously: $$\bar{a} = \frac{\Delta v}{\Delta t}, \quad a(t) = \frac{dv}{dt} = \frac{d^2x}{dt^2}$$

Using the chain rule, $a = \dfrac{dv}{dt} = \dfrac{dv}{dx} \cdot \dfrac{dx}{dt} = v \dfrac{dv}{dx}$. This is the key identity for problems where acceleration depends on position.

Equations of Uniformly Accelerated Motion (constant $a$): Starting from $a = \dfrac{dv}{dt}$ and $v = \dfrac{dx}{dt}$, integration gives the standard kinematic equations:

These hold only for constant acceleration. The last formula gives the distance traversed during the $n$-th second alone (i.e., between $t = n-1$ and $t = n$).

Free Fall: Near Earth's surface, the acceleration due to gravity is $g \approx 9.8 \text{ m/s}^2$ directed downward. With the upward direction positive: $a = -g$. A body projected upward with velocity $u$ reaches a maximum height $H = u^2/(2g)$ in time $t = u/g$, and the total flight time (back to the launch point) is $T = 2u/g$.

Sign Convention: Always fix a positive direction at the start. Displacement, velocity, and acceleration carry signs accordingly. Do not change signs midway through a problem.

Motion graphs encode kinematic information in geometric form. The three primary graphs:

Position-time ($x$-$t$) graph:

• Slope at any point = instantaneous velocity: $v = dx/dt$.
• Straight line ⟹ uniform velocity; curved (parabolic) ⟹ uniform acceleration.

Velocity-time ($v$-$t$) graph:

• Slope at any point = instantaneous acceleration: $a = dv/dt$.
• Area under the curve between $t_1$ and $t_2$ = displacement: $s = \int v\,dt$.
• Straight horizontal line ⟹ constant velocity, zero acceleration.
• Straight inclined line ⟹ constant acceleration.

Acceleration-time ($a$-$t$) graph:

• Area under the curve = change in velocity: $\Delta v = \int a\,dt$.

Important Graph Properties:

• The sign of slope/area carries physical meaning (e.g., negative area under $v$-$t$ ⟹ displacement in the opposite direction).
• A discontinuity in the $v$-$t$ graph implies an infinite (impulsive) acceleration — unphysical for ordinary mechanics but used as idealisation in collisions.
• For SHM, $x(t)$, $v(t)$, $a(t)$ are sinusoidal with $v$ leading $x$ by $\pi/2$ and $a$ leading $x$ by $\pi$.

Non-Uniform Motion Approach (variable $a$):

• If $a = f(t)$: integrate $\int a\,dt$ to get $v$, then $\int v\,dt$ to get $x$.
• If $a = f(v)$: rearrange $dv/dt = f(v)$ → $\int dv/f(v) = \int dt$.
• If $a = f(x)$: use $v\,dv = a\,dx$ → $\int v\,dv = \int f(x)\,dx$.

Example (resistive motion): A particle moving with acceleration $a = -kv^2$ (drag-like) has $\dfrac{dv}{v^2} = -k\,dt$. Integrating: $-1/v + 1/u = -kt$ ⟹ $v = u/(1 + kut)$.

A vector is fully specified by magnitude and direction. The position vector of a particle in 3D is $\vec{r} = x\hat{i} + y\hat{j} + z\hat{k}$ with magnitude $|\vec{r}| = \sqrt{x^2 + y^2 + z^2}$.

Velocity vector: $\vec{v} = \dfrac{d\vec{r}}{dt} = \dfrac{dx}{dt}\hat{i} + \dfrac{dy}{dt}\hat{j} + \dfrac{dz}{dt}\hat{k}$. Its direction is tangent to the trajectory.

Acceleration vector: $\vec{a} = \dfrac{d\vec{v}}{dt}$. Decomposable into tangential (along $\vec{v}$, changes speed) and centripetal (perpendicular to $\vec{v}$, changes direction) components.

Projectile motion is a 2D motion where the only acceleration is gravity, $\vec{a} = -g\hat{j}$ (vertical), with no horizontal acceleration. The motion decouples into independent horizontal (uniform velocity) and vertical (uniformly accelerated) components.

For a projectile launched with speed $u$ at angle $\theta$ above the horizontal from ground level:

The trajectory is a parabola. The range $R$ is maximum when $\sin 2\theta = 1$, i.e., $\theta = 45^\circ$, giving $R_{\max} = u^2/g$.

Two angles giving the same range: Since $\sin 2\theta = \sin(180^\circ - 2\theta)$, the angles $\theta$ and $(90^\circ - \theta)$ yield the same horizontal range. Their times of flight and heights differ though.

Projectile from a height $h$ (horizontal launch with speed $u$):

• Time to land: $t = \sqrt{2h/g}$
• Horizontal distance: $x = u\sqrt{2h/g}$
• Velocity at landing: $|\vec{v}| = \sqrt{u^2 + 2gh}$ at angle $\tan^{-1}(\sqrt{2gh}/u)$ below horizontal.

Projectile on an inclined plane (incline angle $\alpha$, launch angle $\theta$ measured from the incline): Use a tilted coordinate system. The range along the incline: $$R_{\text{incline}} = \frac{2u^2 \sin\theta \cos(\theta + \alpha)}{g \cos^2\alpha}$$ Maximum when $\theta = (90^\circ - \alpha)/2 = 45^\circ - \alpha/2$, giving: $$R_{\max} = \frac{u^2}{g(1 + \sin\alpha)}$$

Relative velocity of body A with respect to body B: $$\vec{v}_{AB} = \vec{v}_A - \vec{v}_B$$ This is the velocity of A as observed from a frame moving with B. In 1D, simply subtract scalar velocities (with signs); in 2D, use vector subtraction.

Magnitude in 2D: If $\vec{v}_A$ and $\vec{v}_B$ make angle $\phi$, $$|\vec{v}_{AB}| = \sqrt{v_A^2 + v_B^2 - 2 v_A v_B \cos\phi}$$

River-Boat (Crossing) Problems: Let $v_b$ = speed of boat in still water, $v_r$ = speed of river current. Two key cases:

Case 1: Crossing the river in shortest time. Boat is steered perpendicular to the bank. Time: $$t_{\min} = \frac{d}{v_b}$$ But the boat is also drifted downstream by $\Delta x = v_r t_{\min} = v_r d / v_b$. The resultant velocity is $\sqrt{v_b^2 + v_r^2}$.

Case 2: Crossing in shortest path (perpendicular landing). Boat is steered at an angle $\theta$ upstream such that the upstream component cancels the current: $v_b \sin\theta = v_r \implies \sin\theta = v_r/v_b$. This requires $v_b > v_r$. The net velocity across the river is $\sqrt{v_b^2 - v_r^2}$. Time: $$t = \frac{d}{\sqrt{v_b^2 - v_r^2}}$$

Rain-Man Problem: If rain falls with velocity $\vec{v}_r$ and a man walks with $\vec{v}_m$, the velocity of rain as observed by the man is $\vec{v}_r - \vec{v}_m$, and the umbrella must be tilted in this direction.

Uniform Circular Motion: A particle moves at constant speed $v$ along a circle of radius $r$. The velocity is always tangent; the acceleration is centripetal (toward the centre): $$a_c = \dfrac{v^2}{r} = \omega^2 r$$ where $\omega = v/r$ is the angular velocity (rad/s). Period of revolution: $T = 2\pi/\omega = 2\pi r/v$.

Non-uniform Circular Motion: When the speed changes, there is also a tangential acceleration $a_t = dv/dt$. Total acceleration: $$|\vec{a}| = \sqrt{a_c^2 + a_t^2}$$

Relationships in Rotation: $\omega = d\theta/dt$ (angular velocity); $\alpha = d\omega/dt$ (angular acceleration). For uniform $\alpha$: $$\omega = \omega_0 + \alpha t, \quad \theta = \omega_0 t + \tfrac{1}{2}\alpha t^2, \quad \omega^2 = \omega_0^2 + 2\alpha\theta$$