Motion in a Plane

Statement: If two vectors are represented in magnitude and direction by the two sides of a triangle taken in the same order, then their resultant is represented by the third side of the triangle taken in opposite order.

If $\vec{A}$ and $\vec{B}$ are two vectors, their sum $\vec{R} = \vec{A} + \vec{B}$

Derivation of Resultant Magnitude and Direction:

Let two vectors $\vec{A}$ and $\vec{B}$ act at a point O at angle $\theta$ between them.

Using parallelogram law, complete parallelogram OPQS. Diagonal OQ represents resultant $\vec{R}$.

Step 1: Find magnitude R

Drop perpendicular from Q to OP extended, meeting at point M.

In triangle OQM:

• $OM = OP + PM = A + B\cos\theta$
• $QM = B\sin\theta$
• $OQ^2 = OM^2 + QM^2$ (Pythagoras theorem)

$R^2 = (A + B\cos\theta)^2 + (B\sin\theta)^2$

$R^2 = A^2 + 2AB\cos\theta + B^2\cos^2\theta + B^2\sin^2\theta$

$R^2 = A^2 + 2AB\cos\theta + B^2(\cos^2\theta + \sin^2\theta)$

Step 2: Find direction α

Let $\alpha$ be angle between $\vec{R}$ and $\vec{A}$.

$\tan\alpha = \frac{QM}{OM} = \frac{B\sin\theta}{A + B\cos\theta}$

Mathematical Formulation:

If vector $\vec{A}$ makes angle $\theta$ with x-axis, then:

• $A_x = A \cos\theta$ (Component along x-axis)
• $A_y = A \sin\theta$ (Component along y-axis)

Reconstruction of vector:

$\vec{A} = A_x\hat{i} + A_y\hat{j}$

Magnitude: $A = \sqrt{A_x^2 + A_y^2}$

Direction: $\tan\theta = \frac{A_y}{A_x}$

Analysis of Motion:

Initial position: Origin O (0, 0)

Initial velocity: $\vec{u} = u\cos\theta \, \hat{i} + u\sin\theta \, \hat{j}$

Acceleration: $\vec{a} = 0\hat{i} - g\hat{j}$ (Only vertical, downward)

Horizontal Motion (x-direction):

• Initial velocity: $u_x = u\cos\theta$
• Acceleration: $a_x = 0$ (No horizontal force)
• Velocity at time t: $v_x = u\cos\theta$ (constant)
• Displacement: $x = u\cos\theta \cdot t$

Vertical Motion (y-direction):

• Initial velocity: $u_y = u\sin\theta$
• Acceleration: $a_y = -g$ (downward)
• Velocity at time t: $v_y = u\sin\theta - gt$
• Displacement: $y = u\sin\theta \cdot t - \frac{1}{2}gt^2$

Goal: Find equation of path y(x)

From horizontal motion: $t = \frac{x}{u\cos\theta}$ ... (i)

From vertical motion: $y = u\sin\theta \cdot t - \frac{1}{2}gt^2$ ... (ii)

Substituting (i) into (ii):

$y = u\sin\theta \left(\frac{x}{u\cos\theta}\right) - \frac{1}{2}g\left(\frac{x}{u\cos\theta}\right)^2$

$y = x\tan\theta - \frac{g x^2}{2u^2\cos^2\theta}$

$y = x\tan\theta - \frac{g x^2}{2u^2}(1 + \tan^2\theta)$ [Using $\sec^2\theta = 1 + \tan^2\theta$]

This is equation of a parabola (form $y = ax + bx^2$)

Definition: Total time for which projectile remains in air.

At landing, projectile returns to ground level: $y = 0$

$0 = u\sin\theta \cdot T - \frac{1}{2}gT^2$

$0 = T(u\sin\theta - \frac{1}{2}gT)$

Either $T = 0$ (at launch) or $u\sin\theta = \frac{1}{2}gT$

Note: Time to reach max height = $T/2 = \frac{u\sin\theta}{g}$

Definition: Greatest vertical distance reached by projectile.

At max height H, vertical velocity becomes zero: $v_y = 0$

Using $v_y^2 = u_y^2 - 2gH$

$0 = (u\sin\theta)^2 - 2gH$

$2gH = u^2\sin^2\theta$

Alternative: Using $v_y = u_y - gt$ at $t = T/2$:

$H = u\sin\theta \cdot \frac{T}{2} - \frac{1}{2}g\left(\frac{T}{2}\right)^2 = \frac{u^2\sin^2\theta}{2g}$ ✓

Definition: Horizontal distance covered during time of flight.

Range R = Horizontal distance at $t = T$

$R = u\cos\theta \cdot T = u\cos\theta \cdot \frac{2u\sin\theta}{g}$

$R = \frac{2u^2\sin\theta\cos\theta}{g}$

Using $\sin 2\theta = 2\sin\theta\cos\theta$:

Maximum Range:

$R_{max}$ occurs when $\sin 2\theta = 1$, i.e., $2\theta = 90°$ → $\theta = 45°$

Important Result: Two angles give same range:

If $\theta$ gives range R, then $(90° - \theta)$ also gives same range R

Because $\sin 2\theta = \sin(180° - 2\theta) = \sin 2(90° - \theta)$

Analysis:

Rain falls vertically downward with velocity $v_r$ (w.r.t. ground)

Man moves horizontally with velocity $v_m$ (w.r.t. ground)

Velocity of rain relative to man:

$\vec{v}_{rm} = \vec{v}_r - \vec{v}_m$

Taking horizontal as x-axis, vertically down as y-axis:

$\vec{v}_r = 0\hat{i} + v_r\hat{j}$, $\vec{v}_m = v_m\hat{i} + 0\hat{j}$

$\vec{v}_{rm} = -v_m\hat{i} + v_r\hat{j}$

Magnitude: $|\vec{v}_{rm}| = \sqrt{v_m^2 + v_r^2}$

Direction: $\tan\theta = \frac{v_m}{v_r}$ from vertical

Practical Note: To avoid getting wet, man should hold umbrella at angle $\theta$ from vertical in forward direction.

Case 1: To reach directly opposite point

Swimmer must swim at angle $\theta$ upstream such that resultant velocity is perpendicular to banks.

Taking perpendicular to bank as y-axis, along flow as x-axis:

$\vec{v}_{river} = v_r\hat{i}$, $\vec{v}_{swimmer} = -v_s\sin\theta\hat{i} + v_s\cos\theta\hat{j}$

For perpendicular crossing, net x-component = 0:

$v_r - v_s\sin\theta = 0$

Net velocity: $v_{net} = v_s\cos\theta = v_s\sqrt{1 - (v_r/v_s)^2} = \sqrt{v_s^2 - v_r^2}$

Time to cross width $d$: $t = \frac{d}{v_{net}} = \frac{d}{\sqrt{v_s^2 - v_r^2}}$

Condition: $v_s > v_r$ (otherwise impossible to reach opposite point)

Case 2: For minimum time crossing

Swimmer should swim perpendicular to banks ($\theta = 90°$ from bank)

Then $t_{min} = \frac{d}{v_s}$

But will drift downstream by drift = $v_r \times t_{min} = \frac{v_r d}{v_s}$

Angular Quantities:

• Angular displacement ($\theta$): Angle swept by radius vector (in radians)
• Angular velocity ($\omega$): $\omega = \frac{d\theta}{dt}$ (rad/s)
• Relation: Arc length $s = R\theta$, so $v = R\omega$

Deriving Centripetal Acceleration:

Consider particle moving from P to Q in small time $\Delta t$.

At P: velocity $\vec{v}_1$ (tangent to circle)

At Q: velocity $\vec{v}_2$ (tangent to circle)

Both have same magnitude $v$, but different directions.

Change in velocity: $\Delta \vec{v} = \vec{v}_2 - \vec{v}_1$

For small $\Delta \theta$, $|\Delta \vec{v}| \approx v \Delta \theta$ (from vector triangle)

Acceleration: $a = \frac{|\Delta \vec{v}|}{\Delta t} = \frac{v \Delta \theta}{\Delta t} = v \omega = v \cdot \frac{v}{R}$

Direction: $\Delta \vec{v}$ points towards center (perpendicular to $\vec{v}$)

Hence Centripetal Acceleration = directed towards center, magnitude $v^2/R$