Motion in a Plane - Comprehensive Study Guide

Introduction to Motion in Two and Three Dimensions

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• Dimensionality and Motion:

  • In one-dimensional motion (motion along a straight line), directional aspects are fully specified using positive ($+$ ) and negative ($-$) signs because only two directions are possible.

  • In two-dimensional motion (motion in a plane) or three-dimensional motion (motion in space), physical quantities such as position, displacement, velocity, and acceleration require vectors to specify both magnitude and direction.

• Scope of Study:

  • Vector definitions and vector algebra (addition, subtraction, and multiplication by real numbers).

  • Resolution of vectors into components along coordinate axes.

  • Kinematics of motion in a plane with constant acceleration.

  • Detailed analysis of projectile motion.

  • Uniform circular motion in a plane.

Scalars and Vectors

• Scalar Quantities:

  • Definition: A scalar quantity is a physical quantity that possesses magnitude only and has no associated direction.

  • Specification: Specified completely by a single real number along with an appropriate unit.

  • Algebraic Rules: Obey ordinary rules of algebra (can be added, subtracted, multiplied, and divided like real numbers).

  • Addition / Subtraction Constraint: Scalar addition and subtraction make sense only for quantities possessing identical physical dimensions and units.

  • Multiplication / Division: Scalars with different units can be multiplied or divided (e.g., mass divided by volume yields density).

  • Examples:

    • Distance between two points.

    • Mass of an object.

    • Temperature of a body.

    • Time at which an event occurs.

    • Concrete Example 1: For a rectangle of length $1.0\,m$ and breadth $0.5\,m$, perimeter is $1.0\,m + 0.5\,m + 1.0\,m + 0.5\,m = 3.0\,m$. Both side lengths and perimeter are scalar quantities.

    • Concrete Example 2: If maximum and minimum daily temperatures are $35.6\,^\circ\text{C}$ and $24.2\,^\circ\text{C}$, the temperature difference is $11.4\,^\circ\text{C}$.

    • Concrete Example 3: A uniform aluminium cube of side $10\,cm$ ($0.1\,m$) with mass $2.7\,kg$ has a volume of $10^{-3}\,m^3$ (scalar) and density of $2.7 \times 10^3\,kg\,m^{-3}$ (scalar).

• Vector Quantities:

  • Definition: A vector quantity is a physical quantity that possesses both magnitude and direction and strictly obeys the triangle law of addition (or equivalently, the parallelogram law of addition).

  • Specification: Specified by giving its magnitude (a number with units) and its direction in space.

  • Examples: Displacement, velocity, acceleration, force, momentum.

  • Notation:

    • Represented in print using boldface type (e.g., $\mathbf{v}$, $\mathbf{A}$, $\mathbf{r}$).

    • Represented in handwriting using an arrow placed over a letter (e.g., $\vec{v}$, $\vec{A}$).

    • Magnitude (absolute value) is represented by lightface type or vertical bars: $|\mathbf{v}| = v$ or |\mathbf{A}| = A$.\n\n* **Position and Displacement Vectors**:\n * **Position Vector**:\n * To describe position in a plane, a reference point O is chosen as origin.\n * The line joining origin O$to object position$P$at time$t$is the position vector$\mathbf{r} = OP$, with its arrow point at$P.\n * Position at time t'$is$P'$, represented by position vector$\mathbf{r'} = OP'.\n * Length of \mathbf{r}$represents magnitude, and arrow direction points toward$P$from$O$.

  • Displacement Vector:

    • The vector $\mathbf{PP'}$ with tail at initial position $P$ (time $t$) and tip at final position $P'$ (time $t'$) is the displacement vector.

    • Displacement is the straight line joining initial and final positions and is independent of the actual path taken between points.

    • For initial point $P$ and final point $Q$, displacement vector $\mathbf{PQ}$ is identical whether the path taken is $PABCQ$, $PDQ$, or $PBEFQ$.

    • Magnitude of displacement is always less than or equal to the actual path length ($|\Delta \mathbf{r}| \le \text{path length}$).

• Equality of Vectors:

  • Two vectors $\mathbf{A}$ and $\mathbf{B}$ are equal ($\mathbf{A} = \mathbf{B}$) if and only if they have identical magnitudes AND identical directions.

  • Free Vectors: Vectors that can be shifted parallel to themselves without changing their physical meaning (location is not fixed).

  • Localised Vectors: Vectors for which line of application or specific point of application matters in physical applications.

  • Vectors $\mathbf{A'}$ and $\mathbf{B'}$ with equal lengths but different directions are NOT equal ($\mathbf{A'} \neq \mathbf{B'}$).

Multiplication of Vectors by Real Numbers

• Multiplication by a Positive Real Number ($\lambda > 0$):

  • Multiplying vector $\mathbf{A}$ by a positive scalar $\lambda$ yields a vector whose magnitude is multiplied by factor $\lambda$ while direction remains unchanged.

  • Equation: $|\lambda \mathbf{A}| = \lambda |\mathbf{A}|$

  • Example: Multiplying $\mathbf{A}$ by $2$ yields resultant $2\mathbf{A}$, pointing in the same direction with magnitude $2|\mathbf{A}|$.

• Multiplication by a Negative Real Number ($-\lambda < 0$):

  • Multiplying vector $\mathbf{A}$ by a negative scalar $-\lambda$ yields a vector pointing in the opposite direction of $\mathbf{A}$ with magnitude $\lambda |\mathbf{A}|$ .

  • Examples: Multiplying $\mathbf{A}$ by $-1$ yields $-\mathbf{A}$ (same magnitude, opposite direction); multiplying by $-1.5$ yields $-1.5\mathbf{A}$ (magnitude $1.5|\mathbf{A}|$, opposite direction).

• Dimensional Multiplication:

  • If scalar multiplier $\lambda$ possesses physical dimensions, the dimension of $\lambda \mathbf{A}$ is the product of dimensions of $\lambda$ and $\mathbf{A}$.

  • Example: Constant velocity vector $\mathbf{v}$ multiplied by time duration $t$ gives displacement vector $\mathbf{r}$.

Addition and Subtraction of Vectors — Graphical Method

• Triangle Method (Head-to-Tail Method):

  • To add vectors $\mathbf{A}$ and $\mathbf{B}$, place the tail of vector $\mathbf{B}$ at the head of vector $\mathbf{A}$.

  • Join the tail of $\mathbf{A}$ to the head of $\mathbf{B}$ to construct line $OQ$, which represents resultant vector $\mathbf{R} = \mathbf{A} + \mathbf{B}$.

  • Commutative Law of Vector Addition:

    • $\mathbf{A} + \mathbf{B} = \mathbf{B} + \mathbf{A}$

  • Associative Law of Vector Addition:

    • $(\mathbf{A} + \mathbf{B}) + \mathbf{C} = \mathbf{A} + (\mathbf{B} + \mathbf{C})$

• Null Vector (Zero Vector):

  • Definition: Sum of two equal and opposite vectors $\mathbf{A}$ and $-\mathbf{A}$.

  • Equation: $\mathbf{A} - \mathbf{A} = \mathbf{0}$, where $|\mathbf{0}| = 0$.

  • Direction: Because magnitude is zero, direction of a null vector is indefinite / unspecified.

  • Properties:

    • $\mathbf{A} + \mathbf{0} = \mathbf{A}$

    • $\lambda \mathbf{0} = \mathbf{0}$

    • $0\mathbf{A} = \mathbf{0}$

  • Physical Meaning: If an object moves from $P$ to $P'$ and returns to $P$, initial and final positions coincide, making net displacement a null vector.

• Vector Subtraction:

  • Difference of vectors $\mathbf{A}$ and $\mathbf{B}$ is defined as the addition of $\mathbf{A}$ and $-\mathbf{B}$.

  • Equation: $\mathbf{A} - \mathbf{B} = \mathbf{A} + (-\mathbf{B})$

  • Constructed graphically by adding vector $-\mathbf{B}$ (equal magnitude, opposite direction to $\mathbf{B}$) to vector $\mathbf{A}$.

• Parallelogram Method:

  • Bring tails of vectors $\mathbf{A}$ and $\mathbf{B}$ to a common origin $O$.

  • Draw a line from head of $\mathbf{A}$ parallel to $\mathbf{B}$, and a line from head of $\mathbf{B}$ parallel to $\mathbf{A}$ to construct parallelogram OQSP$.\n * Resultant vector \mathbf{R}$is directed along diagonal$OS$from origin$O$.

  • Parallelogram method and triangle method are completely equivalent.

Resolution of Vectors and Unit Vectors

• Resolution of a Vector in a Plane:

  • Let $\mathbf{a}$ and $\mathbf{b}$ be non-zero, non-collinear vectors in a plane. Any vector $\mathbf{A}$ in the same plane can be expressed uniquely as:

    • $\mathbf{A} = \lambda \mathbf{a} + \mu \mathbf{b}$

    • where $\lambda$ and $\mu$ are real numbers.

  • $\lambda \mathbf{a}$ and $\mu \mathbf{b}$ are component vectors of $\mathbf{A}$ along $\mathbf{a}$ and $\mathbf{b}$.

• Unit Vectors:

  • Definition: A unit vector is a vector of unit magnitude ($1$) pointing in a specific direction. It has no dimension and no physical unit.

  • Cartesian Coordinate Unit Vectors: Denoted by $\hat{\mathbf{i}}$, $\hat{\mathbf{j}}$, and $\hat{\mathbf{k}}$ along positive $x$-, $y$-, and $z$-axes respectively.

  • Equation: $|\hat{\mathbf{i}}| = |\hat{\mathbf{j}}| = |\hat{\mathbf{k}}| = 1$

  • Unit vectors are mutually perpendicular (orthogonal).

  • General representation: $\mathbf{A} = |\mathbf{A}|\hat{\mathbf{n}}$, where $\hat{\mathbf{n}}$ is unit vector along $\mathbf{A}$.

• Rectangular Components in Two Dimensions:

  • Resolving vector $\mathbf{A}$ in $x\text{-}y$ plane:

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

  • Component scalars $A_x$ and $A_y$ relative to angle $\theta$ with positive $x$-axis:

    • $A_x = A \cos(\theta)$

    • $A_y = A \sin(\theta)$

  • Note: $A_x$ and $A_y$ are real numbers (scalars), whereas $A_x \hat{\mathbf{i}}$ and $A_y \hat{\mathbf{j}}$ are component vectors.

  • Reconstructing vector magnitude and direction:

    • $A = \sqrt{A_x^2 + A_y^2}$

    • $\tan(\theta) = \frac{A_y}{A_x} \implies \theta = \tan^{-1}\left(\frac{A_y}{A_x}\right)$

• Rectangular Components in Three Dimensions:

  • If $\alpha$, $\beta$, and $\gamma$ are angles between vector $\mathbf{A}$ and $x$-, $y$-, and $z$-axes:

    • $A_x = A \cos(\alpha)$

    • $A_y = A \cos(\beta)$

    • $A_z = A \cos(\gamma)$

  • Vector expression:

    • $\mathbf{A} = A_x \hat{\mathbf{i}} + A_y \hat{\mathbf{j}} + A_z \hat{\mathbf{k}}$

  • Magnitude expression:

    • $A = \sqrt{A_x^2 + A_y^2 + A_z^2}$

  • Position vector in 3D:

    • $\mathbf{r} = x \hat{\mathbf{i}} + y \hat{\mathbf{j}} + z \hat{\mathbf{k}}$

Vector Addition — Analytical Method

• Addition of Two Vectors in a Plane:

  • Let $\mathbf{A} = A_x \hat{\mathbf{i}} + A_y \hat{\mathbf{j}}$ and $\mathbf{B} = B_x \hat{\mathbf{i}} + B_y \hat{\mathbf{j}}$.

  • Resultant $\mathbf{R} = \mathbf{A} + \mathbf{B} = (A_x + B_x)\hat{\mathbf{i}} + (A_y + B_y)\hat{\mathbf{j}}$.

  • Components of resultant:

    • $R_x = A_x + B_x$

    • $R_y = A_y + B_y$

• Addition in Three Dimensions:

  • $R_x = A_x + B_x$

  • $R_y = A_y + B_y$

  • $R_z = A_z + B_z$

• General Combination of Vectors:

  • For $\mathbf{T} = \mathbf{a} + \mathbf{b} - \mathbf{c}$:

    • $T_x = a_x + b_x - c_x$

    • $T_y = a_y + b_y - c_y$

    • $T_z = a_z + b_z - c_z$

• Derivation of Law of Cosines and Law of Sines:

  • Consider vectors $\mathbf{A}$ ($OP$) and $\mathbf{B}$ ($OQ$) at angle $\theta$. Parallelogram diagonal $OS$ represents resultant $\mathbf{R}$.

  • Perpendicular $SN$ drawn to $OP$ extended; perpendicular $PM$ drawn to $OS$.

  • From geometry: $ON = OP + PN = A + B \cos(\theta)$ and $SN = B \sin(\theta)$.

  • Using Pythagorean theorem in $\Delta OSN$:

    • $OS^2 = ON^2 + SN^2 = (A + B \cos(\theta))^2 + (B \sin(\theta))^2$

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

    • Law of Cosines: $R = \sqrt{A^2 + B^2 + 2AB \cos(\theta)}$

  • Law of Sines:

    • In $\Delta OSN$, $SN = R \sin(\alpha)$. In $\Delta PSN$, $SN = B \sin(\theta)$.

    • Equating gives $R \sin(\alpha) = B \sin(\theta) \implies \frac{R}{\sin(\theta)} = \frac{B}{\sin(\alpha)}$.

    • Similarly, $PM = A \sin(\alpha) = B \sin(\beta) \implies \frac{A}{\sin(\beta)} = \frac{B}{\sin(\alpha)}$.

    • Combined Law of Sines: $\frac{R}{\sin(\theta)} = \frac{A}{\sin(\beta)} = \frac{B}{\sin(\alpha)}$

  • Direction angle $\alpha$ relative to vector $\mathbf{A}$:

    • $\tan(\alpha) = \frac{B \sin(\theta)}{A + B \cos(\theta)}$

General Motion in a Plane

• Position Vector and Displacement:

  • Position vector at time $t$: $\mathbf{r} = x \hat{\mathbf{i}} + y \hat{\mathbf{j}}$, where $x$ and $y$ are coordinates.

  • Displacement between position $P$ (time $t$) and $P'$ (time $t'$):

    • $\Delta \mathbf{r} = \mathbf{r'} - \mathbf{r} = (x' - x)\hat{\mathbf{i}} + (y' - y)\hat{\mathbf{j}} = \Delta x \hat{\mathbf{i}} + \Delta y \hat{\mathbf{j}}$

• Velocity:

  • Average Velocity:

    • $\overline{\mathbf{v}} = \frac{\Delta \mathbf{r}}{\Delta t} = \frac{\Delta x}{\Delta t}\hat{\mathbf{i}} + \frac{\Delta y}{\Delta t}\hat{\mathbf{j}} = \overline{v}_x \hat{\mathbf{i}} + \overline{v}_y \hat{\mathbf{j}}$

    • Direction of average velocity is identical to direction of displacement $\Delta \mathbf{r}$.

  • Instantaneous Velocity:

    • Limiting value of average velocity as time interval approaches zero:

    • $\mathbf{v} = \lim_{\Delta t \rightarrow 0} \frac{\Delta \mathbf{r}}{\Delta t} = \frac{d\mathbf{r}}{dt} = v_x \hat{\mathbf{i}} + v_y \hat{\mathbf{j}}$

    • Components: $v_x = \frac{dx}{dt}$ and $v_y = \frac{dy}{dt}$.

  • Direction of Velocity:

    • As $\Delta t \rightarrow 0$, displacement $\Delta \mathbf{r}$ becomes chord aligned with tangent to trajectory.

    • The direction of instantaneous velocity at any point on a particle path is always tangential to the path at that point in the direction of motion.

  • Magnitude and Direction Angle:

    • $v = \sqrt{v_x^2 + v_y^2}$

    • $\tan(\theta) = \frac{v_y}{v_x} \implies \theta = \tan^{-1}\left(\frac{v_y}{v_x}\right)$

• Acceleration:

  • Average Acceleration:

    • $\overline{\mathbf{a}} = \frac{\Delta \mathbf{v}}{\Delta t} = \frac{\Delta v_x}{\Delta t}\hat{\mathbf{i}} + \frac{\Delta v_y}{\Delta t}\hat{\mathbf{j}} = \overline{a}_x \hat{\mathbf{i}} + \overline{a}_y \hat{\mathbf{j}}$

    • Direction of average acceleration is identical to direction of velocity change $\Delta \mathbf{v}$.

  • Instantaneous Acceleration:

    • $\mathbf{a} = \lim_{\Delta t \rightarrow 0} \frac{\Delta \mathbf{v}}{\Delta t} = \frac{d\mathbf{v}}{dt} = a_x \hat{\mathbf{i}} + a_y \hat{\mathbf{j}}$

    • Components: $a_x = \frac{dv_x}{dt} = \frac{d^2 x}{dt^2}$ and $a_y = \frac{dv_y}{dt} = \frac{d^2 y}{dt^2}$.

  • Angular Relationship between Velocity and Acceleration:

    • In 1D motion, velocity and acceleration are collinear (angle $0^\circ$ or $180^\circ$).

    • In 2D or 3D motion, velocity and acceleration vectors may form any angle between $0^\circ$ and $180^\circ$.

Motion in a Plane with Constant Acceleration

• Vector Equations for Constant Acceleration:

  • If acceleration $\mathbf{a}$ is constant over time interval $t$:

  • Velocity vector at time $t$:

    • $\mathbf{v} = \mathbf{v}_0 + \mathbf{a}t$

  • Position vector at time $t$:

    • $\mathbf{r} = \mathbf{r}_0 + \mathbf{v}_0 t + \frac{1}{2}\mathbf{a}t^2$

• Component Equations:

  • Velocity components:

    • $v_x = v_{0x} + a_x t$

    • $v_y = v_{0y} + a_y t$

  • Position components:

    • $x = x_0 + v_{0x} t + \frac{1}{2}a_x t^2$

    • $y = y_0 + v_{0y} t + \frac{1}{2}a_y t^2$

• Independence of Perpendicular Motions:

  • Motion in a plane (two dimensions) with constant acceleration is exactly equivalent to two separate, simultaneous, independent one-dimensional motions along perpendicular axes.

Projectile Motion

• Definition & Historical Context:

  • A projectile is an object in flight after being thrown or projected into space (e.g., football, cricket ball, baseball).

  • Galileo Galilei first stated the independence of horizontal and vertical components of projectile motion in Dialogue on the Great World Systems (1632).

• Assumptions & Acceleration Components:

  • Air resistance force is assumed to be negligible.

  • Free fall acceleration due to gravity acts downward: $\mathbf{a} = -g \hat{\mathbf{j}}$.

  • Horizontal acceleration component: $a_x = 0$

  • Vertical acceleration component: $a_y = -g$

• Initial Velocity Components:

  • Launched at speed $v_0$ at angle $\theta_0$ with positive $x$-axis:

  • $v_{0x} = v_0 \cos(\theta_0)$

  • $v_{0y} = v_0 \sin(\theta_0)$

• Kinematic Equations of Projectile Motion (with origin $x_0 = 0$, $y_0 = 0$):

  • Horizontal position: $x = (v_0 \cos(\theta_0)) t$

  • Vertical position: $y = (v_0 \sin(\theta_0)) t - \frac{1}{2}g t^2$

  • Horizontal velocity: $v_x = v_{0x} = v_0 \cos(\theta_0)$ (constant throughout flight)

  • Vertical velocity: $v_y = v_0 \sin(\theta_0) - g t$

• Equation of Trajectory / Path:

  • Eliminating time $t = \frac{x}{v_0 \cos(\theta_0)}$ from coordinate equations yields:

    • $y = \tan(\theta_0) x - \left(\frac{g}{2(v_0 \cos(\theta_0))^2}\right) x^2$

  • This equation has the mathematical form $y = ax + bx^2$ (where $a$ and $b$ are constants), proving that the path of a projectile is a parabola.

• Time of Maximum Height ($t_m$):

  • At peak height $h_m$, vertical velocity $v_y = 0$ .

  • Equation: $v_0 \sin(\theta_0) - g t_m = 0 \implies t_m = \frac{v_0 \sin(\theta_0)}{g}$

• Total Time of Flight ($T_f$):

  • Total duration during which projectile remains in air (found by setting $y = 0$):

  • $T_f = \frac{2v_0 \sin(\theta_0)}{g} = 2 t_m$

• Maximum Height ($h_m$):

  • Substituting $t = t_m$ into vertical position equation:

  • $h_m = v_0 \sin(\theta_0) \left(\frac{v_0 \sin(\theta_0)}{g}\right) - \frac{1}{2}g \left(\frac{v_0 \sin(\theta_0)}{g}\right)^2$

  • $h_m = \frac{(v_0 \sin(\theta_0))^2}{2g}$

• Horizontal Range ($R$):

  • Total horizontal distance covered during time of flight $T_f$:

  • $R = (v_0 \cos(\theta_0)) T_f = (v_0 \cos(\theta_0)) \left(\frac{2v_0 \sin(\theta_0)}{g}\right)$

  • $R = \frac{v_0^2 \sin(2\theta_0)}{g}$

  • Maximum Horizontal Range ($R_m$):

    • Range is maximum when $\sin(2\theta_0) = 1 \implies 2\theta_0 = 90^\circ \implies \theta_0 = 45^\circ$.

    • $R_m = \frac{v_0^2}{g}$

Uniform Circular Motion

• Definition:

  • When an object traverses a circular path of radius $R$ at a constant speed $v$, its motion is termed uniform circular motion.

• Velocity and Acceleration:

  • Velocity vector continuously changes direction along the tangent to the circle, creating acceleration despite constant speed.

  • Direction of acceleration is always directed towards the center of the circular path at every point.

  • Termed centripetal acceleration ($\mathbf{a}_c$) from Greek meaning "center-seeking". First published analysis by Christiaan Huygens (1673).

• Derivation of Centripetal Acceleration Magnitude:

  • Consider position vectors $\mathbf{r}$, $\mathbf{r'}$ and velocity vectors $\mathbf{v}$, $\mathbf{v'}$ subtending angle $\Delta \theta$.

  • Triangle formed by position vectors $\Delta \mathbf{r}$ and triangle formed by velocity vectors $\Delta \mathbf{v}$ are similar:

    • $\frac{|\Delta \mathbf{v}|}{v} = \frac{|\Delta \mathbf{r}|}{R} \implies |\Delta \mathbf{v}| = \frac{v}{R} |\Delta \mathbf{r}|$

  • Centripetal acceleration magnitude:

    • $a_c = \lim_{\Delta t \rightarrow 0} \frac{|\Delta \mathbf{v}|}{\Delta t} = \frac{v}{R} \lim_{\Delta t \rightarrow 0} \frac{|\Delta \mathbf{r}|}{\Delta t} = \frac{v^2}{R}$

  • Magnitude $a_c = \frac{v^2}{R}$ is constant, but centripetal acceleration vector $\mathbf{a}_c$ is NOT constant because its direction changes continuously.

• Angular Speed ($\omega$) and Frequency Relations:

  • Angular displacement $\Delta \theta$ over time interval $\Delta t$.

  • Angular speed: $\omega = \frac{\Delta \theta}{\Delta t}$

  • Arc length $\Delta s = R \Delta \theta \implies v = \frac{\Delta s}{\Delta t} = R \frac{\Delta \theta}{\Delta t} = R \omega$

  • Centripetal acceleration in terms of $\omega$:

    • $a_c = \frac{(R\omega)^2}{R} = \omega^2 R$

  • Time period $T$ (time for 1 revolution) and frequency $\nu = \frac{1}{T}$ (revolutions per second):

    • Speed: $v = \frac{2\pi R}{T} = 2\pi R \nu$

    • Angular speed: $\omega = \frac{2\pi}{T} = 2\pi \nu$

    • Centripetal acceleration: $a_c = 4\pi^2 \nu^2 R$

Textbook Examples and Solved Problems

• Example 3.1:

  • Problem: Rain falls vertically at $35\,m\,s^{-1}$. Wind blows from east to west at $12\,m\,s^{-1}$. Find the direction to hold an umbrella.

  • Solution:

    • Rain velocity $\mathbf{v}_r = 35\,m\,s^{-1}$ downward; wind velocity $\mathbf{v}_w = 12\,m\,s^{-1}$ westward.

    • Resultant speed $R = \sqrt{v_r^2 + v_w^2} = \sqrt{35^2 + 12^2} = \sqrt{1225 + 144} = \sqrt{1369} = 37\,m\,s^{-1}$.

    • Angle $\theta$ with vertical: $\tan(\theta) = \frac{v_w}{v_r} = \frac{12}{35} \approx 0.343 \implies \theta = \tan^{-1}(0.343) \approx 19^\circ$.

    • Conclusion: Hold umbrella at an angle of about $19^\circ$ with the vertical towards the east.

  • 
    

  

• Example 3.3:

  • Problem: Motorboat races north at $25\,km/h$; water current is $10\,km/h$ in direction $60^\circ$ east of south. Find resultant velocity.

  • Solution:

    • Angle between boat velocity $\mathbf{v}_b$ (North) and current $\mathbf{v}_c$ ($60^\circ$ E of S) is $\theta = 180^\circ - 60^\circ = 120^\circ$.

    • Law of Cosines: $R = \sqrt{v_b^2 + v_c^2 + 2v_b v_c \cos(120^\circ)} = \sqrt{25^2 + 10^2 + 2(25)(10)(-0.5)} = \sqrt{625 + 100 - 250} = \sqrt{475} \approx 22\,km/h$.

    • Law of Sines for angle $\phi$ with North: $\frac{R}{\sin(120^\circ)} = \frac{v_c}{\sin(\phi)} \implies \sin(\phi) = \frac{10 \sin(120^\circ)}{21.8} = \frac{10 \times 0.866}{21.8} \approx 0.397 \implies \phi \approx 23.4^\circ$.

    • Conclusion: Resultant velocity is $22\,km/h$ at $23.4^\circ$ east of north.

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• Example 3.4:

  • Problem: Position $\mathbf{r}(t) = 3.0t \hat{\mathbf{i}} + 2.0t^2 \hat{\mathbf{j}} + 5.0 \hat{\mathbf{k}}$. (a) Find $\mathbf{v}(t)$ and $\mathbf{a}(t)$. (b) Find magnitude and direction of $\mathbf{v}(t)$ at t = 1.0\,s$.\n * *Solution*:\n * (a) \mathbf{v}(t) = \frac{d\mathbf{r}}{dt} = 3.0 \hat{\mathbf{i}} + 4.0t \hat{\mathbf{j}}$;$\mathbf{a}(t) = \frac{d\mathbf{v}}{dt} = 4.0 \hat{\mathbf{j}}\,m\,s^{-2}.\n * (b) At t = 1.0\,s$,$\mathbf{v} = 3.0 \hat{\mathbf{i}} + 4.0 \hat{\mathbf{j}}$. Magnitude$v = \sqrt{3.0^2 + 4.0^2} = 5.0\,m\,s^{-1}$. Direction$\theta = \tan^{-1}\left(\frac{4.0}{3.0}\right) \approx 53^\circ relative to positive x-axis.\n\n* **Example 3.5**:\n * *Problem*: Particle starts from origin at t=0$with$\mathbf{v}_0 = 5.0 \hat{\mathbf{i}}\,m/s$, moving under constant acceleration$\mathbf{a} = (3.0 \hat{\mathbf{i}} + 2.0 \hat{\mathbf{j}})\,m/s^2$. (a) Find y-coordinate when $x = 84\,m$. (b) Find speed at this instant.

  • Solution:

    • Position: $\mathbf{r}(t) = (5.0t + 1.5t^2)\hat{\mathbf{i}} + (1.0t^2)\hat{\mathbf{j}}$.

    • (a) Set $x(t) = 84 \implies 1.5t^2 + 5.0t - 84 = 0 \implies t = 6\,s$. At $t = 6\,s$, y = 1.0(6)^2 = 36.0\,m$.\n * (b) Velocity at t = 6\,s$:$\mathbf{v} = (5.0 + 3.0(6))\hat{\mathbf{i}} + (2.0(6))\hat{\mathbf{j}} = 23.0 \hat{\mathbf{i}} + 12.0 \hat{\mathbf{j}}\,m/s$. Speed$v = \sqrt{23^2 + 12^2} = \sqrt{673} \approx 26\,m\,s^{-1}.\n\n* **Example 3.6**:\n * *Problem*: Prove Galileo's statement that ranges are equal for elevation angles exceeding or falling short of 45^\circ$by equal amounts$\alpha$.

  • Solution:

    • Angles $\theta_1 = 45^\circ + \alpha$ and $\theta_2 = 45^\circ - \alpha$.

    • $2\theta_1 = 90^\circ + 2\alpha \implies \sin(2\theta_1) = \sin(90^\circ + 2\alpha) = \cos(2\alpha)$.

    • $2\theta_2 = 90^\circ - 2\alpha \implies \sin(2\theta_2) = \sin(90^\circ - 2\alpha) = \cos(2\alpha)$.

    • Since $\sin(2\theta_1) = \sin(2\theta_2)$, horizontal ranges $R = \frac{v_0^2 \cos(2\alpha)}{g}$ are equal.

• Example 3.7:

  • Problem: Stone thrown horizontally from cliff height $490\,m$ with initial speed $15\,m\,s^{-1}$. Find time to reach ground and impact speed ($g = 9.8\,m\,s^{-2}$).

  • Solution:

    • Vertical motion: -490 = -\frac{1}{2}(9.8)t^2 \implies 4.9t^2 = 490 \implies t^2 = 100 \implies t = 10\,s$.\n * Velocity components at impact: v_x = 15\,m\,s^{-1}$;$v_y = 0 - (9.8)(10) = -98\,m\,s^{-1}.\n * Speed v = \sqrt{v_x^2 + v_y^2} = \sqrt{15^2 + (-98)^2} = \sqrt{225 + 9604} = \sqrt{9829} \approx 99\,m\,s^{-1}.\n\n* **Example 3.8**:\n * *Problem*: Cricket ball thrown at 28\,m\,s^{-1}$at$30^\circ above horizontal. Calculate (a) maximum height, (b) time of flight, (c) horizontal range.\n * *Solution*:\n * (a) h_m = \frac{(28 \sin(30^\circ))^2}{2(9.8)} = \frac{14^2}{19.6} = 10.0\,m$.

    • (b) T_f = \frac{2(28)\sin(30^\circ)}{9.8} = \frac{28}{9.8} \approx 2.9\,s$.\n * (c) R = \frac{28^2 \sin(60^\circ)}{9.8} = \frac{784 \times 0.866}{9.8} = 80 \times 0.866 \approx 69\,m$.

• Example 3.9:

  • Problem: Insect in circular groove radius $12\,cm$ completes 7 revolutions in $100\,s$. (a) Find angular and linear speed. (b) Is acceleration vector constant? Find its magnitude.

  • Solution:

    • (a) Angular speed \omega = \frac{2\pi \times 7}{100} \approx 0.44\,rad/s$. Linear speed v = \omega R = (0.44)(12) = 5.3\,cm\,s^{-1}.\n * (b) Acceleration vector is NOT constant because its direction changes continuously (always towards center). Magnitude a_c = \omega^2 R = (0.44)^2 (12) \approx 2.3\,cm\,s^{-2}.\n\n# Summary and Key Conceptual Points\n\n* **Path Length vs Displacement Magnitude**:\n * Path length depends on actual path traversed; displacement depends strictly on initial and final positions.\n * Path length equals displacement magnitude ONLY if motion occurs along a straight line without changing direction; in all other cases, path length > displacement magnitude.\n\n* **Average Speed vs Average Velocity Magnitude**:\n * Average speed (path length divided by time) \ge magnitude of average velocity (displacement divided by time).\n\n* **Applicability of Kinematic Equations**:\n * Kinematic equations for uniform acceleration DO NOT apply to uniform circular motion because the direction of centripetal acceleration changes continuously (acceleration vector is not constant).\n\n* **Resultant vs Relative Velocity**:\n * Resultant velocity of an object subjected to two velocity vectors \mathbf{v}1$and$\mathbf{v}_2$:$\mathbf{v} = \mathbf{v}_1 + \mathbf{a}_2\n * Relative velocity of object 1 with respect to object 2: \mathbf{v}{12} = \mathbf{v}_1 - \mathbf{v}_2\n\n* **Determination of Trajectory Shape**:\n * The shape of a particle trajectory is NOT determined by acceleration alone; it depends on both acceleration and initial conditions (initial position \mathbf{r}_0$and initial velocity$\mathbf{v}_0).\n\n# Key Exercises and Conceptual Applications\n\n* **Classification of Physical Quantities**:\n * *Scalars*: Volume, mass, speed, density, number of moles, angular frequency, work, current, pressure, time, power, total path length, energy, gravitational potential, coefficient of friction, charge.\n * *Vectors*: Acceleration, velocity, displacement, angular velocity, force, angular momentum, linear momentum, electric field, magnetic moment, relative velocity, impulse.\n\n* **Vector Triangle Inequalities**:\n * |\mathbf{a} + \mathbf{b}| \le |\mathbf{a}| + |\mathbf{b}|$(Equality holds when$\mathbf{a}$and$\mathbf{b} are collinear in the same direction).\n * |\mathbf{a} + \mathbf{b}| \ge ||\mathbf{a}| - |\mathbf{b}||$(Equality holds when$\mathbf{a}$and$\mathbf{b} are collinear in opposite directions).\n * |\mathbf{a} - \mathbf{b}| \le |\mathbf{a}| + |\mathbf{b}|$(Equality holds when$\mathbf{a}$and$\mathbf{b} are collinear in opposite directions).\n * |\mathbf{a} - \mathbf{b}| \ge ||\mathbf{a}| - |\mathbf{b}||$(Equality holds when$\mathbf{a}$and$\mathbf{b} are collinear in the same direction).\n\n* **Properties of Four Vectors Summing to Zero (\mathbf{a} + \mathbf{b} + \mathbf{c} + \mathbf{d} = \mathbf{0})**:\n * Magnitude of (\mathbf{a} + \mathbf{c})$equals magnitude of$(\mathbf{b} + \mathbf{d}).\n * Magnitude of \mathbf{a}$cannot exceed the sum of magnitudes of$\mathbf{b}$,$\mathbf{c}$, and$\mathbf{d}.\n * \mathbf{b} + \mathbf{c}$must lie in the plane of$\mathbf{a}$and$\mathbf{d}$if$\mathbf{a}$and$\mathbf{d}$$ are non-collinear.