Comprehensive Study Guide: Motion in a Plane

Introduction to Motion in Two Dimensions

Motion in one dimension requires only a straight line, where directional aspects are fully specified using positive ( $+\mathbf{}$ ) and negative ( $-\mathbf{}$ ) signs.
Motion in two dimensions (in a plane) or three dimensions (in space) requires a formal mathematical language to express directional physical quantities, which is provided by vector algebra.
Vectors are used to define kinematic quantities such as position, displacement, velocity, and acceleration in two and three dimensions.
Key special applications of two-dimensional kinematics include:
- Motion with constant acceleration.
- Projectile motion.
- Uniform circular motion.
The kinematic equations formulated for two-dimensional motion can be readily extended to three dimensions.

Scalars and Vectors

Physical quantities in physics are broadly classified into two categories: scalars and vectors.
Scalar Quantities:
- A scalar quantity is a physical quantity that possesses magnitude only and has no direction associated with it.
- It is completely specified by a single real number along with an appropriate unit of measurement.
- Examples of scalar quantities include:
- Distance between two points.
- Mass of an object.
- Temperature of a body.
- Time of an event.
- Volume and density of a substance.
- Number of moles.
- Angular frequency.
- Algebraic Rules for Scalars:
- Scalars obey the ordinary rules of algebra for addition, subtraction, multiplication, and division.
- Addition and subtraction of scalar quantities are meaningful only if the quantities have identical units.
- Multiplication and division can be performed between scalars possessing different units.
- Examples of Scalar Calculations:
- For a rectangle of length $1.0\,m$ and breadth $0.5\,m$ , the perimeter is the scalar sum $1.0\,m + 0.5\,m + 1.0\,m + 0.5\,m = 3.0\,m$ .
- If maximum and minimum daily temperatures are $35.6\,^{\circ}\text{C}$ and $24.2\,^{\circ}\text{C}$ , the temperature difference is $35.6\,^{\circ}\text{C} - 24.2\,^{\circ}\text{C} = 11.4\,^{\circ}\text{C}$ .
- A solid aluminium cube of side $10\,cm$ ( $0.1\,m$ ) with mass $2.7\,kg$ has a volume of $(0.1\,m)^3 = 10^{-3}\,m^3$ (a scalar) and a mass density of $\frac{2.7\,kg}{10^{-3}\,m^3} = 2.7 \times 10^3\,kg\,m^{-3}$ (a scalar).
Vector Quantities:
- A vector quantity is a physical quantity that possesses both a magnitude and a direction, and strictly obeys the vector addition laws (the triangle law or the parallelogram law).
- Examples of vector quantities include displacement, velocity, acceleration, force, linear momentum, angular momentum, and electric field.
- Vector Representation and Notation:
- In printed text, vectors are represented by boldface letters (e.g., $\mathbf{v}$ , $\mathbf{A}$ , $\mathbf{r}$ ).
- In handwritten form, vectors are depicted with an arrow placed above the letter symbol (e.g., $\vec{v}$ , $\vec{A}$ , $\vec{r}$ ).
- The magnitude (or absolute value) of a vector $\mathbf{v}$ is denoted by $|\mathbf{v}|$ or simply by the lightface character $v$ . Magnitude is always a non-negative scalar quantity.

Position and Displacement Vectors

Position and displacement vectors

To specify the position of an object in a plane, a convenient reference point is selected as the origin $\text{O}$ .
Position Vector:
- If an object is located at point $\text{P}$ at time $t$ , the straight line directed from $\text{O}$ to $\text{P}$ forms the position vector $\mathbf{r} = \mathbf{OP}$ .
- The length of the arrow representing $\mathbf{r}$ corresponds to the magnitude of the vector, and the arrowhead points in the direction of $\text{P}$ relative to $\text{O}$ .
- If the object moves to point $\text{P}'$ at time $t'$ , its new position vector is $\mathbf{r}' = \mathbf{OP}'$ .
Displacement Vector:
- The vector $\mathbf{PP}'$ directed from initial position $\text{P}$ (tail) to final position $\text{P}'$ (head) is defined as the displacement vector $\Delta \mathbf{r}$ .
- The displacement vector is independent of the actual path traversed by the object between $\text{P}$ and $\text{P}'$ . Whether an object travels along path $\text{PABCQ}$ , $\text{PDQ}$ , or $\text{PBEFQ}$ , the displacement vector between initial position $\text{P}$ and final position $\text{Q}$ remains identically $\mathbf{PQ}$ .
- The magnitude of the displacement vector is always less than or equal to the total path length traversed between two points.

Equality and Classification of Vectors

Equal Vectors:
- Two vectors $\mathbf{A}$ and $\mathbf{B}$ are defined as equal ( $\mathbf{A} = \mathbf{B}$ ) if and only if they possess identical magnitudes ( $|\mathbf{A}| = |\mathbf{B}|$ ) and identical directions.
- If vector $\mathbf{B}$ is shifted parallel to itself until its tail coincides with the tail of $\mathbf{A}$ , their heads will also coincide completely.
- Two vectors having equal length but different directions are not equal ( $\mathbf{A}' \neq \mathbf{B}'$ ).
Free Vectors vs. Localized Vectors:
- Free Vectors: Vectors that have no fixed location in space. Translating a free vector parallel to itself leaves the vector completely unchanged.
- Localized Vectors: Vectors whose physical effects depend on their specific point of application or line of action.

Multiplication of Vectors by Real Numbers

Multiplication by a Positive Scalar ( $\lambda > 0$ ):
- Multiplying vector $\mathbf{A}$ by a positive real number $\lambda$ produces a vector $\lambda \mathbf{A}$ whose magnitude is scaled by $\lambda$ :

$|\lambda \mathbf{A}| = \lambda |\mathbf{A}|$

The direction of $\lambda \mathbf{A}$ remains identical to that of $\mathbf{A}$ . For instance, $2\mathbf{A} $points in the same direction as$ \mathbf{A} with twice its length.\n\n- **Multiplication by a Negative Scalar (-\lambda < 0):**\n - Multiplying vector \mathbf{A} $by a negative real number$ -\lambda $yields a vector whose magnitude is$ \lambda |\mathbf{A}| $and whose direction is exactly opposite ($ 180^\circ $rotated) to$ \mathbf{A}.\n - Examples include -\mathbf{A} $(same magnitude, opposite direction) and$ -1.5\mathbf{A} $($ 1.5 times the magnitude, opposite direction).\n\n- **Physical Dimensions of Scaled Vectors:**\n - If the multiplier \lambda $is a scalar quantity with physical dimensions, the dimension of the resulting vector$ \lambda \mathbf{A} $is the product of the dimensions of$ \lambda $and$ \mathbf{A}.\n - For example, multiplying a constant velocity vector \mathbf{v} $($ \text{L}\text{T}^{-1} $) by time duration$ t $($ \text{T} $) yields a displacement vector$ \mathbf{r} $($ \text{L}).\n\n# Graphical Addition and Subtraction of Vectors\n\n- **Triangle Method (Head-to-Tail Method):**\n - To sum two vectors \mathbf{A} $and$ \mathbf{B} graphically:\n 1. Draw vector \mathbf{A} to scale.\n 2. Place the tail of vector \mathbf{B} $at the head (tip) of vector$ \mathbf{A}.\n 3. Join the tail of \mathbf{A} $to the head of$ \mathbf{B} $with a line segment. This directed segment represents the resultant vector$ \mathbf{R} = \mathbf{A} + \mathbf{B}.\n - **Commutative Law:** Vector addition is commutative:\n\n\mathbf{A} + \mathbf{B} = \mathbf{B} + \mathbf{A}\n\n - **Associative Law:** Vector addition obeys the associative property:\n\n(\mathbf{A} + \mathbf{B}) + \mathbf{C} = \mathbf{A} + (\mathbf{B} + \mathbf{C})\n\n- **Null Vector (Zero Vector):**\n - The sum of two equal and opposite vectors \mathbf{A} $and$ -\mathbf{A} $produces a resultant with zero magnitude, termed the null vector or zero vector ($ \mathbf{0}):\n\n\mathbf{A} - \mathbf{A} = \mathbf{0}\n\n|\mathbf{0}| = 0\n\n - Since the magnitude of \mathbf{0} is zero, its direction is arbitrary or undefined.\n - Multiplying any vector by the scalar number 0 $, or multiplying the null vector by any scalar$ \lambda, results in a null vector:\n\n\mathbf{A} + \mathbf{0} = \mathbf{A}\n\n\lambda \mathbf{0} = \mathbf{0}\n\n0 \mathbf{A} = \mathbf{0}\n\n - **Physical Significance:** If an object moves from position \text{P} $to$ \text{P}' $and returns to$ \text{P}, its initial and final positions coincide, making its net displacement vector a null vector.\n\n- **Vector Subtraction:**\n - Subtraction of vector \mathbf{B} $from vector$ \mathbf{A} $is defined as the addition of vector$ -\mathbf{B} $to$ \mathbf{A}:\n\n\mathbf{A} - \mathbf{B} = \mathbf{A} + (-\mathbf{B})\n\n- **Parallelogram Method:**\n - To add two vectors \mathbf{A} $and$ \mathbf{B} using the parallelogram law:\n 1. Bring the tails of \mathbf{A} $and$ \mathbf{B} $to a common origin$ \text{O}.\n 2. Construct parallel lines passing through the head of each vector to complete a parallelogram.\n 3. The diagonal line drawn from the common origin \text{O} $to the opposite intersection corner represents the resultant vector$ \mathbf{R} = \mathbf{A} + \mathbf{B}.\n - The parallelogram method is geometrically equivalent to the triangle method.\n\n# Resolution of Vectors and Unit Vectors\n\n- **Resolution in a Plane:**\n - Any vector \mathbf{A} $lying in a plane can be uniquely resolved into two component vectors along any two non-zero, non-collinear vectors$ \mathbf{a} $and$ \mathbf{b} in the same plane:\n\n\mathbf{A} = \lambda \mathbf{a} + \mu \mathbf{b}\n\n where \lambda $and$ \mu are real numbers.\n\n- **Unit Vectors:**\n - A unit vector is a vector having a magnitude of exactly 1, directed along a specific direction.\n - Unit vectors are dimensionless and carry no units; their sole function is to specify direction.\n - Unit vectors along the rectangular Cartesian coordinate axes x $,$ y $, and$ z $are designated as$ \hat{\mathbf{i}} $,$ \hat{\mathbf{j}} $, and$ \hat{\mathbf{k}}, respectively:\n\n|\hat{\mathbf{i}}| = |\hat{\mathbf{j}}| = |\hat{\mathbf{k}}| = 1\n\n - Unit vectors \hat{\mathbf{i}} $,$ \hat{\mathbf{j}} $, and$ \hat{\mathbf{k}} are mutually perpendicular (orthogonal).\n - Any vector \mathbf{A} $can be written as the product of its magnitude$ |\mathbf{A}| $and a unit vector$ \hat{\mathbf{n}} $pointing along$ \mathbf{A}:\n\n\mathbf{A} = |\mathbf{A}|\hat{\mathbf{n}}\n\n- **2D Rectangular Components:**\n - A vector \mathbf{A} $in the$ x-y $plane can be resolved into orthogonal components along$ \hat{\mathbf{i}} $and$ \hat{\mathbf{j}}:\n\n\mathbf{A} = A_x \hat{\mathbf{i}} + A_y \hat{\mathbf{j}}\n\n - A_x $and$ A_y $are scalar real numbers representing the$ x $- and$ y $-components of$ \mathbf{A} $.$ A_x \hat{\mathbf{i}} $and$ A_y \hat{\mathbf{j}} are vector components.\n - Trigonometric relationships with the angle \theta $formed by$ \mathbf{A} $relative to the positive$ x-axis:\n\nA_x = A \cos(\theta)\n\nA_y = A \sin(\theta)\n\n - Conversion from rectangular components A_x, A_y $to magnitude$ A $and direction angle$ \theta:\n\nA = \sqrt{A_x^2 + A_y^2}\n\n\tan(\theta) = \frac{A_y}{A_x} \implies \theta = \tan^{-1}\left(\frac{A_y}{A_x}\right)\n\n- **3D Rectangular Components:**\n - A vector \mathbf{A} in three dimensions is represented as:\n\n\mathbf{A} = A_x \hat{\mathbf{i}} + A_y \hat{\mathbf{j}} + A_z \hat{\mathbf{k}}\n\n - If \alpha $,$ \beta $, and$ \gamma $are the spatial angles between$ \mathbf{A} $and the$ x $-,$ y $-, and$ z-axes, respectively:\n\nA_x = A \cos(\alpha)\n\nA_y = A \cos(\beta)\n\nA_z = A \cos(\gamma)\n\n - The magnitude of \mathbf{A} in 3D is:\n\nA = \sqrt{A_x^2 + A_y^2 + A_z^2}\n\n - A general position vector \mathbf{r} in 3D is expressed as:\n\n\mathbf{r} = x \hat{\mathbf{i}} + y \hat{\mathbf{j}} + z \hat{\mathbf{k}}\n\n# Analytical Method of Vector Addition\n\n- Adding vectors analytically via components is more precise and efficient than graphical methods.\n- **Addition in Two Dimensions:**\n - For two vectors \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}} $, the resultant$ \mathbf{R} = \mathbf{A} + \mathbf{B} is:\n\n\mathbf{R} = (A_x + B_x)\hat{\mathbf{i}} + (A_y + B_y)\hat{\mathbf{j}}\n\nR_x = A_x + B_x\n\nR_y = A_y + B_y\n\n- **Addition in Three Dimensions:**\n - For \mathbf{A} = A_x \hat{\mathbf{i}} + A_y \hat{\mathbf{j}} + A_z \hat{\mathbf{k}} $and$ \mathbf{B} = B_x \hat{\mathbf{i}} + B_y \hat{\mathbf{j}} + B_z \hat{\mathbf{k}} $, the components of$ \mathbf{R} are:\n\nR_x = A_x + B_x\n\nR_y = A_y + B_y\n\nR_z = A_z + B_z\n\n- **Multiple Vector Combinations:**\n - For vectors \mathbf{a}, \mathbf{b}, \mathbf{c} $, a combined vector$ \mathbf{T} = \mathbf{a} + \mathbf{b} - \mathbf{c} has scalar components:\n\nT_x = a_x + b_x - c_x\n\nT_y = a_y + b_y - c_y\n\nT_z = a_z + b_z - c_z\n\n- **Law of Cosines and Law of Sines:**\n - For two vectors \mathbf{A} $and$ \mathbf{B} $separated by an angle$ \theta:\n - **Law of Cosines (Resultant Magnitude):**\n\nR = \sqrt{A^2 + B^2 + 2AB \cos(\theta)}\n\n - **Law of Sines:**\n\n\frac{R}{\sin(\theta)} = \frac{A}{\sin(\beta)} = \frac{B}{\sin(\alpha)}\n\n - **Direction Angle \alpha $relative to$ \mathbf{A}:**\n\n\tan(\alpha) = \frac{B \sin(\theta)}{A + B \cos(\theta)}\n\n\sin(\alpha) = \frac{B}{R}\sin(\theta)\n\n# Motion in a Plane: Kinematic Vectors\n\n- **Position Vector and Displacement:**\n - The position vector \mathbf{r} $of a particle$ \text{P} $in the$ x-y plane is:\n\n\mathbf{r} = x \hat{\mathbf{i}} + y \hat{\mathbf{j}}\n\n - If the particle moves from \text{P} $at time$ t $to$ \text{P}' $at time$ t' $, the displacement vector$ \Delta \mathbf{r} is:\n\n\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}}\n\n- **Average and Instantaneous Velocity:**\n - **Average Velocity (\bar{\mathbf{v}} $):** The ratio of displacement$ \Delta \mathbf{r} $to time interval$ \Delta t:\n\n\bar{\mathbf{v}} = \frac{\Delta \mathbf{r}}{\Delta t} = \frac{\Delta x}{\Delta t}\hat{\mathbf{i}} + \frac{\Delta y}{\Delta t}\hat{\mathbf{j}} = \bar{v}x \hat{\mathbf{i}} + \bar{v}_y \hat{\mathbf{j}}\n\n - The direction of \bar{\mathbf{v}} $is identical to the direction of$ \Delta \mathbf{r}.\n - **Instantaneous Velocity (\mathbf{v} $):** The limiting value of average velocity as$ \Delta t \to 0:\n\n\mathbf{v} = \lim{\Delta t \to 0} \frac{\Delta \mathbf{r}}{\Delta t} = \frac{\mathrm{d}\mathbf{r}}{\mathrm{d}t}\n\n\mathbf{v} = \left(\lim_{\Delta t \to 0}\frac{\Delta x}{\Delta t}\right)\hat{\mathbf{i}} + \left(\lim_{\Delta t \to 0}\frac{\Delta y}{\Delta t}\right)\hat{\mathbf{j}} = \frac{\mathrm{d}x}{\mathrm{d}t}\hat{\mathbf{i}} + \frac{\mathrm{d}y}{\mathrm{d}t}\hat{\mathbf{j}} = v_x \hat{\mathbf{i}} + v_y \hat{\mathbf{j}}\n\n - **Direction of Velocity:** As \Delta t \to 0 $,$ \Delta \mathbf{r} aligns with the tangent to the path. Therefore, instantaneous velocity at any point along a path is directed tangentially to the trajectory at that point.\n - **Magnitude and Direction:**\n\nv = \sqrt{v_x^2 + v_y^2}\n\n\tan(\theta) = \frac{v_y}{v_x} \implies \theta = \tan^{-1}\left(\frac{v_y}{v_x}\right)\n\nv_x = v \cos(\theta)\n\nv_y = v \sin(\theta)\n\n- **Average and Instantaneous Acceleration:**\n - **Average Acceleration (\bar{\mathbf{a}} $):** The change in velocity$ \Delta \mathbf{v} $divided by time interval$ \Delta t:\n\n\bar{\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}} = a_x \hat{\mathbf{i}} + a_y \hat{\mathbf{j}}\n\n - **Instantaneous Acceleration (\mathbf{a} $):** The limiting value of average acceleration as$ \Delta t \to 0:\n\n\mathbf{a} = \lim_{\Delta t \to 0} \frac{\Delta \mathbf{v}}{\Delta t} = \frac{\mathrm{d}\mathbf{v}}{\mathrm{d}t} = a_x \hat{\mathbf{i}} + a_y \hat{\mathbf{j}}\n\n - Component derivative formulas:\n\na_x = \frac{\mathrm{d}v_x}{\mathrm{d}t} = \frac{\mathrm{d}^2 x}{\mathrm{d}t^2}\n\na_y = \frac{\mathrm{d}v_y}{\mathrm{d}t} = \frac{\mathrm{d}^2 y}{\mathrm{d}t^2}\n\n - In 1D motion, velocity and acceleration vectors are strictly collinear (0^\circ $or$ 180^\circ $angle). In 2D or 3D motion, velocity and acceleration vectors can form any angle between$ 0^\circ $and$ 180^\circ.\n\n# Motion in a Plane with Constant Acceleration\n\n- If an object moves in the x-y $plane with constant acceleration$ \mathbf{a} $, its average acceleration equals$ \mathbf{a} over any time interval.\n- Let initial velocity at t = 0 $be$ \mathbf{v}0 $and velocity at time$ t $be$ \mathbf{v}. Then:\n\n\mathbf{a} = \frac{\mathbf{v} - \mathbf{v}_0}{t} \implies \mathbf{v} = \mathbf{v}_0 + \mathbf{a}t\n\n - Component equations:\n\nv_x = v{0x} + a_x t\n\nv_y = v_{0y} + a_y t\n\n- **Position Equations:**\n - Average velocity over interval t $is$ \frac{\mathbf{v}0 + \mathbf{v}}{2} $. Displacement$ \mathbf{r} - \mathbf{r}_0 is:\n\n\mathbf{r} - \mathbf{r}_0 = \left(\frac{\mathbf{v}_0 + \mathbf{v}}{2}\right)t = \left(\frac{\mathbf{v}_0 + \mathbf{v}_0 + \mathbf{a}t}{2}\right)t = \mathbf{v}_0 t + \frac{1}{2}\mathbf{a}t^2\n\n\mathbf{r} = \mathbf{r}_0 + \mathbf{v}_0 t + \frac{1}{2}\mathbf{a}t^2\n\n - Component equations:\n\nx = x_0 + v{0x}t + \frac{1}{2}a_x t^2\n\ny = y_0 + v_{0y}t + \frac{1}{2}a_y t^2\n\n- **Independence of Perpendicular Motions:** Two-dimensional motion under constant acceleration can be decoupled and analyzed as two independent, simultaneous one-dimensional motions along orthogonal directions.\n\n# Projectile Motion\n\n- **Definition:** A projectile is any object in flight after being launched or thrown into space (e.g., a thrown cricket ball, baseball, or kicked football).\n- **Galileo's Principle of Independence:** In 1632 (*Dialogue on the Great World Systems*), Galileo Galilei stated that projectile motion consists of two independent, simultaneous motion components:\n 1. Horizontal component at constant velocity (zero acceleration).\n 2. Vertical component at constant acceleration due to gravity.\n- **Assumptions:** Air resistance effects are negligible.\n\n- **Kinematic Equations of Projectile Motion:**\n - A projectile is launched from origin \mathbf{r}0 = (0,0) $at time$ t = 0 $with initial speed$ v_0 $at projection angle$ \theta_0 $relative to the horizontal$ x-axis.\n - Acceleration components:\n\na_x = 0\n\na_y = -g\n\n - Initial velocity components:\n\nv{0x} = v_0 \cos(\theta_0)\n\nv_{0y} = v_0 \sin(\theta_0)\n\n - Velocity components at time t:\n\nv_x = v_{0x} = v_0 \cos(\theta_0) \quad \text{(constant throughout flight)}\n\nv_y = v_0 \sin(\theta_0) - gt\n\n - Position coordinates at time t:\n\nx = (v_0 \cos(\theta_0)) t\n\ny = (v_0 \sin(\theta_0)) t - \frac{1}{2}gt^2\n\n- **Equation of the Path (Trajectory):**\n - Substituting t = \frac{x}{v_0 \cos(\theta_0)} $into the$ y-position equation:\n\ny = \tan(\theta_0)x - \left(\frac{g}{2 v_0^2 \cos^2(\theta_0)}\right)x^2\n\n - This equation is of the form y = ax + bx^2 $(where$ a = \tan(\theta_0) $and$ b = -\frac{g}{2 v_0^2 \cos^2(\theta_0)} are constants), which defines a **parabola**. Thus, the trajectory of a projectile is parabolic.\n\n- **Key Parameters of Projectile Motion:**\n - **Time of Maximum Height (t_m):**\n - At peak height h_m $, vertical velocity$ v_y = 0:\n\nv_y = v_0 \sin(\theta_0) - g t_m = 0 \implies t_m = \frac{v_0 \sin(\theta_0)}{g}\n\n - **Time of Flight (T_f):**\n - Total duration in air until the projectile lands (y = 0):\n\nT_f = \frac{2 v_0 \sin(\theta_0)}{g} = 2 t_m\n\n - **Maximum Height (h_m):**\n - Substituting t = t_m $into$ y(t):\n\nh_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\n\nh_m = \frac{(v_0 \sin(\theta_0))^2}{2g}\n\n - **Horizontal Range (R):**\n - Total horizontal distance covered during time of flight T_f:\n\nR = v_x T_f = (v_0 \cos(\theta_0)) \left(\frac{2 v_0 \sin(\theta_0)}{g}\right) = \frac{v_0^2 (2 \sin(\theta_0) \cos(\theta_0))}{g}\n\nR = \frac{v_0^2 \sin(2\theta_0)}{g}\n\n - **Maximum Horizontal Range (R_m):**\n - Range R $is maximized when$ \sin(2\theta_0) = 1 $, which occurs at launch angle$ \theta_0 = 45^\circ:\n\nR_m = \frac{v_0^2}{g}\n\n# Uniform Circular Motion\n\n- **Definition:** When an object moves along a circular path of radius R $at a constant speed$ v, its motion is called uniform circular motion.\n- **Centripetal Acceleration (\mathbf{a}c):**\n - Although speed v $is constant, velocity vector$ \mathbf{v} changes continuously in direction. Thus, an object in uniform circular motion is constantly accelerating.\n - Direction: Directed radially inward toward the center of the circular trajectory ("center-seeking"). First thoroughly published in 1673 by Christiaan Huygens.\n - Because its direction changes continuously as the object moves, centripetal acceleration is **not** a constant vector, even though its magnitude is constant.\n - Derivation via similar triangles:\n - Position vector triangle \Delta \mathbf{r} $and velocity vector triangle$ \Delta \mathbf{v} are similar:\n\n\frac{|\Delta \mathbf{v}|}{v} = \frac{|\Delta \mathbf{r}|}{R} \implies |\Delta \mathbf{v}| = \frac{v}{R}|\Delta \mathbf{r}|\n\n - Dividing by \Delta t $and evaluating limit$ \Delta t \to 0:\n\na_c = \lim{\Delta t \to 0}\frac{|\Delta \mathbf{v}|}{\Delta t} = \frac{v}{R} \left(\lim_{\Delta t \to 0}\frac{|\Delta \mathbf{r}|}{\Delta t}\right) = \frac{v}{R}(v)\n\na_c = \frac{v^2}{R}\n\n- **Angular Speed (\omega):**\n - Rate of change of angular displacement \Delta \theta with time:\n\n\omega = \frac{\Delta \theta}{\Delta t}\n\n - Relation between linear arc distance \Delta s = R \Delta \theta $and linear speed$ v:\n\nv = \frac{\Delta s}{\Delta t} = R \frac{\Delta \theta}{\Delta t} = R \omega\n\n - Centripetal acceleration in terms of \omega:\n\na_c = \frac{(R \omega)^2}{R} = \omega^2 R\n\n- **Period (T $) and Frequency ($ u):**\n - Time period T $is the duration for one complete revolution. Frequency$ u = \frac{1}{T} is the number of revolutions per unit time.\n - Linear speed v $over distance$ 2\pi R $in time$ T:\n\nv = \frac{2\pi R}{T} = 2\pi R u\n\n\omega = \frac{2\pi}{T} = 2\pi u\n\n - Centripetal acceleration in terms of u:\n\na_c = 4 \pi^2 u^2 R\n\n# Fully Transcribed Step-by-Step Examples\n\n- **Example 3.1:**\n - *Problem:* Rain falls vertically with speed 35\,m\,s^{-1} $. Wind blows from east to west at$ 12\,m\,s^{-1}. Find the direction to hold an umbrella.\n - *Solution:* Rain velocity \mathbf{v}_r = 35\,m\,s^{-1} $downward, wind velocity$ \mathbf{v}_w = 12\,m\,s^{-1} westward.\n - Resultant velocity magnitude:\n\nR = \sqrt{v_r^2 + v_w^2} = \sqrt{35^2 + 12^2} = \sqrt{1225 + 144} = \sqrt{1369} = 37\,m\,s^{-1}\n\n - Angle \theta with vertical:\n\n\tan(\theta) = \frac{v_w}{v_r} = \frac{12}{35} \approx 0.343\n\n\theta = \tan^{-1}(0.343) \approx 19^\circ\n\n - The boy should hold the umbrella at an angle of about 19^\circ with the vertical towards the east.\n\n- **Example 3.2:**\n - *Problem:* Find the magnitude and direction of the resultant of two vectors \mathbf{A} $and$ \mathbf{B} $in terms of their magnitudes and angle$ \theta between them.\n - *Solution:* Geometry of parallelogram \text{OS} $with$ \mathbf{OS} = \mathbf{R} = \mathbf{A} + \mathbf{B} $.$ \text{SN} $perpendicular to$ \text{OP}.\n - \text{ON} = \text{OP} + \text{PN} = A + B \cos(\theta)\n - \text{SN} = B \sin(\theta)\n - \text{OS}^2 = \text{ON}^2 + \text{SN}^2 = (A + B \cos(\theta))^2 + (B \sin(\theta))^2\n - R^2 = A^2 + 2AB \cos(\theta) + B^2 \cos^2(\theta) + B^2 \sin^2(\theta) = A^2 + B^2 + 2AB \cos(\theta)\n\nR = \sqrt{A^2 + B^2 + 2AB \cos(\theta)} \quad \text{(Law of Cosines)}\n\n - Direction angle \alpha $relative to$ \mathbf{A}:\n\n\tan(\alpha) = \frac{\text{SN}}{\text{ON}} = \frac{B \sin(\theta)}{A + B \cos(\theta)}\n\n - Law of sines derivation:\n\n\frac{R}{\sin(\theta)} = \frac{A}{\sin(\beta)} = \frac{B}{\sin(\alpha)}\n\n- **Example 3.3:**\n - *Problem:* Motorboat speed 25\,km/h $north. Water current$ 10\,km/h $at$ 60^\circ east of south. Find resultant boat velocity.\n - *Solution:* Boat velocity \mathbf{v}_b = 25\,km/h $north. Current$ \mathbf{v}_c = 10\,km/h $at$ 60^\circ $east of south. Angle between$ \mathbf{v}_b $and$ \mathbf{v}_c $is$ \theta = 180^\circ - 60^\circ = 120^\circ.\n - Magnitude of resultant R:\n\nR = \sqrt{v_b^2 + v_c^2 + 2v_b v_c \cos(120^\circ)} = \sqrt{25^2 + 10^2 + 2(25)(10)(-0.5)}\n\nR = \sqrt{625 + 100 - 250} = \sqrt{475} \approx 21.8\,km/h \approx 22\,km/h\n\n - Angle \phi with north using Law of Sines:\n\n\frac{R}{\sin(120^\circ)} = \frac{v_c}{\sin(\phi)} \implies \sin(\phi) = \frac{v_c \sin(120^\circ)}{R} = \frac{10 \times 0.866}{21.8} \approx 0.397\n\n\phi = \sin^{-1}(0.397) \approx 23.4^\circ \quad \text{east of north}\n\n- **Example 3.4:**\n - *Problem:* Position \mathbf{r} = 3.0t \hat{\mathbf{i}} + 2.0t^2 \hat{\mathbf{j}} + 5.0 \hat{\mathbf{k}}.\n - (a) Find \mathbf{v}(t) $and$ \mathbf{a}(t).\n - (b) Find magnitude and direction of \mathbf{v}(t) $at$ t = 1.0\,s$.
Solution:
- (a) $\mathbf{v}(t) = \frac{\mathrm{d}\mathbf{r}}{\mathrm{d}t} = 3.0 \hat{\mathbf{i}} + 4.0t \hat{\mathbf{j}}$
  - $\mathbf{a}(t) = \frac{\mathrm{d}\mathbf{v}}{\mathrm{d}t} = 4.0 \hat{\mathbf{j}}\,m\,s^{-2}$ (along positive $y$ -direction).
- (b) At $t = 1.0\,s$ , $\mathbf{v} = 3.0 \hat{\mathbf{i}} + 4.0 \hat{\mathbf{j}}$ .
  - Speed $v = \sqrt{3.0^2 + 4.0^2} = 5.0\,m\,s^{-1}$ .
  - Direction $\theta = \tan^{-1}\left(\frac{4.0}{3.0}\right) \approx 53^\circ$ with positive $x$ -axis.
- Example 3.5:
Problem: Particle starts from origin at $t = 0$ with $\mathbf{v}_0 = 5.0 \hat{\mathbf{i}}\,m/s$ , constant acceleration $\mathbf{a} = (3.0 \hat{\mathbf{i}} + 2.0 \hat{\mathbf{j}})\,m/s^2$ .\n - (a) What is y$-coordinate when $x = 84\,m$?\n - (b) What is particle speed at this time?\n - *Solution:*\n - Position \mathbf{r}(t) = \mathbf{v}_0 t + \frac{1}{2}\mathbf{a}t^2 = (5.0t + 1.5t^2)\hat{\mathbf{i}} + (1.0t^2)\hat{\mathbf{j}}.\n - (a) Set x(t) = 84\,m:\n\n1.5t^2 + 5.0t - 84 = 0 \implies t = 6\,s\n\n - y(6) = 1.0(6)^2 = 36.0\,m.\n - (b) Velocity \mathbf{v}(t) = (5.0 + 3.0t)\hat{\mathbf{i}} + (2.0t)\hat{\mathbf{j}}$.
- At $t = 6\,s$ , $\mathbf{v} = (5.0 + 3.0(6))\hat{\mathbf{i}} + (2.0(6))\hat{\mathbf{j}} = 23.0 \hat{\mathbf{i}} + 12.0 \hat{\mathbf{j}}$ .
- Speed $v = \sqrt{23^2 + 12^2} = \sqrt{529 + 144} = \sqrt{673} \approx 26\,m\,s^{-1}$ .
  - Example 3.6:
Problem: Prove Galileo's statement that ranges are equal for elevation angles exceeding or falling short of $45^\circ$ by equal amounts \alpha$.\n - *Solution:* Launch angles \theta_1 = 45^\circ + \alpha $and$ \theta_2 = 45^\circ - \alpha$.
- Double angles: $2\theta_1 = 90^\circ + 2\alpha$ and 2\theta_2 = 90^\circ - 2\alpha$.\n - Sine values: \sin(90^\circ + 2\alpha) = \cos(2\alpha) $and$ \sin(90^\circ - 2\alpha) = \cos(2\alpha).\n - Since \sin(2\theta_1) = \sin(2\theta_2) $, horizontal ranges$ R = \frac{v_0^2 \sin(2\theta_0)}{g} are equal.\n\n- **Example 3.7:**\n - *Problem:* Cliff height 490\,m $, horizontal initial stone velocity$ 15\,m\,s^{-1} $,$ g = 9.8\,m\,s^{-2}. Find time to reach ground and impact speed.\n - *Solution:* Origin at cliff edge, y-axis positive upward.\n - Vertical position y(t) = -\frac{1}{2}gt^2.\n - At ground, y = -490\,m:\n\n-490 = -\frac{1}{2}(9.8)t^2 \implies t^2 = 100 \implies t = 10\,s\n\n - Velocity components at impact:\n - v_x = v_{0x} = 15\,m\,s^{-1}\n - v_y = -gt = -9.8 \times 10 = -98\,m\,s^{-1}\n - Impact speed:\n\nv = \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 $above horizontal. Calculate maximum height, time of flight, and horizontal range ($ g = 9.8\,m\,s^{-2}).\n - *Solution:*\n - (a) Max height:\n\nh_m = \frac{(v_0 \sin(\theta_0))^2}{2g} = \frac{(28 \sin(30^\circ))^2}{2(9.8)} = \frac{14^2}{19.6} = \frac{196}{19.6} = 10.0\,m\n\n - (b) Time of flight:\n\nT_f = \frac{2 v_0 \sin(\theta_0)}{g} = \frac{2(28)\sin(30^\circ)}{9.8} = \frac{28}{9.8} \approx 2.9\,s\n\n - (c) Range:\n\nR = \frac{v_0^2 \sin(2\theta_0)}{g} = \frac{28^2 \sin(60^\circ)}{9.8} = \frac{784 \times 0.866}{9.8} \approx 69\,m\n\n- **Example 3.9:**\n - *Problem:* Insect in circular groove radius 12\,cm $, completes$ 7 $revolutions in$ 100\,s$.
- (a) Find angular speed $\omega$ and linear speed v$.\n - (b) Is acceleration vector constant? Find its magnitude.\n - *Solution:*\n - (a) Angular speed \omega:\n\n\omega = \frac{2\pi u}{1} = \frac{2\pi \times 7}{100} \approx 0.44\,\text{rad/s}\n\n - Linear speed v:\n\nv = \omega R = 0.44\,\text{s}^{-1} \times 12\,cm = 5.3\,cm\,s^{-1}\n\n - (b) Acceleration vector points toward center; its direction changes continuously, so it is **not** a constant vector.\n - Acceleration magnitude:\n\na_c = \omega^2 R = (0.44)^2 \times 12 \approx 2.3\,cm\,s^{-2}\n\n# Points to Ponder\n\n- Path length traversed by an object between two points is in general not equal to the magnitude of displacement. They are equal only if motion occurs along a straight line without changing direction.\n- Average speed is always greater than or equal to the magnitude of average velocity over any given time interval.\n- Vector kinematic equations are independent of coordinate axis choices; resolving into components is merely a convenient analytical tool.\n- Kinematic equations for uniform acceleration cannot be applied to uniform circular motion because the direction of acceleration changes continuously.\n- An object subjected to two velocities \mathbf{v}1 $and$ \mathbf{v}_2 $has a resultant velocity$ \mathbf{v} = \mathbf{v}_1 + \mathbf{v}_2 $. This is distinct from relative velocity$ \mathbf{v}{12} = \mathbf{v}_1 - \mathbf{v}_2.\n- Acceleration in circular motion points strictly to the center only if speed v$$ is constant. If speed varies, there is also a tangential acceleration component.
- Trajectory shape is determined jointly by acceleration and initial conditions (initial position and velocity). Under gravity, motion can be a vertical straight line or a parabola depending on initial velocity direction.