Chapter 4 — Motion In A Plane

Fig 1.1 — Educational Diagram: Projectile Motion Trajectory & Kinematic Equations

Introduction (Motion in a plane)
Motion in a plane (two‑dimensional motion) describes the change of position of a particle on a flat surface. The particle's position is specified by a position vector r = xi + yj (in Cartesian coordinates), where x and y are coordinates measured from an origin and i, j are unit vectors along x and y axes respectively. The path traced by the particle in the xy‑plane is called its trajectory.

Displacement and Distance
Distance is the total length of the path traveled (scalar, always positive). Displacement is the straight‑line vector from initial to final position: Δr = r_final − r_initial. Displacement has magnitude and direction.

Scalars vs Vectors
A scalar is a physical quantity described by magnitude only (e.g., mass, temperature, speed, time). A vector requires magnitude and direction (e.g., displacement, velocity, acceleration, force). Important vector properties: equality (same magnitude and direction), addition, subtraction and multiplication by scalars.

Vector representation and components
A vector A in the plane can be written as A = A_x i + A_y j. Its magnitude is |A| = sqrt(A_x^2 + A_y^2). The direction (angle with x‑axis) θ satisfies tan θ = A_y / A_x. To find components: A_x = |A| cos θ, A_y = |A| sin θ. A unit vector along A is â = A/|A|.

Vector addition
Vectors add according to the triangle or parallelogram law. If A and B are two vectors, the resultant R = A + B. In components: R_x = A_x + B_x, R_y = A_y + B_y, and |R| = sqrt(R_x^2 + R_y^2). Subtraction: A − B = A + (−B) (reverse direction of B).

Speed and Velocity
Speed (scalar) = total distance / time; average velocity (vector) = total displacement / time: v_avg = Δr / Δt. Instantaneous velocity is the derivative v = dr/dt; instantaneous speed is |v|. In plane motion the x and y components of velocity are v_x = dx/dt, v_y = dy/dt — motion in two dimensions splits into two independent one‑dimensional motions.

Why vectors matter
Many physical quantities (displacement, velocity, acceleration) depend on direction. Vector methods let us resolve problems into components, add contributions from different directions, and handle relative motion in a clear way.

Fig 2.1 — Educational Diagram: Projectile Motion Trajectory & Kinematic Equations

Vectors vs Scalars: Scalars are quantities described by magnitude only (e.g., temperature, mass). Vectors have both magnitude and direction (e.g., displacement, velocity, force). A vector is usually depicted as an arrow: length ∝ magnitude, arrowhead gives direction.

Representation and Components: A vector A in 2D can be written in component form as A = A_x i + A_y j. In 3D: A = A_x i + A_y j + A_z k, where i, j, k are unit vectors along x, y, z axes. Magnitude: |A| = sqrt(A_x^2 + A_y^2 (+ A_z^2)). Direction angle(s): θ = arctan(A_y / A_x) (choose correct quadrant).

Unit Vector: Unit vector in the direction of A: â = A / |A|. Unit vectors have magnitude 1 and indicate direction.

Resolution of a Vector: Any vector can be resolved into orthogonal components. For a vector of magnitude A making angle θ with +x axis: A_x = A cosθ, A_y = A sinθ. This is used to convert between magnitude-angle form and component form.

Vector Addition and Subtraction: Addition: triangle, parallelogram or polygon rules. If A and B are vectors, resultant R = A + B. In components: R_x = A_x + B_x, R_y = A_y + B_y (and R_z for 3D). Subtraction: A - B = A + (−B) (reverse direction of B and add). Properties: A + B = B + A (commutative), A + (B + C) = (A + B) + C (associative).

Multiplication by a Scalar: For scalar k and vector A, kA has magnitude |k||A| and direction same as A if k>0, opposite if k<0. Distributive laws hold: k(A + B) = kA + kB.

Dot (Scalar) Product: A · B = |A||B| cosθ, where θ is the angle between A and B. In components: A · B = A_xB_x + A_yB_y (+ A_zB_z). The dot product is a scalar and gives projection information: component of A along B = (A · B)/|B|.

Cross (Vector) Product: A × B is a vector perpendicular to the plane of A and B with magnitude |A||B| sinθ and direction given by the right-hand rule. It is anti-commutative: A × B = −(B × A). In component form (using determinant): A × B = (A_yB_z − A_zB_y) i + (A_zB_x − A_xB_z) j + (A_xB_y − A_yB_x) k.

Position and Displacement Vectors: Position vector r locates a point: r = x i + y j (+ z k). Displacement between two points r_1 and r_2: Δr = r_2 − r_1 (a vector). Path length (distance) is not the same as displacement (vector).

Key Properties (summary): Equality of vectors requires same magnitude and direction. Vector algebra follows linear rules: addition commutative/associative, scalar multiplication distributive, dot product is bilinear and scalar, cross product is bilinear and anti-commutative.

Fig 3.1 — Educational Diagram: Projectile Motion Trajectory & Kinematic Equations

Introduction
In two- or three-dimensional motion, any vector (displacement, velocity, acceleration, force) can be described by its components along mutually perpendicular axes. Resolution of a vector means splitting it into two or more component vectors whose vector sum equals the original vector. Unit vectors are vectors of length one that give direction.

Components and Resolution (Orthogonal axes)
For a vector V in the xy-plane that makes angle θ with the +x-axis, the rectangular (orthogonal) components are found by forming a right triangle:

• V_x = V cos θ (component along x-axis)
• V_y = V sin θ (component along y-axis)

The vector can be written in component form using unit vectors i and j:
V = V_x i + V_y j

Magnitude and direction from components:

• |V| = sqrt(V_x² + V_y²)
• θ = arctan(V_y/V_x) (take care of the quadrant)

Negative components
Sign of a component indicates direction: e.g., V_x < 0 means the x-component points toward negative x direction.

Vector addition using components
To add vectors A and B, add corresponding components:
A + B = (A_x + B_x) i + (A_y + B_y) j. This is the simplest way to perform vector addition in coordinate form.

Unit Vectors
Unit vectors have magnitude 1 and indicate direction. Standard unit vectors in Cartesian coordinates are i (along +x), j (along +y) and k (along +z). Any vector R can be written as R = R_x i + R_y j + R_z k. The unit vector along R is û = R / |R|.

Projection on an arbitrary direction
The scalar projection (component) of V along a unit vector û is V·û (dot product). The vector projection is (V·û) û. This allows resolution along non-orthogonal or rotated axes.

Important properties

• Components are independent: motion along x does not affect motion along y (useful in projectile motion).
• Resolution is unique only when axes are fixed (especially orthogonal axes give simple formulas).
• Unit vectors are dimensionless and used to express direction; multiplying a unit vector by a scalar gives a vector with that magnitude in the given direction.

Summary
Resolution into components + unit vectors provide a systematic way to represent, add, and manipulate vectors in mechanics problems, simplifying analysis of motion and forces.

Fig 4.1 — Educational Diagram: Projectile Motion Trajectory & Kinematic Equations

Position vector: The position vector of a particle (often denoted by r) is a vector drawn from a fixed origin O to the particle's instantaneous location P in a coordinate system. It gives both the location of the particle and the direction from the origin. In two dimensions, if the coordinates of P are (x, y), the position vector is written as:

r = xi + yj.

The magnitude (distance from origin) is |r| = sqrt(x² + y²) and its direction (angle with x-axis) is θ = arctan(y/x) (with quadrant checks).

Displacement vector: The displacement vector Δr between two events (or instants) is the vector change in position. If the position at time t₁ is r₁ and at time t₂ is r₂, then

Δr = r₂ − r₁.

In components, if r₁ = x₁i + y₁j and r₂ = x₂i + y₂j, then

Δr = (x₂ − x₁)i + (y₂ − y₁)j = Δxi + Δyj,

and its magnitude |Δr| = sqrt((Δx)² + (Δy)²).

Key distinctions:

• Position is measured from a chosen origin; displacement is the change in position between two points in time.
• Both position and displacement are vectors (have magnitude and direction). Distance (path length) is a scalar and is generally ≥ |Δr|; equality holds only for straight-line motion between the two points.
• Displacement depends only on initial and final positions — it is independent of the path taken.

Vector operations and unit vectors: The unit vector along r is r̂ = r/|r|. Position and displacement vectors add/subtract using usual vector rules (graphical or component-wise).

Time dependence: If the particle's position varies with time, write r(t). The displacement over time interval Δt = t₂ − t₁ is Δr = r(t₂) − r(t₁). Instantaneous velocity is v = dr/dt (vector), introduced naturally from the position vector.

Fig 5 — Educational Diagram: Scalar (Dot) Product and Projection

Overview: The scalar (dot) product is an operation that takes two vectors and returns a single number (scalar). It quantifies how much one vector goes in the direction of another. Projection uses this idea to find the component of one vector along the direction of another.

Geometric definition: For two vectors a and b with angle θ between them, the dot product is
a · b = |a| |b| cosθ. This gives a scalar; its sign tells whether vectors point generally in the same (positive), opposite (negative) or perpendicular (zero) directions.

Algebraic (component) definition: If a = (a_x, a_y) and b = (b_x, b_y) in 2‑D (or include z components in 3‑D), a · b = a_x b_x + a_y b_y ( + a_z b_z). This formula is used for calculations in coordinate form.

Scalar projection (component): The scalar projection (also called component) of a on b is the signed length of a along the direction of b: comp_b(a) = |a| cosθ = (a · b) / |b|. It is positive when a has a component in the direction of b, negative if opposite, zero if perpendicular.

Vector projection: The vector projection of a onto b (the actual vector along b) is proj_b(a) = ((a · b) / |b|^2) b. This yields a vector parallel to b whose magnitude equals the scalar projection.

Key properties (for vectors a, b, c and scalar k):

• a · b = b · a (commutative)
• (ka) · b = k(a · b)
• a · (b + c) = a · b + a · c (distributive)
• a · a = |a|^2
• a · b = 0 ⇔ a ⟂ b (orthogonality)

Physical meaning and use:

• Work: If a constant force F acts and an object is displaced by s, the work done W = F · s = |F||s|cosθ (θ is angle between force and displacement). Only the component of force along the displacement does work.
• Component of velocity along a direction (useful on inclines or along roads): use scalar projection to find how fast you move along that direction.
• Computer graphics and lighting: brightness depends on the dot product of surface normal and light direction (cosine law).

How to compute:

1. If vectors are given in components, use a_x b_x + a_y b_y (+ a_z b_z).
2. If magnitudes and angle given, use |a||b|cosθ.
3. To get projection length of a on b: (a · b)/|b|. To get projection vector: ((a · b)/|b|^2) b.

Quick example (conceptual): A force of 10 N acts at 30° to the horizontal while an object is displaced 5 m horizontally. Work = F · s = (10 N)(5 m)cos30° = 50 × (√3/2) ≈ 43.3 J.

Summary: The dot product connects geometry (angle) and algebra (components); projection extracts the part of one vector along another and is found directly from the dot product.

Fig 6 — Educational Diagram: Velocity in Two Dimensions

Introduction: Velocity in two dimensions describes how fast and in what direction a particle moves on a plane. Position is a vector r = x i + y j. Motion in a plane is treated by resolving vectors into orthogonal components (usually x and y) which evolve independently (if forces separate similarly).

Displacement and Average Velocity: For a time interval Δt the displacement is Δr = r(t + Δt) − r(t). The average velocity is a vector: v_avg = Δr / Δt. Its direction is the direction of the net displacement.

Instantaneous Velocity: The instantaneous velocity is the time derivative of position: v(t) = dr/dt = v_x(t) i + v_y(t) j, where v_x = dx/dt and v_y = dy/dt. The speed is the magnitude: v = |v| = sqrt(v_x^2 + v_y^2). The direction (measured from +x) is θ = arctan(v_y / v_x) (take quadrant into account).

Component Independence: In many problems (e.g., projectile motion under uniform gravity), x- and y-components are independent: x-motion may be uniform (no acceleration) while y-motion may be uniformly accelerated. This allows solving two 1D problems and recombining their vector results.

Relative Velocity: Velocity is frame-dependent. The velocity of A relative to B is v_{A/B} = v_A − v_B. This is used for problems like a boat in a flowing river or an airplane in wind.

Geometric picture: Instantaneous velocity is tangent to the trajectory. For circular motion the velocity vector rotates while its magnitude may be constant (uniform circular motion) — the tip of the velocity vectors plotted head-to-tail traces a curve called a hodograph.

Fig 7 — Educational Diagram: Acceleration in Two Dimensions

Definition: In two dimensions, acceleration is the rate of change of the velocity vector with time. If the position vector is r(t) = x(t) i + y(t) j, then the velocity v = dr/dt and the acceleration a = dv/dt = d^2r/dt^2.

Component form: Acceleration is naturally expressed in components: a = a_x i + a_y j where a_x = d^2x/dt^2 and a_y = d^2y/dt^2. The magnitude and direction are

• Magnitude: a = sqrt(a_x^2 + a_y^2)
• Direction (angle with x-axis): tan(theta) = a_y / a_x

Vector relation with position and speed: a = dv/dt = d^2r/dt^2. This means acceleration can change the speed (magnitude of v) and/or the direction of v.

Tangential and normal (centripetal) components: For motion along a curved path it is useful to decompose a into two orthogonal components:

• a_t (tangential) = dv/dt — the component of acceleration along the instantaneous velocity direction, responsible for change in speed.
• a_n (normal) = v^2 / R — the component perpendicular to velocity (toward center of curvature), responsible for change in direction. R is the radius of curvature of the path.

So a = a_t t_hat + a_n n_hat and |a| = sqrt(a_t^2 + a_n^2). In uniform circular motion a_t = 0 and a_n = v^2/R (directed to the centre).

Special case — projectile motion: For a projectile near Earth (neglecting air resistance) acceleration is constant and vertical: a = 0 i + (−g) j where g ≈ 9.8 m/s^2. Hence a_x = 0 (horizontal velocity constant) and a_y = −g (vertical velocity changes uniformly).

Interpretation: In two dimensions acceleration describes how both components of velocity change. A nonzero a_x changes horizontal speed; a nonzero a_y changes vertical speed. A vector a not parallel to v produces curvature by changing direction.

How to solve problems: 1) Write r(t) or v(t) in components. 2) Differentiate components to get a_x and a_y. 3) If asked, compute magnitude and angle, or decompose into tangential/normal using v and its derivative.

Fig 8.1 — Educational Diagram: Projectile Motion Trajectory & Kinematic Equations

What it describes: Kinematic equations in vector form give the position, velocity and acceleration of a particle moving in a plane while keeping track of both magnitude and direction. They are especially useful when acceleration is constant (uniform acceleration) and when motion is treated as two independent component motions (usually x and y).

Definitions and relations (vector form):

• Position vector r(t) = x(t) i + y(t) j.
• Velocity vector v(t) = dr/dt.
• Acceleration vector a(t) = dv/dt = d^2r/dt^2.

For constant acceleration a (a is a constant vector):

• Integrating a = dv/dt gives v(t) = v₀ + a t. (vector equation)
• Integrating v = dr/dt gives r(t) = r₀ + v₀ t + (1/2) a t². (vector equation)
• Using dot product one gets v·dv/dt = a·v ⇒ v² = v₀² + 2 a·(r − r₀). This is the vector form of the v² relation.

Component form (useful in a plane):

• x(t) = x₀ + v_0x t + (1/2) a_x t², v_x(t) = v_0x + a_x t
• y(t) = y₀ + v_0y t + (1/2) a_y t², v_y(t) = v_0y + a_y t

Key points to remember: (1) For constant a, motion in x and y are independent and can be solved separately. (2) The vector v is tangent to the trajectory; acceleration may have tangential (change speed) and normal (change direction) components. (3) If acceleration direction changes (e.g., circular motion), use instantaneous vectors and decompose a into a_tangential and a_normal. (4) For non-constant a(t), use v(t)=v₀+∫a(t) dt and r(t)=r₀+∫v(t) dt.

Fig 9.1 — Educational Diagram: Projectile Motion Trajectory & Kinematic Equations

What is a projectile? A projectile is any object that, after an initial launch, moves only under the influence of gravity (neglecting air resistance). Projectile motion is the two-dimensional motion of such an object under constant acceleration g directed vertically downward.

Key idea — independence of motions: Projectile motion can be treated as two independent motions: (1) uniform motion in the horizontal direction with constant velocity, and (2) uniformly accelerated motion in the vertical direction with acceleration −g.

Initial conditions and components: If a projectile is launched from origin with initial speed u at angle θ above horizontal, resolve initial velocity into components: u_x = u cosθ (constant), u_y = u sinθ (initial vertical component).

Equations of motion (vector form):

x(t) = u cosθ · t
y(t) = u sinθ · t − (1/2) g t^2
v_x(t) = u cosθ
v_y(t) = u sinθ − g t

Trajectory (eliminate t):

y(x) = x tanθ − (g x^2) / (2 u^2 cos^2θ)

This is a parabola opening downward.

Important derived results (for landing at same vertical level from which it was launched):

• Time of flight (total time in air): T = 2 u sinθ / g
• Maximum height (highest y reached): H_max = (u^2 sin^2θ) / (2 g)
• Horizontal range (horizontal distance covered): R = (u^2 sin 2θ) / g

Remarks and properties:

• Trajectory is symmetric: the ascent and descent times about the peak are equal.
• Horizontal velocity is constant; vertical velocity changes linearly with time (v_y = u sinθ − g t).
• Range is maximum when sin 2θ = 1 → 2θ = 90° → θ = 45° (for level launch and landing).
• Two different angles θ and (90° − θ) give the same range for a given speed u (complementary angles).
• All formulas above assume uniform gravity, negligible air resistance, and a flat horizontal landing plane.

Velocity at any time: The instantaneous speed is the magnitude of velocity vector:

v(t) = sqrt[(u cosθ)^2 + (u sinθ − g t)^2]
θ_v(t) = arctan[(u sinθ − g t) / (u cosθ)]  (angle of velocity with horizontal)

Extensions (brief): If launch and landing heights differ, solve y(t) = y_final to get the flight time(s) and then x = u cosθ · t to get range. Air resistance, wind, or non-uniform g require more advanced treatment.

Typical derivations (outline): For T use y(T)=0 → u sinθ T − 1/2 g T^2 = 0 → T( u sinθ − 1/2 g T) = 0 → T = 2 u sinθ/g. For R use R = u cosθ · T → R = u cosθ · (2 u sinθ/g) = u^2 sin2θ/g.

Fig 10.1 — Educational Diagram: Newton's Laws of Motion & Free Body Diagrams

Overview
Motion on an inclined plane in 2D covers two related situations: (A) an object moving (or sliding) on the plane under gravity (useful for resolving g into components), and (B) projectile motion where the trajectory is relative to an inclined plane (projectile launched from or toward a slope). For CBSE Class 11 the important concept is choosing convenient axes (usually one axis along the plane and the other perpendicular), resolving velocity and acceleration into those axes, and using kinematic equations to get range, time of flight and velocity components.

Key idea — axis rotation
If the incline makes an angle β with the horizontal, it is often easiest to rotate axes so x' is along the plane (up the slope) and y' is perpendicular to it (out of the plane). Gravity g (vertically downward) then has components along these axes:

• Along the plane (down slope): a_x' = −g sinβ
• Perpendicular to the plane (into plane): a_y' = −g cosβ

Projectile launched from the plane
Suppose a projectile is launched from the plane with speed u at an angle φ above the plane (so the angle with horizontal is α = φ + β). Initial components in the rotated axes are u_x' = u cosφ, u_y' = u sinφ. Using kinematic equations:

• y'(t) = u sinφ · t − (1/2) g cosβ · t²
• x'(t) = u cosφ · t − (1/2) g sinβ · t²

For launch from the plane, the projectile returns to the plane when y'(t) = 0 (nonzero t), giving time of flight t = 2u sinφ / (g cosβ). Plugging into x'(t) gives the range along the plane.

Range along the plane (measured along slope)
If the same launch is described by α measured from horizontal, the range s (distance along the plane from launch point to impact on the plane) is:

s = (2 u² cosα · sin(α − β)) / (g cos²β)

Equivalently, in terms of φ (angle above the plane):

s = (2 u² cos(φ + β) · sinφ) / (g cos²β)

Time of flight (when launched from plane)
t = 2 u sinφ / (g cosβ)

Maximum range on an inclined plane
For fixed speed u and incline β, s is maximized when the launch angle α (w.r.t horizontal) satisfies

α_max = 45° + β/2

Substituting gives maximum range along the plane

s_max = (u² / (g cos²β)) · (1 − sinβ)

Alternate viewpoint (intersection method)
Treat projectile in standard horizontal/vertical coordinates: x = u cosα t, y = u sinα t − (1/2) g t². The inclined plane (through origin) has equation y = x tanβ. Equate y and x tanβ, solve for t (nonzero root), then compute distance along plane s = x / cosβ. This yields the same range formula above.

Important practical notes

• When launching uphill (α close to β), sin(α − β) becomes small and range short; launching too steep reduces horizontal component.
• If launching from a height above or below the plane (not at plane origin), include the plane offset in the intersection equation and solve the quadratic for t.
• For block or object constrained to move on the incline (no projectile), you only need the scalar components of acceleration: along-plane acceleration = g sinβ (downwards if no other forces).

Fig 11.1 — Educational Diagram: Newton's Laws of Motion & Free Body Diagrams

Definition: Uniform circular motion (UCM) is motion of a particle along a circular path with constant speed. Although the speed is constant, the velocity is not constant because its direction continuously changes.

Key features:

• Path: a circle of fixed radius r.
• Speed v: constant in magnitude.
• Velocity v⃗ : always tangent to the circle (perpendicular to the radius at each point).
• Acceleration a⃗ : directed toward the centre of the circle (centripetal), with constant magnitude but continuously changing direction.
• Angular speed ω: constant (rate of change of angular position θ).

Position, velocity and acceleration (parametric form):

Choose origin at centre and let r be the radius. For constant angular speed ω, the angular position is θ(t)=ωt+θ₀ and the position vector is

r⃗(t)=r(cosωt, sinωt).

Velocity: v⃗ (t)=d r⃗/dt = r(−ω sinωt, ω cosωt) = ω×r⃗ (magnitude v=ωr, direction tangent).

Acceleration: a⃗ (t)=d v⃗/dt = r(−ω² cosωt, −ω² sinωt) = −ω² r⃗ (directed radially inward). Its magnitude a_c = ω² r = v²/r.

Derivation (sketch) of centripetal acceleration: For a small angular change Δθ, the change in velocity |Δv| ≈ vΔθ (because velocity vector rotates by Δθ). Then average acceleration ≈ |Δv|/Δt = v(Δθ/Δt) = vω. Since ω = v/r, a = v²/r.

Physical points:

• Centripetal acceleration requires a net inward force F_c = m a_c (provided by tension, gravity component, friction, normal force, etc.).
• No work is done by centripetal force (force is perpendicular to instantaneous displacement), so kinetic energy and speed remain constant.
• In a rotating (non-inertial) frame a fictitious outward 'centrifugal' force m v²/r can be introduced to explain apparent effects.

Fig 12.1 — Educational Diagram: Newton's Laws of Motion & Free Body Diagrams

What it is
Relative motion in two dimensions studies how the position, velocity and acceleration of one object appear when observed from another moving object, using vector subtraction. Observations depend on the chosen reference frame (observer). For two observers A and B, the position of A relative to B is r_{A/B} = r_A - r_B. Differentiating with respect to time gives relative velocity and acceleration.

Key vector relations

• Relative position: r_{A/B}(t) = r_A(t) - r_B(t)
• Relative velocity: v_{A/B} = v_A - v_B
• Relative acceleration: a_{A/B} = a_A - a_B

Galilean transformation
If frame B moves with constant velocity V relative to an inertial frame S, then coordinates transform as r_S = r_B + r_{S/B}. For constant V the time derivatives are simple differences of velocities. (Rotating frames are not covered here.)

Component method
Resolve vectors into x and y components and subtract components: v_{A/B,x} = v_{A,x} - v_{B,x}, v_{A/B,y} = v_{A,y} - v_{B,y}. Then |v_{A/B}| = sqrt((v_{A/B,x})^2 + (v_{A/B,y})^2).

Magnitude formula (using law of cosines)
If the speeds v_A and v_B make angle θ between them, then |v_{A/B}| = sqrt(v_A^2 + v_B^2 - 2 v_A v_B cos θ).

Solving strategy (stepwise)

1. Choose a reference frame (often ground or one of the objects).
2. Draw vector diagram of velocities; label magnitudes and angles.
3. Resolve into components or form velocity triangle.
4. Compute relative components: subtract corresponding components.
5. Find magnitude and direction of relative vector if needed.

Important notes
- If the reference frame moves with constant velocity (inertial shift), accelerations are identical in both frames (a_A = a_{A/B} + a_B).
- Relative motion is purely vector subtraction; order matters (v_{A/B} = v_A - v_B, not vice versa).
- For problems like crossing a river or an airplane in wind, use component subtraction to find ground velocity and drift.

Fig 13.1 — Educational Diagram: Newton's Laws of Motion & Free Body Diagrams

Overview: Graphical analysis of plane motion studies how position, velocity and acceleration of a particle moving in a plane (two dimensions) are represented and interpreted using graphs. Plane motion is treated component-wise: break vectors into x- and y-components, draw separate time-graphs for each component, and reconstruct vector behaviour from those component graphs.

Vectors and components: Represent the position vector r = xi + yj as an arrow from origin to point (x,y). Its x- and y-projections are the component vectors. Graphical vector addition uses tip-to-tail or parallelogram rules.

From graphs to kinematics:

• x(t), y(t) (position vs time): slope at a time = instantaneous velocity component (dx/dt or dy/dt). Area under these curves is not commonly used; rather, integration of velocity components gives position change.
• vx(t), vy(t) (velocity components vs time): slope = acceleration component (dvx/dt or dvy/dt). Area under vx(t) between t1 and t2 = Δx; area under vy(t) = Δy.
• a_x(t), a_y(t) (acceleration components vs time): area under acceleration curve = change in corresponding velocity component.
• Speed |v|(t): magnitude of velocity; equals sqrt(vx² + vy²). Its graph often differs from component graphs; its slope is not simply component acceleration.

Parametric and trajectory plots: Many plane motions are parametric: x = x(t), y = y(t). Eliminating t yields trajectory y(x) (e.g., a parabola for projectile motion). The shape of y vs x gives the path; tangent to the path gives direction of velocity vector.

Special cases:

• Projectile motion (no air resistance): x(t) = v₀ cosθ · t (linear), y(t) = v₀ sinθ · t − (1/2) g t² (parabolic). So vx = constant, vy = v₀ sinθ − g t (linear).
• Uniform circular motion (in plane): x(t) = R cos(ωt), y(t) = R sin(ωt). x and y vary sinusoidally; vx and vy are also sinusoidal and shifted in phase by 90°; acceleration components are −Rω² cos(ωt) and −Rω² sin(ωt).

How to analyze a given plane-motion graph:

• Separate motion into x and y components and draw/inspect their individual graphs.
• Find instantaneous component velocities from slopes of x(t) and y(t).
• Find component accelerations from slopes of vx(t) or by differentiating vx, vy.
• Compute displacement in each direction from area under vx(t) and vy(t) curves (or integrate analytically).
• Get vector magnitude and direction: |v| = sqrt(vx²+vy²), angle φ = arctan(vy/vx).

Practical tips for graphs: Use parametric plotting (x vs t and y vs t) and trajectory plotting (y vs x). Always mark axes, scales, and sign conventions (up/down, right/left). When reading instantaneous values on curves, draw a tangent; slope of tangent gives the instantaneous rate.

Fig 14 — Educational Diagram: Problem-Solving Strategies and Applications

Overview: Problem-solving in motion in a plane reduces two-dimensional motion to one-dimensional component problems using vectors. Treat position, velocity and acceleration as vectors, pick convenient axes, resolve vectors into components, apply kinematic equations separately in x- and y-directions, and combine results.

Step-by-step strategy:

• Read the problem carefully and identify what is given and what is required.
• Draw a clear diagram showing the path (trajectory), initial and final points, and all vectors (u, v, a, r). Mark angles and choose axes (usually x horizontal, y vertical).
• Resolve initial velocity and acceleration into components: u = u_x i + u_y j, a = a_x i + a_y j.
• Write kinematic equations for each component (x and y) independently, since for constant acceleration components evolve independently.
• Use symmetry, eliminate the time variable when needed (e.g., to get trajectory y(x)), and check limiting cases (θ = 0, small/large u) to validate answers.
• Consider alternate approaches: energy methods (when heights/speeds matter), centripetal/rotational relations (for circular motion), or relative velocity (moving frames).
• Always check units and signs, and sketch the motion and vector directions to ensure consistency.

Key ideas to apply:

• Independent-component treatment: motion in x and y governed by the same one-dimensional formulas.
• Eliminate time by solving one component for t and substituting into the other to obtain trajectory equations.
• Use geometry or symmetry to find maxima, ranges, and times of flight for projectiles.
• For curved motion, velocity is tangent to the path and acceleration has tangential (change of speed) and normal/centripetal (change of direction) components.

Common pitfalls: forgetting to resolve vectors, mixing signs for acceleration (take g = +9.8 m/s^2 downward or use -g in y-equations), using equations valid only for constant acceleration, and ignoring horizontal vs vertical independence.