Motion in a Plane Motion in a Plane Think of motion in a plane not as one complicated curve, but as two simple stories running in parallel. When a ball is thrown, the horizontal story is calm: in ideal conditions there is no horizontal acceleration, so the horizontal speed stays the same. The vertical story is lively: gravity pulls downward, changing the vertical speed every moment. The curved path (a parabola) that your eyes see is just the overlay of these two independent stories. Change the vertical story and you change how high it goes, but the horizontal speed at that instant is unaffected; change the horizontal speed and you stretch the arc without altering how gravity acts. This independence is the master key for the entire chapter. Once you write vectors as components along fixed axes x ,( i ) and y ,( j ) , every kinematics tool from 1D carries over cleanly: add displacements componentwise, differentiate to get velocity components, differentiate again for acceleration components. Real life constantly mixes motions like this: a drone moving north while the wind blows east, or a floodlight mounted on a rotating arm while the base translates. The math says: solve each perpendicular direction as if it were a separate 1D problem and then combine using vectors. With this mindset, projectile motion, river-boat navigation, rain-umbrella orientation, and uniform circular motion all become straightforward applications rather than new worlds to memorize. Crossing a moving river in a boat: the engine drives you straight across, the current drags you downstream, and the actual track is the diagonal sum. The river doesn’t affect your engine’s northward push; the engine doesn’t affect the current’s pull. Together they decide where you land. remember A quantity with both magnitude and direction; represented by components along chosen axes, e.g., A =A x i +A y j . Vector Vector from origin to the particle: r =x i +y j . Position vector Change in position: r = r 2- r 1=( x) i +( y) j . Displacement Velocity Rate of change of position; average v avg = r t , instantaneous v = d r dt =v x i +v y j . Rate of change of velocity; a = d v dt =a x i +a y j . Acceleration Projectile An object projected into air moving only under gravity (air resistance neglected). Inward acceleration needed to keep motion on a circle: a c= v 2 r = 2 r . Centripetal acceleration In 2D kinematics we choose fixed perpendicular axes and resolve every vector into components. This is more than a trick: Newton’s second law and the calculus definitions are vector equations, so equality must hold componentwise. If a is constant, the x - and y -motions follow the familiar 1D formulas independently with a x and a y plugged in. The scalar speed is v= v x 2+v y 2 and the motion’s direction at any instant is given by = v y v x . A common student relief: you rarely need to compute vector magnitudes early—work with components all the way and combine only at the end. Unit vectors i and j point along the positive x and y directions and have unit length. Choosing axes cleverly simplifies algebra. For projectiles on level ground, pick x horizontal and y vertical with +y upward and take g downward as -g j . For rivers, let x be across the river and y be along the current; then one component solves crossing time, the other tracks drift. Clear axis choice and sign convention eliminate 80% of mistakes. Position vector Coordinates describe the instantaneous location in the plane. Used to define the position of a point object in a two-dimensional Cartesian coordinate system. Displacement Vector difference between final and initial positions. Instantaneous velocity Time derivative of position; components v x , v y . Instantaneous acceleration Time derivative of velocity; second derivative of position. Pick axes and write all given vectors as components. Write x(t) and y(t) using 1D kinematics separately. Use a common t to connect the components. Combine at the end for speed v or angle if required. How to analyze any 2D motion Projectile motion is the cleanest showcase of component independence. Launch with speed u at an angle above horizontal. The horizontal component u x=u never changes if air resistance is ignored, so v x stays constant. The vertical component u y=u fights gravity, so v y decreases linearly with time, becomes zero at the top, then becomes downward. The top point is where v y=0 but v x is still u ; hence the projectile is never at rest at the top—its motion is purely horizontal there. Taking +y upward and g downward. Parametric equations of a projectile Valid when landing height equals launch height. Time of flight Reached when v y=0 . Maximum height This calculation determines the maximum vertical height reached by a projectile due to its initial upward velocity component. Maximum at =45 ; complementary angles give equal R . Horizontal range Calculate the total horizontal distance traveled by the projectile when launched and landing at the same height. Equation of trajectory y(x) for a projectile y=x - g x 2 2u 2 2 Substitute t from the first equation into the second. Uniform gravity; air resistance neglected Flat ground; same launch and landing level Axes: x horizontal, y upward; g acts downward Parabolic path of a projectile obtained by eliminating time between x(t) and y(t) . Use the three standard projectile results with care. They apply only when launch and landing are at the same vertical level and g is constant. If the projectile lands at a different height (cliff problems), keep x(t) and y(t) and solve for t from the quadratic in y(t) . Complementary angles and 90 - yield the same range for a fixed u , but different flight times and peak heights. This symmetry is a frequent exam shortcut. T, H, R A ball is projected with speed u=20 , m/s at =30 . Find time of flight, maximum height, and range. Take g=9.8 , m/s 2 . s, m, m u = 20 m/s θ = 30° g = 9.8 m/s² Standard level-ground projectile formulas. easy Stroboscopic projectile: horizontal component arrows are equal (constant v x ), while vertical arrows shrink to zero at the top and grow downward after (gravity). Projectile arc in a grid room with red horizontal and vertical component arrows marking decomposed vectors. tip Independence of perpendicular motions: Solve horizontal and vertical as separate 1D problems with the same time t . Conditions: uniform g , negligible air drag, flat Earth over the range of motion. Relative motion formalizes how one moving observer sees another. The velocity of A relative to B is v AB = v A- v B . In river problems, think in the ground (shore) frame first: the man’s velocity relative to water plus the river’s velocity relative to ground gives the man’s velocity relative to ground. For shortest crossing time, aim perpendicular to the banks so the across-river component is maximized, accepting downstream drift. For zero drift (shortest path), tilt upstream so that the across component remains but the along-current components cancel exactly. Relative velocity definition Velocity of A as seen from B. This formula determines the observed velocity of an object by subtracting the velocity of the observer from the object's velocity. Hold umbrella at angle with the vertical when you walk with speed v p and rain falls with speed v r vertically. Rain-umbrella condition Vector addition in ground frame: v MG = v MW + v WG . medium s, m, degree Width d = 200 m vr = 3 m/s (east) vm = 5 m/s (a) t and drift; (b) angle and t A river is 200 m wide and flows east at v r=3 , m/s . A swimmer can swim at v m=5 , m/s in still water. (a) If aiming straight across, find crossing time and drift. (b) To land directly opposite, find the steering angle upstream and the crossing time. Overhead neon boat with three vectors: heading across, current downstream, resultant diagonal. Boat crossing: white arrow shows swimmer’s heading across, blue arrow the downstream current, and the dotted yellow arrow the actual ground-track resultant. Shortest time ≠ shortest path. Aiming straight across gives minimum time but nonzero drift; aiming upstream to cancel current gives zero drift but takes longer. Mixing these up is a common loss of 1–2 marks. neet-alert Uniform circular motion (UCM) is 2D motion with constant speed and continuously turning direction. Even though v is constant in magnitude, the velocity vector changes because its direction changes, so there must be an acceleration. This acceleration is purely radial and points toward the center, keeping the object on the circular path. No separate “centripetal force” exists as a new kind of force; it is just the name for the net inward force supplied by tension, gravity, friction, or a normal reaction, depending on context. Radial inward acceleration for motion on a circle of radius r at speed v . Centripetal acceleration In polar unit vectors e r (radial) and e (transverse). Radial acceleration for circular motion a c= v 2 r ; (towards center) Particle moves on a circle of radius r with speed v Uniform speed (UCM) so | v | is constant Inward acceleration required for circular motion; does not change speed, only direction. Use a c= v 2 r and T= 2 r v , f=1/T . medium m/s², s, Hz r = 50 m v = 20 m/s a c , T, f A car moves on a circular track of radius 50 , m at 20 , m/s . Find a c , the time for one lap, and the frequency. Projectiles from elevated or depressed points use the same component method with a nonzero initial height y 0 . Write y(t)=y 0+u ,t- 1 2 gt 2 and set y=0 at the ground to solve for flight time. The physically meaningful root is the positive time. Range then follows from x(t)=u ,t . Avoid the level-ground shortcuts T= 2u g and R= u 2 2 g in this case—they no longer apply. hard Use y=0=y 0+u ,t- 1 2 gt 2 and R=u ,T . A stone is thrown from a cliff 45 , m high with u=30 , m/s at =37 above horizontal. Find time of flight and horizontal range. Take g=9.8 , m/s 2 , 37 0.601 , 37 0.799 . T and R y0 = 45 m u = 30 m/s θ = 37° s, m remember For level-ground projectiles: (i) t up =t down = u g , (ii) H 2 , (iii) R 2 so complementary angles give equal range. Projectile Motion Vectors Component Horizontal ( x ) Vertical ( y ) Acceleration Source Velocity at Max Height Horizontal is Happy and Constant; Vertical is Variable and Veers due to Gravity. Displacement (x = (u )t ) (y = (u )t - 1 2 gt 2 ) No horizontal force after launch (x max ) depends on (u ) Initial Velocity (u x = u ) (u y = u ) Projection impulse (u x ) (Remains constant throughout) Acceleration (a x = 0 ) (a y = -g ) Earth's gravitational pull (a y = -g ) (Always constant) Velocity at time (t ) (v x = u ) (v y = u - gt ) Acceleration due to gravity ((g) ) (v y = 0 ) (Only horizontal exists) Force acting (F x = 0 ) (F y = -mg ) Weight of the projectile (F net = mg ) (downwards) Kinetic Energy (K x = 1 2 m(u ) 2 ) (K y = 1 2 m(u - gt) 2 ) Energy conversion (Potential to Kinetic) (K min = 1 2 mu 2 2 ) Momentum (p x = mu ) (p y = m(u - gt) ) Impulse due to weight (p = mu ) Path Equation Linear in (x ) Quadratic in (y ) Parabolic trajectory relation Vertex of the parabola projectile motion vectors For a projectile on level ground: v x is a horizontal line at u cosθ; v y is a straight line decreasing from u sinθ to −u sinθ with slope −g. Launch vertical speed u sinθ Top point T/2 Just before landing −u sinθ v x dependent dependent v y The horizontal component vₓ = u·cosθ stays constant; vy falls linearly through zero at the apex. 2D PLOT Launch speed theta Launch angle (rad) vy = u sin(theta) - g t Projectile velocity component vy vs time vy vt m/s −u to +u Velocity components 0 to T Time Synthesis diagram: vertical free fall, horizontal uniform motion, and their combination forming a projectile arc with dashed vertical time markers. Split-panel: vertical drop, horizontal roll, and central projectile parabola. Direction of the resultant when adding two vectors using parallelogram/triangle law: = B A+B . Average velocity is displacement over time: v avg = r t ; for components, compute x and y separately. Average speed is total distance over total time v avg = d t ; it is always nonnegative and differs from average velocity. Only the vertical component is zero there; the horizontal component remains u , so the projectile still moves horizontally. At the highest point of a projectile, velocity is zero. Centrifugal force pushes objects outward in a circle. In an inertial frame there is no outward force; the net real force must point inward (centripetal) to change direction. “Centrifugal” appears only as a fictitious force in a rotating frame. THR for level-ground projectiles: T = 2u g , H = u 2 2 2g , R = u 2 2 g . Think “Time–Height–Range” in that order. Resolve u into u x=u and u y=u . Write x(t) and y(t) ; choose a common origin and time t=0 at launch. If landing at a different height, solve y(t)=0 (or the target’s y ) for t . Use that t in x(t) to get range or in v y(t) for impact speed/angle. Combine components at the end for v or the trajectory angle. Typical steps for projectile targets neet-alert Units and angles: Always keep g in m/s 2 and angles in radians inside trigonometric functions if you’re using a calculator set to radian mode. Mixing degrees and radians silently breaks answers. tip Boundary checks: 0 gives T 0 , H 0 , but R u 2 0 g =0 —a flat skid. 90 gives R 0 with maximum H . In circular motion, r 0 with finite v makes a c= v 2 r (physically impossible). Vector addition in planes often reduces to a right triangle. When two velocities are perpendicular, the resultant speed is v R= A 2+B 2 and the direction relative to A is = B A . For oblique vectors, the parallelogram rule with = B A+B is handy. In exams, sketch the head-to-tail triangle and mark known angles before writing any numbers—diagrams prevent sign mistakes and clarify which component solves the asked quantity (time, drift, or direction). Impact speed and angle for projectiles are frequent asks. Compute v x=u (constant) and v y=u -gt at the impact time. Then v= v x 2+v y 2 and the angle of impact below the horizontal is = | v y v x | at that instant. For level-ground landings, |v y| regains the launch magnitude but with negative sign; thus the impact speed equals the launch speed u (energy symmetry) though the direction is mirrored. In non-uniform circular motion, the tangential acceleration a t= dv dt changes the speed while the radial component a r= v 2 r changes only direction. The total acceleration is the vector sum a =a r( inward )+a t( tangent ) . Many real systems (e.g., a car speeding up around a bend) have both components. NEET typically focuses on UCM, but recognizing the two-component picture avoids conceptual traps. River–swimmer and rain–umbrella are the same vector subtraction idea presented in different clothing. For rivers, the ground-frame resultant determines where you land; for rain, you move the origin into the walker’s frame so the umbrella must align opposite to the relative velocity of rain seen by the walker. Drawing the relative velocity arrow first, then orienting the umbrella along its opposite, is the quickest reliable method under time pressure. Key terms recap Projectile motion Parabolic motion 2D motion under uniform gravity with no air drag. Time of flight Total airborne time from launch to landing. Maximum height Largest vertical displacement from launch level. Range Horizontal distance between launch and landing levels. Centripetal acceleration Inward acceleration keeping circular motion possible. a c Relative velocity Velocity of one object as measured from another moving frame. v AB