Relative Motion 1D & 2D Relative Motion 1D & 2D Motion is always seen from somewhere: a road, a train, a boat on a river, or a person running in rain. Relative motion gives a clean rule to switch between these viewpoints. In 1D (a straight line), deciding who is “fast” depends on who is watching. Two cars moving east at 60 km/h and 50 km/h: from the ground they both move east, but from the slower car the faster one seems to pull away at 10 km/h. In 2D (a plane), the idea is the same, but you must use vector subtraction and components. A swimmer’s actual path across a river is the vector sum of his velocity relative to water and the river’s flow. Rain that falls vertically in the ground frame appears slanted to a running person, because the runner adds a horizontal component to the rain’s relative velocity. Why this matters: the exam often hides a simple subtraction inside a story. “Who meets whom, when, and where?” becomes a relative speed along the line that connects them. “What angle to hold the umbrella?” becomes “what is the direction of rain relative to the runner?” “How to reach the point directly opposite across a river?” becomes “cancel the downstream current by aiming upstream.” The math is short if the diagram is right: subtract vectors, resolve into perpendicular components, and apply 1D kinematics independently along each axis. Think in frames: pick a convenient observer (ground, train, boat, runner). Then write velocity of A relative to B as v AB = v A − v B $. In 1D, this is simple sign-aware addition/subtraction. In 2D, write i- and j-components, subtract component-wise, and reassemble magnitude and angle. Edges and traps: the method assumes a single common time for both objects and that both velocities are measured in the same units and axes. In river–swimmer, the shortest time path is perpendicular to the banks (use full swimming speed across), but this drifts downstream; the no-drift path is angled upstream so the upstream component cancels the current—possible only if the swimmer is at least as fast as the current. In rain–man, the umbrella points opposite to the apparent rain direction; the tangent of the umbrella angle is runner speed divided by vertical rain speed (with the right reference line!). When ideas are framed this way, even “scary” 2D motion becomes two friendly 1D problems placed side by side. remember Big-picture analogies: (1) Train-platform: a ball rolling inside a moving train has one velocity relative to the train and another relative to the platform—add/subtract the train’s speed. (2) Rain-man: running adds horizontal velocity to otherwise vertical rain; the apparent rain direction tilts. (3) River-swimmer: the swimmer’s velocity relative to water combines with current to give the ground-track. Frame of reference The chosen observer and coordinate axes with respect to which motion is measured (e.g., ground, train, boat). Relative velocity Velocity of one object as observed from another moving object; in vectors, v A/B = v A - v B . Velocity of A relative to B ( v A/B ) The instantaneous rate of change of A’s position as seen by B. In 1D, v A/B = v A - v B with sign convention; in 2D, subtract components. Ground frame (inertial approximation) The Earth-fixed frame used for most NEET problems where accelerations due to Earth’s rotation are negligible. Component method Resolve vectors into perpendicular axes (e.g., i, j ), do algebra in 1D independently, then recombine for magnitude and direction. Sideways displacement caused by a transverse velocity component (e.g., downstream shift due to river current during a river crossing). Drift Closing speed The component of relative velocity along the line joining two objects that reduces their separation. Velocity of A as seen from B equals A’s velocity minus B’s velocity. Vector law of relative velocity In 1D, choose a positive direction. If car A has v A = +20 , m/s and car B has v B = +15 , m/s , then v A/B = +5 , m/s : A pulls ahead at 5 m/s. If B moves toward A with v B = -15 , m/s , then v A/B = 35 , m/s : they approach rapidly. In 2D, subtract components: if v A = (v Ax , v Ay ) and v B = (v Bx , v By ) , then v A/B = (v Ax - v Bx , ; v Ay - v By ) . The magnitude is | v A/B | = (v Ax - v Bx ) 2 + (v Ay - v By ) 2 and the direction angle from the x -axis satisfies = v Ay - v By v Ax - v Bx , with quadrant decided by signs. Treat each perpendicular direction as a separate 1D problem. Component subtraction (2D) Used to calculate the components of the velocity of object A relative to object B in two-dimensional motion. Independence of perpendicular components is the engine under most plane-motion problems. A river flows east while the swimmer targets north; the north-travel time depends only on the north component of the swimmer’s speed, not on the east current. Likewise, in rain–man, the umbrella direction depends on the horizontal-to-vertical ratio, independent of how big each is individually, as long as both are measured in the same frame. Always label axes and define angles with respect to a clear reference (horizontal vs vertical) to avoid sign and tangent mistakes. tip Validity of vector subtraction: Both velocities must be measured at the same instant, in the same units, and in the same coordinate axes. Perpendicular component independence holds only for orthogonal axes and constant axes in time (non-rotating frames). Finite time interval t > 0 Position x(t) is defined and single-valued over the interval Average velocity v avg = x t Define displacement over the interval. Define the corresponding time duration. Average velocity is displacement per time. v avg = x t Average velocity equals total displacement divided by total time. It can be negative in 1D depending on direction. s, m Closing speed along the line. Time to close the gap. Distance from A’s start. easy Relative speed method in 1D: closing speed = v A − v B . Catch-up time t and distance x from A’s start Two cars A and B move along a straight road in the same direction with speeds 25 m/s and 15 m/s. Initially, B is 200 m ahead of A. Find the time and distance (from A’s start) when A catches B. v A = 25 m/s (forward) v B = 15 m/s (forward) Initial separation d = 200 m Meeting conditions in 1D: If initial positions are x A(0) and x B(0) with velocities v A, v B (constant), then they meet when x A(0) + v A t = x B(0) + v B t . Solving gives t = x B(0) - x A(0) v A - v B , provided the numerator and denominator have the same sign to yield t > 0 . If v A = v B and initial separation is nonzero, they never meet. Runner speed u horizontal, rain speed v r vertical. Angle is measured from the vertical towards the backward direction. Umbrella angle for vertical rain = B A + B Resolve B into components along and perpendicular to A. Angle is with respect to A. Direction of resultant: = B A + B Vectors A and B lie in a plane Angle between A and B is Resultant R = A + B Gives the angle of the resultant when adding two vectors with included angle θ; used for quick direction finding before magnitude. In rain–man, the apparent rain relative to the runner is v R/M = v R - v M . If rain is vertical ( v R downward) and the man runs horizontally ( v M forward), the apparent rain tilts backward. The umbrella should be oriented exactly opposite to v R/M , so that drops strike the umbrella normal and not the face. Relative velocity v R/M = ( -u, -v r) with forward as +x and downward as +y. Umbrella tilt angle α from vertical (backwards) and | v R/M | Rain falls vertically at 12 m/s. A person runs horizontally at 6 m/s. Find (i) the angle to hold the umbrella with respect to vertical, and (ii) the apparent speed of rain relative to the runner. v r = 12 m/s (downward) u = 6 m/s (forward) degree, m/s Angle from vertical. Tilt backward by about 26.6°. Apparent rain speed. medium Vector diagram for rain-man showing vertical rain velocity and horizontal runner velocity forming a right triangle; umbrella oriented opposite to the resultant. Rain-man vector triangle with umbrella opposite to apparent rain direction. Apparent rain direction is the vector sum of vertical rain and horizontal running (relative to runner). River–swimmer model: Let the river flow east with speed v r . The swimmer can swim with speed v s relative to water at an angle from the north (perpendicular to banks), positive towards west (upstream). Components of swimmer’s velocity relative to ground: north v Ny = v s , east v Ex = v r - v s . If the river width is w , then crossing time t = w v s , provided > 0 (heading has a north component). Drift downstream is x = v Ex , t = (v r - v s ) w v s . Two special cases: (1) Shortest time: set = 0 so t min = w v s and drift = v r w v s . (2) No drift: require v Ex = 0 v s = v r , so = -1 ( v r v s ) (possible only if v s v r ) and t = w v s 2 - v r 2 . River-swimmer crossing time and drift Angle is measured from the perpendicular-to-banks (north) towards upstream (west). Geometric view: The ground-track velocity is the vector sum v G = v SW + v W , where v SW is swimmer relative to water, and v W is water relative to ground. Minimum time is obtained by maximizing the north component v Ny , achieved when all of v s points north. Zero drift makes the east component zero by aiming upstream enough to cancel the current. Both are direct consequences of independent 1D motions in x and y . w = 120 m v r = 2 m/s (east) v s = 3 m/s A river is 120 m wide and flows east at 2 m/s. A swimmer can swim at 3 m/s relative to water. (i) Minimum time to cross and downstream drift. (ii) To land directly opposite, what heading angle from north and crossing time? (i) t min, drift; (ii) θ for no drift, and t Use component method with t = w/(v s ) and drift criterion. medium Shortest time when = 0 . Downstream shift during minimum-time crossing. Aim upstream to cancel current. Longer than minimum-time case. s, m, degree North-South (m) Velocity triangle showing swimmer’s heading relative to water and river current adding vectorially to give ground velocity; resultant north component sets crossing time; east component sets drift. East-West (m) control v r v s control control theta v r t Landing point with drift custom Vector addition diagram for river–swimmer. Angle reference trap: In rain–man, = u/v r if is measured from vertical. If you measure from horizontal, use = v r/u . Write the reference line on your diagram to avoid inverted tangents. neet-alert Always subtract vectors component-wise. Magnitudes are computed only after components are subtracted and the direction is found with correct quadrant. Relative motion in 2D means subtracting magnitudes and then assigning an angle. Only the component of relative velocity along the line joining the objects changes their separation. Closing speed along line of sight Using relative velocity simplifies the separation vector needed to find the minimum distance between two objects. Closest approach problems: Let r A/B (t) = r A(t) - r B(t) . The distance squared is D 2 = r A/B r A/B . Differentiate: dD 2 dt = 2 , r A/B v A/B . At minimum distance, this derivative is zero, so r A/B v A/B . Solve for time of closest approach using this orthogonality, then substitute back for the minimum distance. t min and D min Use relative position and velocity: r A/B = r A − r B , v A/B = v A − v B . Closest approach when r A/B ⋅ v A/B = 0. Particle A starts at origin with velocity 4 i m/s. Particle B starts at position (0, 30) m with velocity 3 j m/s. Find the time of closest approach and the minimum separation. A: r A (0) = (0, 0), v A = (4, 0) m/s B: r B (0) = (0, 30) m, v B = (0, 3) m/s s, m Relative motion from B’s frame. Set derivative of distance squared to zero. Closest approach occurred in the past; in the future they move apart. Since t min <0 , minimal future separation is the initial one. hard Uniform circular motion with constant speed v Radius r is constant; motion in a plane Centripetal acceleration a c = v 2 r Polar unit vectors e r, e with r=0 . For uniform speed, =0 . Since v = r , radial acceleration is inward. a c = v 2 r (directed towards the center) In circular motion, acceleration points radially inwards with magnitude v 2 /r; speed changes direction, not magnitude. Link to relative motion: If you observe a bead on a rotating disc from a co-rotating frame, the bead appears at rest while the disc provides an inward interaction (e.g., friction or tension) producing the centripetal acceleration seen in the ground frame. The relative velocity between neighboring points on the rim is always tangential, explaining why v is perpendicular to the radius and why only direction changes in UCM. River–swimmer strategies compared. Goal Heading Crossing Time Downstream Drift Feasibility Shortest time Perpendicular to banks ( = 0 ) t min = w v s x = v r w v s Always No drift (land opposite) Aim upstream so v s = v r t = w v s 2 - v r 2 v s v r only Minimum drift for fixed speed Aim upstream, > 0 but < -1 (v r/v s) t = w v s x = (v r - v s ) w v s Always; reduces drift as increases Independence principle: Solve plane motion as two 1D problems—one along x, one along y. Time is common to both. remember Write RAB on your vector triangle to keep subtraction order straight. Relative = A minus B (RAB): Remember A as seen by B is vA − vB. Relative speed is the difference (with sign) of velocities along the line of motion, not the average. Use v A/B = v A - v B . Average of speeds gives relative speed in 1D. Space-time (x–t) diagram of two cars: two straight lines intersecting at the meeting event; slope indicates velocity. x–t graph with intersecting world-lines of two cars. Meeting time is where the x–t lines intersect; relative slope reflects relative speed. vt Visualizing relative velocity as vertical separation on a v–t graph. Time (s) v A control control v B v rel derived Horizontal lines for constant velocities of A and B; the difference gives relative velocity in 1D. Velocity (m/s) General 1D meet formula with initial separation: If A starts behind B by distance d ( x B(0) - x A(0) = d > 0 ), then t meet = d v A - v B provided v A > v B . If v A v B , A never catches B. If moving in opposite directions, use algebraic signs consistently. Requires t meet > 0 for a physical meeting in the future. Meeting time (1D, constant velocities) t meet and position from A’s start Relative speed is v A − v B = 35 m/s; closing from opposite directions. v A = +20 m/s v B = -15 m/s Initial separation d = 700 m Two trains on parallel tracks: Train A moves east at 20 m/s; Train B moves west at 15 m/s. At t = 0, they are 700 m apart. How long until they cross, and where relative to A’s starting point? s, m easy Time to meet. Position from A’s start. Feasibility edge cases: (1) For no-drift river crossing, v s < v r means impossible—aiming upstream cannot fully cancel current. (2) For umbrella tilt, if the runner stops ( u=0 ), =0 ; if the rain is a horizontal spray ( v r 0 ), the umbrella must be held nearly horizontal backward. tip y = x - g x 2 2u 2 2 Parametric equations. Eliminate time using x(t). Substitute t into y(t). Projectile trajectory y = x - g x 2 2u 2 2 Uniform gravity g downward No air resistance Launch from origin with speed u at angle Eliminating time between x(t) and y(t) yields a parabolic path; a template for eliminating time in relative-motion setups too. Why include projectile trajectory here? The technique—eliminate time between independent 1D motions—also drives relative-motion diagrams. In rain–man, eliminating t between x = -u t and y = -v r t gives a straight line (constant angle). In river–swimmer, the landing point is set by x = (v r - v s ) t and y = v s t = w . Master this pattern: write two 1D equations sharing the same time, then eliminate t to get a direct relation between the coordinates. Tangent confusion: = opposite adjacent depends on how you defined . Write a small right triangle with labeled legs on every diagram. Many sign and angle errors vanish if the reference line is stated. neet-alert Relative acceleration: When accelerations are constant (or zero), you can subtract them too: a A/B = a A - a B . For constant accelerations, the relative motion follows the same equations of motion as any single particle: along each axis, x A/B = u A/B t + 1 2 a A/B t 2 . Use this to handle pursuits with accelerations (e.g., a police car accelerating to catch a speeding car). Bike’s position after police starts. Police covers t 2 meters. Meeting condition. Discard negative root. medium Use relative displacement with synchronized time origin at police start (t = 0). Bike already 60 m ahead. Catch-up time after police starts A motorbike passes a traffic camera at 20 m/s. 3 s later, a police car starts from rest at the camera and accelerates at 2 m/ s 2 . After how much time from the police start does it catch the bike? Bike: u = 20 m/s, a = 0 Police: u = 0, a = 2 m/ s 2 Start delay for police: 3 s If two objects never meet, their relative speed must be zero. They may have nonzero relative speed but move such that their paths never intersect in time (e.g., parallel lines with same direction and equal speed but separated, or perpendicular motions that miss in time). Meeting needs both positions equal at the same instant. Frames and transformations: Switching frames is algebra. If frame S' moves with velocity u relative to ground frame S, then any object with velocity v in S has velocity v' = v - u in S'. This is the Galilean velocity transformation, valid at everyday speeds (much less than the speed of light). It preserves the linearity of addition and underpins all relative motion problems in this unit. Galilean velocity transformation Velocity in a frame moving with velocity u relative to ground is reduced by u . Vector direction via components: Given R = A + B with A along the x-axis and the angle from A to B, the resultant direction is governed by = B A + B . Check for A + B = 0 (vertical resultant) and choose the correct quadrant with signs of R x, R y . Dimensional and unit sanity checks: Relative velocity has the same units as velocity. Ratios like = u/v r are dimensionless. When converting km/h to m/s, use the standard factor (multiply by 5/18). Always keep track of which axis is horizontal or vertical to avoid mixing components. neet-alert Do not mix distance and displacement in average formulas. Average speed uses total path length; average velocity uses net displacement. In relative motion, meeting condition uses displacement equality, not path lengths. Average speed = total distance / total time. Useful to estimate time budgets but not a vector; do not apply sign. Extended rain–man: If rain has an initial horizontal component due to wind (say v Rx 0 ), and the runner has v Mx = u , then apparent horizontal component is v R/M,x = v Rx - u . The umbrella angle from vertical then satisfies = |v R/M,x | |v R/M,y | with v R/M,y = v Ry (downward). The sign decides whether to tilt backward or forward; physically, tilt opposite to the apparent rain direction. degree, m/s medium Apparent horizontal component eastward. Downward component unchanged by horizontal running. Tilt about 35° towards west (opposite to eastward apparent rain). Apparent rain speed. Angle α from vertical and | v R/M | Compute v R/M = v R - v M component-wise. v R = (+4, +10) m/s (east, down) v M = (-3, 0) m/s (west) Wind-blown rain: Rain has horizontal eastward speed 4 m/s and downward speed 10 m/s. A runner moves west at 3 m/s. Find the apparent rain direction (angle from vertical) and speed relative to the runner. Relative position method summary: For two moving objects, write r A/B (t) = r A(0) - r B(0) + ( v A - v B) t + 1 2 ( a A - a B) t 2 . Meeting requires r A/B (t) = 0 in both components at the same time t. Closest approach uses d dt ( r A/B r A/B ) = 0 . Key terms recap Relative velocity Velocity of one object as observed from another: v A/B = v A - v B . Chosen observer and axes for measuring motion; often the ground is used as an inertial frame. Frame of reference Sideways displacement accumulated due to a transverse velocity component. Drift Component of relative velocity along the line of sight that reduces separation. Closing speed Relation between velocities in two frames moving at constant relative velocity: v' = v - u . Galilean transformation neet-alert Sign convention discipline: Pick axes and stick to them. Write a small legend: +x east, +y north (or down). Many wrong answers in NEET come from flipping a sign halfway. Worked template (build your own): 1) Choose frame and axes; 2) Draw vectors to scale; 3) Write component equations; 4) Apply conditions (meet, no drift, shortest time); 5) Solve for the required time/angle/speed; 6) Unit and feasibility check (e.g., | | 1 ). This checklist prevents most mistakes. easy Component-wise sum. 13-5-12 triangle. Track angle. m/s, degree v DA = (0, +12) m/s v wind = (+5, 0) m/s A drone flies at 12 m/s due north relative to air. A steady wind blows 5 m/s from west to east. Find the drone’s ground speed and track angle. Ground velocity is vector sum: v G = v DA + v wind . Ground speed | v G | and angle east of north Quality-of-solution cues for exams: A computed or larger than 1 signals a setup error or an infeasible demand (e.g., trying no drift with v s < v r ). A negative meeting time indicates the event would have occurred in the past if extrapolated; re-check if the problem asks for future meeting. Always fix quadrant using signs of R x, R y . Magnitude and direction from components When analyzing relative motion, the dot product determines the component of one relative vector along a specific direction or line of sight. Final consolidation: Relative motion is vector subtraction dressed in story form. Anchor every problem with a neat diagram, an axis choice, and a subtraction v A - v B . Convert the story to components, decide the goal (meet, minimize time, zero drift, closest approach), and the algebra will be short and robust.