Equations of Motion 1D Think of velocity as how fast you are going, and acceleration as the speed of your speed. In uniformly accelerated motion, you are changing your speed by the exact same amount every single second. Picture a speedometer where the needle is not steady, yet it is not jerking either. It glides at a perfectly steady pace. With every passing second you cover more ground than the second before, because you are now faster than a moment ago. The direction of acceleration does not wobble; it stays fixed, and its magnitude is constant. This regularity is what lets us write compact relations between displacement, velocity, acceleration, and time. Equations of Motion 1D Interpreting Motion Graphs Graph Type Slope Meaning Area Under Curve Curvature Implication Slope slides you forward from x v a , while Area pulls you back from a v x . Velocity-Time ( v-t ) Instantaneous Acceleration ( a = dv dt ) Displacement (Area = v dt ); Distance (Total Magnitude) Linear Constant a ; Upward curve Increasing a Position-Time ( x-t ) Instantaneous Velocity ( v = dx dt ) No direct physical significance in basic kinematics (Concave up) a > 0 ; (Concave down) a < 0 Acceleration-Time ( a-t ) Jerk ( j = da dt ) Change in Velocity ( v = v f - v i ) Horizontal Constant a ; Slope Variable a Acceleration-Position ( a-x ) No standard term (Rate of change of a w.r.t x ) Change in Kinetic Energy per unit mass ( a dx = v f 2 - v i 2 2 ) Slope indicates the spatial variation of the field or force Speed-Time ( |v|-t ) Magnitude of Tangential Acceleration ( a t ) Total Distance traveled Cannot be negative; Always remains on or above the t -axis Force-Position ( F-x ) Negative of the Spring Constant (if F = -kx ) Work Done ( W = F dx ) Area represents energy transfer (Potential/Kinetic conversion) interpreting motion graphs Uniform acceleration is the simplest non-trivial motion: there is change, but the rate of change is perfectly steady. In one dimension, every quantity can carry a plus or minus sign to encode direction. The three famous relations connect five symbols: u (initial velocity at t=0 ), v (velocity after time t ), a (constant acceleration), s (displacement in the chosen positive direction), and t (elapsed time). Why do we care? Because a large fraction of real motions are either approximately uniform in acceleration (car starting or braking, stone under gravity ignoring air) or can be broken into short time slots that are. These equations also power graph shortcuts: the slope of a v – t line is a , and the area under it is s . With thoughtful sign choices, the same formulas handle speeding up, slowing down, going upward against gravity, or dropping freely. remember Rolling a marble down a long smooth ramp: gravity gives a nearly constant pull. Near the top the marble creeps; halfway it is brisk; at the bottom it is fast. The steady "push" makes speed rise evenly each second. Uniformly Accelerated Motion (UAM) 1D motion where acceleration a is constant in magnitude and direction for the entire interval considered. Velocity at t=0 along the chosen positive axis. Can be positive, negative, or zero. Initial velocity ( u ) Velocity at time t along the same axis. Sign encodes direction. Final velocity ( v ) Rate of change of velocity. For UAM, a is constant. Positive means pushing in the positive axis direction. Acceleration ( a ) Displacement ( s ) Vector change in position over the interval; in 1D it is signed distance along the axis. Elapsed duration of motion; t 0 with SI unit s . Time ( t ) Sign convention is the heart of reliability. Choose the positive direction once (usually upward or to the right) and never change it inside a solution. Substitute every known with its sign: for free fall if upward is positive, then a=-g . Displacement s is signed too: upward +s , downward -s under the same convention. A neat test: when you plug numbers into the equations, you should not need to manually switch signs later—the algebra will take care of direction. tip Under gravity problems, fix one convention early. If you take upward as positive, always use a=-g with g 9.8 , m/s 2 (or 10 , m/s 2 for neat numbers). If downward is positive, then a=+g and displacements downward are positive. Velocity grows linearly with time if acceleration is constant. First equation Useful for finding the final velocity when acceleration is constant and the time interval is known. Meaning: every second, velocity gains a . If a>0 and u>0 , speed rises; if a<0 while u>0 , the object slows and may reverse after some time. The instant of stopping (if it happens) is when v=0 , giving t stop =-u/a for u ,a<0 . Displacement is the sum of a uniform part ut and a quadratic part from acceleration. Second equation The total displacement achieved during constant acceleration can be found by calculating the area under the velocity-time graph. Use definitions and the first equation. Separate variables. Integrate with limits. Evaluate integrals. s = ut + 1 2 a t 2 s = ut + 1 2 a t 2 1D motion Acceleration a is constant At t=0 , position is the origin or displacement is measured from the start Geometric view on a v – t graph: the area under the straight line from 0 to t equals the displacement. That area is a trapezium with parallel sides u and v , giving s= (u+v) 2 t . Substituting v=u+at reproduces s=ut+ 12 at 2 . Velocity–displacement link independent of time. Third equation Use this time-free relation when calculating displacement or final velocity is required, but the time interval is unknown. 1D motion with constant a v 2 = u 2 + 2 a s v 2 = u 2 + 2 a s Chain rule with v=ds/dt . Separate variables. Integrate with limits. Evaluate the integrals. The time-free form is powerful for braking and stopping-distance questions, or fall-from-rest cases where u or s is fixed and time is not asked. Boundary check: if a 0 , the third equation reduces to v u unless s grows unbounded; the equation is intended only when a is a fixed non-zero constant over s . Average-velocity form in UAM Valid only when acceleration is constant so velocity changes linearly. Calculates the velocity of one object as measured from the moving reference frame of a second object. When a is constant, instantaneous velocity varies linearly from u to v in time t . The average over that interval is the arithmetic mean, (u+v)/2 . Do not use this when acceleration is not constant; in that case you must integrate the actual v(t) . Displacement in the nth second Displacement between t=n-1 and t=n for constant a . Total displacement after time t . Displacement in the nth second is the difference of totals. Simplify the quadratic terms. s n = u + a 2 (2n-1) s n = u + a 2 (2n-1) 1D motion Constant acceleration Time steps of exactly 1 second Interpretation: s n tells the displacement only during the nth one‑second slot. For a>0 and u 0 , these slots increase linearly. In gravity problems, s n sequences like 1,3,5, times a scale appear naturally. Always check whether a turn-around (velocity becoming zero and reversing) occurs inside that 1 s window; if it does, split the window. Average velocity over any interval is displacement divided by time: v avg = x t . Vector in 1D with sign. Finite non-zero interval t Any motion (no need for constant a ) Definitions of changes. Ratio definition; direction by sign of x . v avg = x t v avg = x t For constant a , the displacement during the nth second is s n = u + a 2 (2n-1) . Use carefully if motion reverses mid-second. Projectile path y = x - g x 2 2 u 2 2 . In 1D vertical projection, set x=0 to recover UAM in y . Uniform g downward No air resistance Origin at launch point y = x - g x 2 2 u 2 2 y = x - g x 2 2 u 2 2 Independent 1D UAM along axes. Eliminate time. Substitute and simplify to parabola. neet-alert Displacement in the nth second s n is NOT the total in n seconds. It is the difference s(n)-s(n-1) . Many lose marks by plugging n into s=ut+ 12 at 2 and calling that s n . neet-alert Distance vs displacement: area under a v – t graph gives signed displacement. If velocity becomes negative, the area below the axis subtracts. For total distance, take absolute area pieces and add. Stroboscopic drop: equal time gaps, increasing spacing 1, 4, 9, 16… showing s t 2 in free fall. Multiple bright spheres at t=0,1,2,3,4 falling with distances labeled 0,1,4,9,16 on a grid titled Acceleration due to Gravity. The drop image encodes the quadratic law. In each 1 s slot the vertical displacements are in the ratio 1:3:5:7: when starting from rest, because s n = g 2 (2n-1) downward. The cumulative distances at t=1,2,3,4 , s scale like 1,4,9,16 , matching the growing gaps you see. Car in 1D: green velocity arrows grow, red acceleration arrows are equal—constant a . Row of cars at t=0 to t=4 with longer green velocity vectors and equal red acceleration vectors underneath. Vector snapshots make the meaning of constant acceleration tangible. The red vectors all have the same length and direction: a is fixed. Velocity vectors increase by the same amount each second, so their tips lie on a straight line when plotted against time. If a were zero, all green arrows would be equal. Transparent dashboard showing a velocity-time graph as a straight line from 0 to about 150 m/s in 10 s with a constant acceleration indicator. Cockpit HUD: a straight rising v – t line with a steady acceleration bar—textbook v=u+at . On a v – t plot, the slope equals acceleration. A straight line means a is constant and equals the slope. The area under that line from 0 to t is displacement: if the line starts above zero, the area is a rectangle plus a triangle; if it passes through zero, the triangular area alone gives s= 12 at 2 from rest. 0 to t Time Velocity u to v m/s vt control control v=u+at derived Area under the line from 0 to t equals displacement s. t=0, v=u t, v u + a t Straight line with slope a, intercept u. Area method recap: s = rectangle area ut plus triangle area 12 (v-u)t = 12 at t . This gives s=ut+ 12 at 2 . If u=0 , the area reduces to a single triangle and s= 12 at 2 . For a braking line declining to zero, the area is still positive until the line dips below the axis; then signed area turns negative (displacement reverses). Slope at any point on s–t is velocity; curvature reflects sign of acceleration. control control s=ut+1/2 at 2 derived Parabola opening up if a>0 and down if a<0; tangent slope equals instantaneous velocity. Displacement depends on u,a Time 0 to t st The s – t curve is quadratic. If a>0 , the slope keeps increasing: the curve bends upward. If a<0 , slope decreases and the curve bends downward, possibly turning flat at the top when v=0 before descending. At any instant, a tangent line slope equals v ; the steeper the tangent, the larger the speed. Decision guide for constant-a problems u, a, t v, s v=u+at; s=ut+ 12 at 2 Direct substitution u, v, a v 2 = u 2 +2as Time not needed u, s, t s=ut+ 12 at 2 Solve for a v, s, t s= (u+v) 2 t Arithmetic mean only for constant a Start from rest (u=0), a s in each second s n = a 2 (2n-1) Sequence of odd multiples Knowns Wanted Best starting equation(s) Notes SUVAT: S (displacement), U (initial velocity), V (final velocity), A (acceleration), T (time). Think: Smart Users Verify All Timings. Take g=+9.8 , m/s 2 always; signs can be fixed later. Choose an axis first. If upward is positive, use a=-g throughout. Inconsistent signs make correct equations give wrong numbers. Average velocity in UAM equals (u+v)/2 for any motion. That mean works only when acceleration is constant. In general, v avg = x/ t ; if acceleration varies, integrate v(t) . Negative acceleration always means the object is slowing down. Speed changes depend on the relative signs of v and a . If v<0 and a<0 , speed increases in magnitude. easy Use first and second equations. A scooter starts from rest and accelerates uniformly at 2.0 , m/s 2 for 6.0 , s . Find its velocity and displacement. v and s after 6.0 s u = 0 a = 2.0 , m/s 2 t = 6.0 , s SI The example shows how area under the v – t triangle also gives s= 12 at 2=36 , m . If you plotted the line from 0 to 12 m/s in 6 s, the triangle’s area would match exactly. SI u = 25 , m/s a = -2.5 , m/s 2 A car moving at 25 , m/s uniformly decelerates at -2.5 , m/s 2 . Find (i) time to stop and (ii) stopping distance. t stop and s stop Use v=0 at stop; then use time-free equation for distance. medium Note the convenience of v 2 =u 2 +2as : it avoids an extra step with time. The sign of a takes care of the positive distance automatically. At top, v=0 so use v 2 = u 2 +2as. By t=3 s the ball is at top and then descending. Position at 4 s relative to start. Negative means 15 m downward during the 5th second. H, T, and s 5 A ball is thrown vertically upward with u=30 , m/s . Taking upward as positive and g=10 , m/s 2 , find (i) maximum height, (ii) total time of flight, (iii) displacement during the 5th second. SI u = +30 , m/s a = -10 , m/s 2 Use UAM equations and piecewise logic for the nth second if turn-around occurs. hard The 5th-second result highlights the turn-around trap. The formula s n =u+ a 2 (2n-1) with u=30 and a=-10 gives s 5 =30-5 9=-15 , m , which matches the careful difference y(5)-y(4) . This works only because we measured upward positive and kept a=-g consistently. tip If a reversal happens inside an nth one‑second window, the single s n formula still works provided your signs are consistent. If you are in doubt, compute y(n) and y(n-1) separately and subtract. Choose axis and fix signs; write u, a with signs. Sketch a quick v–t line; mark known points. Decide: do I need time? If not, try v 2 = u 2 +2as. If time is known or asked, pair v=u+at with s=ut+1/2 at 2. Cross-check with units and limiting cases (a→0, t→0). Strategy for any constant-a problem Quick checks before final answer Does the sign of displacement match the physical direction? If stopping, is t stop positive (u and a must be opposite in sign)? For free fall, do the odd-number gaps appear as expected? Have you rounded to 2 significant figures for NEET-style numeric outputs? Exam speed tips: write the knowns with units in a clean list, choose the minimal equation that eliminates unknowns fastest, and keep mental checks: if braking distance scales like u 2 , doubling speed quadruples stopping distance. Graph sketches prevent sign errors and provide instant estimates even before calculation. Dimensional sense-making keeps you safe. In s=ut+ 12 at 2 , both terms carry units of length: (m/s) s=m and (m/s 2 ) s 2 =m . In the nth‑second expression the hidden factor is the 1 s window: u 1 , s and a (1 , s ) 2 /2 (2n-1) . If a result shows mixed units, re-check algebra. Common free-fall modeling choices: either take upward positive with a=-g or downward positive with a=+g . Both give the same physics when used consistently. For quick mental math NEET allows g=10 , m/s 2 , but if answers are close, prefer 9.8 , m/s 2 . Graph-to-equation translations: a straight rising v – t line through (0,u) has equation v(t)=u+at where a is the slope. If the line crosses the time axis at t=-u/a , that is the hypothetical time when velocity would have been zero if extended backward. The trapezium area formula becomes a quick mental shortcut once you visualize these shapes. Turning-point logic: set v=0 to find the instant of reversal. After that instant, velocity changes sign but acceleration does not. Split the motion into phases on either side if a question spans across the turning point. This avoids mixing displacements of opposite signs unintentionally. When solving piecewise, write positions using the same origin. Example: for a vertical throw with origin at launch, y(t)=ut- 12 gt 2 works for all t (both ascent and descent) as long as you keep the sign of a fixed. Then any per-second displacement is just a difference y(n)-y(n-1) . Average speed versus average velocity: on a round trip of equal path out and back at different speeds, the displacement is zero so average velocity is zero, but the distance is non-zero so average speed is positive. This contrast shows why you must read the question word: speed or velocity. Limiting and extreme cases provide confidence. As t 0 , s ut , reflecting the initial straight-line tangency of the s – t graph. As a over a tiny t , the quadratic term dominates, producing a strongly curved start. If u=0 and a>0 , the motion is purely quadratic in time with v t and s t 2 . Units and rounding: NEET numerical answers typically accept 2 significant figures unless options dictate otherwise. Keep track of m , s , and m/s carefully; copy them across steps so unit cancellations show plainly. This habit also catches algebra slips early. Uniformly Accelerated Motion; a is constant. UAM Initial velocity (u) Velocity at the start ( t=0 ). Velocity at time t . Final velocity (v) Rate of change of velocity; constant in UAM. Acceleration (a) Signed change in position. Displacement (s) Time (t) Elapsed duration. Displacement between t=n-1 and t=n . Nth-second displacement Average velocity Total displacement over total time; v avg = x/ t . Recap of key terms