When acceleration is constant, three equations tie together the initial velocity u, final velocity v, acceleration a, time t and displacement s. They are the workhorses of JEE Main kinematics — the whole skill is picking the right one and keeping the signs honest.
The three equations (uniform acceleration only):
v = u + at — no ss = ut + ½at² — no vv² = u² + 2as — no t (use this whenever time is neither given nor asked)
Plus the distance in the nth second: sₙ = u + a(n − ½) — note this is a distance covered during one second, not a total distance (a favourite trap). And the average velocity under uniform acceleration is (u + v)/2.
1 — Signs are everything. Fix a positive direction first. A body slowing down has a opposite to v (negative if v is positive). Stopping distance comes from v² = u² + 2as with v = 0: s = u²/(2a) — so it scales as u² (double the speed → four times the stopping distance). Total stopping distance with reaction time = u·t_react + u²/(2a).
2 — Motion under gravity is just constant acceleration with a = g (take g = 10 m/s² unless told otherwise). Key results, with up taken positive for a body thrown up at speed u:
- maximum height
H = u²/2g; time to the top = u/g; time up = time down; total flight = 2u/g;
- the speed on returning to the launch level equals
u (same magnitude, opposite direction);
- at the highest point the velocity is zero but the acceleration is still
g downward.
For a body dropped from rest: h = ½gt², v = gt, v² = 2gh, and the distances in successive seconds are in the ratio 1 : 3 : 5 : 7 … (Galileo's odd-number rule; cumulative distances go as t², i.e. 1 : 4 : 9).
3 — Thrown from a height / released from a moving carrier. Set the net displacement to −h (ground below the start) and solve the quadratic −h = ut − ½gt². A body released from a rising balloon keeps the balloon's upward velocity as its own initial velocity (it first goes up, then falls); from a descending lift it starts downward. Thrown up vs thrown down from the same height give the same landing speed (v² = u² + 2gh) but different times.
4 — Two-body and multi-stage problems. For two bodies under gravity, write each one's position on a common clock and set them equal to find where/when they meet; since both share the same g, their relative acceleration is zero, so their separation changes at the constant relative speed. For multi-stage motion (accelerate → cruise → brake), carry the end velocity of each phase into the next. A body passing a height twice on the way up and down has the two times as the roots of h = ut − ½gt² (t₁ + t₂ = 2u/g, t₁t₂ = 2h/g).
How this concept is examined. JEE Main asks for a missing kinematic variable by direct substitution, an nth-second distance, a stopping distance (mind the u² scaling), a free-fall time/speed, a max height or time of flight, a meeting point of two bodies, or a Galileo-ratio result. The equations are few; the marks reward choosing the time-free equation when appropriate and never dropping a sign.
Equations of motion are introduced in the NCERT Class 11 Physics chapter on Motion in a Straight Line, and are among the most searched board-revision topics under queries like 'equations of motion formula class 11 physics' and 'kinematics important questions.' Because nearly every JEE Main and NEET mechanics numerical starts from these three relations, mastering the sign conventions here pays off well beyond the CBSE syllabus.