Q.Derive the equation S = ut + (1/2)at^2 from v-t graph.

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 s
• s = ut + ½at² — no v
• v² = 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. …