Physics · Motion In A Straight Line · NEET
No. All three equations (v = u + at, s = ut + ½at², v² = u² + 2as) are valid ONLY when acceleration a is constant (uniform). NCERT calls them 'kinematic equations for uniformly accelerated motion'. If a changes with time or position, you must use calculus (integration), not these formulas. Free fall near Earth (a = g = constant) and a car braking at a steady rate are fine; a car whose acceleration keeps rising is not.
u = initial velocity (velocity at t = 0), v = final velocity (velocity after time t), a = acceleration (constant), t = time taken, s = displacement (NCERT writes this as x). Note: s is displacement, NOT distance — for straight-line motion in one direction they are equal, but if the object reverses, use displacement with correct signs. These 5 quantities are what the three equations connect.
Yes. NCERT uses x for displacement and writes the third equation as v² = v0² + 2ax, where v0 is the initial velocity. Most NEET books use s for displacement and u for initial velocity, so v² = u² + 2as. They mean exactly the same thing — just different symbols. If the starting position is not zero, NCERT writes it as (x − x0), the displacement from the start point.
Use the third equation, v² = u² + 2as. It is the only one of the three with NO time in it, so it directly links velocity and displacement. This is why stopping-distance problems (find how far a car travels before stopping) and 'find final speed after falling height h' problems almost always use v² = u² + 2as.
Starting from rest means u = 0. The equations simplify to v = at, s = ½at², and v² = 2as. A body dropped from a height (free fall, u = 0) uses exactly these with a = g. This is the most common exam shortcut — always check whether the object 'starts from rest' or is 'dropped', which both mean u = 0.
A ball is thrown vertically downward with a velocity of 20 m/s from the top of a tower. It hits the ground with a velocity of 80 m/s. The height of the tower is (g = 10 m/s²):
A ball is thrown vertically upward with a velocity of 4 m/s from a bridge. The ball strikes the water after 4 s. The height of the bridge above the water is (g = 10 m/s²):
A bullet enters a wooden block with speed u and, after travelling 24 cm inside, its speed reduces to u/3. Assuming uniform retardation, the further distance it travels before stopping is:
Try the real previous-year questions from this chapter — each with the answer and a full solution.
v = u + at (velocity–time), s = ut + ½at² (position–time), and v² = u² + 2as (velocity–position). Here u = initial velocity, v = final velocity, a = constant acceleration, t = time, s = displacement.
Only for uniformly accelerated motion, i.e. when acceleration is constant in magnitude and direction. For changing acceleration you must integrate (use calculus) instead.
Look at what is given and what is asked. If time is missing, use v² = u² + 2as. If final velocity is missing, use s = ut + ½at². If displacement is missing, use v = u + at. Each equation drops exactly one quantity.
Yes. Free fall is uniformly accelerated motion with a = g (about 9.8 m/s², often taken as 10 m/s² in NEET). Just replace a with g and choose a sign convention for up and down.
The distance in the nth second, Sn = u + a(2n − 1)/2, is derived from s = ut + ½at² (it is s at time n minus s at time n − 1). It is a handy result, not a separate law.
