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Equations of motion look like the easiest chapter in Class 11 physics: three or four formulas, a few substitutions. Yet they cause a steady stream of lost marks in school exams and JEE Main, mostly from two errors: picking an equation that needs a quantity you do not have, and mixing up signs when something moves up and then down. This post shows a quick way to choose the equation, a sign rule that always works, three worked examples and a practice routine.
Which equations are we talking about?
The NCERT Class 11 Physics chapter on motion in a straight line derives the kinematic equations for uniformly accelerated motion, in its notation:
v = v₀ + at
x = v₀t + ½at²
v² = v₀² + 2ax
It also shows that the displacement equals the average of the initial and final velocities times the time, x = ½(v₀ + v)t, for constant acceleration only. The JEE (Main) 2026 syllabus lists uniformly accelerated motion, relations for uniformly accelerated motion, and velocity-time and position-time graphs under Unit 2, Kinematics. Check the syllabus for your own exam year, since NTA publishes a new one each cycle.
Below, u is the initial velocity (NCERT's v₀) and s the displacement.
How do you choose the right equation?
There are five quantities: u, v, a, t and s. Every equation contains four of them. So list what you know and what you want, find the one quantity that is neither, and use the equation without it.
No s: v = u + at
No v: s = ut + ½at²
No t: v² = u² + 2as
No a: s = ½(u + v)t
No u: s = vt − ½at²
The last one is not in NCERT's list, but it follows from the others and saves time when the final velocity is given, such as a car that ends at rest.
What is the one sign rule that always works?
Choose a positive direction before you write anything, and write it on the diagram. Then every vector, u, v, a and s, gets a sign relative to that direction. Never put g = 9.8 into an equation without asking which way it points. NCERT does exactly this in its worked examples: it takes upward as positive and writes a = −g.
Worked example 1: a ball thrown up from a balcony
A ball is thrown straight up at 10 m/s from a balcony 15 m above the ground. Take g = 10 m/s². When does it hit the ground, and how fast?
Take upward as positive and the origin at the balcony. Then u = +10 m/s, a = −10 m/s², and the ground is at s = −15 m.
We want t, and we do not need v. Use s = ut + ½at²: −15 = 10t − 5t².
Rearrange: 5t² − 10t − 15 = 0, so t² − 2t − 3 = 0, giving (t − 3)(t + 1) = 0. Take t = 3 s; the negative root has no meaning here.
For the impact speed, use v² = u² + 2as = 100 + 2(−10)(−15) = 400, so v = ±20 m/s. The ball is moving down, so v = −20 m/s.
Check with v = u + at = 10 − 30 = −20 m/s. The answers agree.
There is no need to split the motion into "up" and "down" parts. As NCERT notes for its own version of this problem, one equation covers the whole flight because the acceleration is constant throughout.
Worked example 2: a braking car
A car moving at 72 km/h brakes uniformly and stops in 40 m. Find the deceleration, the stopping time and the distance covered in the third second of braking.
Convert first: 72 km/h = 72 × 1000/3600 = 20 m/s.
No t needed: 0 = 20² + 2a(40), so a = −400/80 = −5 m/s².
No s needed: 0 = 20 + (−5)t, so t = 4 s.
Third second: s(3) − s(2) = (60 − 22.5) − (40 − 10) = 37.5 − 30 = 7.5 m.
The shortcut sₙ = u + a(2n − 1)/2 gives 20 − 5 × 5/2 = 7.5 m too, but subtracting two displacements is safer to remember.
Worked example 3: distance is not displacement
A ball is thrown up at 20 m/s. Take g = 10 m/s². Find its displacement and the distance travelled in the first 3 s.
Displacement: s = 20 × 3 − 5 × 9 = 60 − 45 = +15 m.
But the ball turned round. It stops when v = 0, at t = 2 s, having risen 20 × 2 − 5 × 4 = 20 m. In the next second it falls 5 × 1² = 5 m.
Distance = 20 + 5 = 25 m.
Whenever v changes sign during the interval, find the turning point first.
Why do students mix up these ideas?
Physics education research found that the trouble is often conceptual, not algebraic. Trowbridge and McDermott (1980) interviewed students about two objects moving at once and found that almost every failure to compare their velocities came from using position to judge velocity: if two objects were side by side, students assumed they had the same speed. A follow-up on acceleration (1981) found related difficulties with acceleration as the rate of change of velocity, Δv/Δt. Test yourself on a classic case: a ball at the top of its flight is momentarily at rest, but its acceleration is not zero. It is still −g, because its velocity is changing from upward to downward.
Our guide to reading physics graphs covers slope and area on position-time and velocity-time graphs.
Common mistakes and how to avoid them
Using these equations when acceleration is not constant. If a depends on time or position, use calculus, such as v dv = a dx.
Forgetting units. Convert km/h to m/s before anything else.
Taking the wrong root. A negative time or a speed with the wrong sign should be rejected using the physical situation, not dropped automatically.
A practice routine for this week
Day 1: Ten problems. For each, write the five quantities, circle the one not needed, and name the equation before solving.
Day 2: Five vertical-throw problems, each solved with one equation and upward positive, then again with downward positive. Both must give the same physical answer.
Day 3: Five problems that ask for distance, so you practise finding turning points.
Day 4: A mixed timed set of eight, including two from projectile motion, where the same equations apply to each component.
If you get stuck on the set-up rather than the algebra, the Fermi Problem Library has worked problems with a step-by-step solver that explains the reasoning, so you can compare your choice of equation with the logic of a full solution.
Start with worked example 1: redo it taking downward as positive and confirm you get the same time and speed.
