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JEE and NEET papers like questions such as "a body covers 12 m in the 3rd second and 20 m in the 5th". The equations of motion can do these, but slowly. The nth-second formula does them in two lines, if you know what it means and where it breaks. This post derives it, explains it with a graph, and works four problems including one where it gives the wrong answer.
Is this in the syllabus?
Yes. The JEE (Main) 2026 syllabus and the NEET (UG) 2026 syllabus both list "uniformly accelerated motion" and "relations for uniformly accelerated motion" in Unit 2, Kinematics. The NCERT Class 11 Physics chapter Motion in a Straight Line (Reprint 2026-27) does not print the sₙ formula, but its Example 2.5 proves Galileo's law of odd numbers, which is the same idea for a body falling from rest.
Where does the formula come from?
Use x = ut + ½at², the displacement from t = 0. The displacement in the nth second is what happens between t = n − 1 and t = n:
sₙ = x(n) − x(n − 1) = [un + ½an²] − [u(n − 1) + ½a(n − 1)²]
Expanding and simplifying:
sₙ = u + ½a(2n − 1) = u + a(n − ½)
What does sₙ mean on a graph?
With constant acceleration, the velocity-time graph is a straight line, and displacement is the area under it. Over any one-second strip, the area is the average velocity times one second. For a straight line, the average velocity is the velocity at the middle of the strip, t = n − ½. And u + a(n − ½) is exactly the velocity at that instant.
Worked example 1: two seconds given, find u and a
A body with uniform acceleration covers 12 m in the 3rd second and 20 m in the 5th. Find u and a.
s₃ = u + 2.5a = 12 and s₅ = u + 4.5a = 20.
Subtract: 2a = 8, so a = 4 m/s².
Then u = 12 − 2.5 × 4 = 2 m/s.
Check: s₅ = 2 + 4.5 × 4 = 20 m. Correct.
A shortcut follows: consecutive one-second distances differ by a, so seconds that are two apart differ by 2a.
Worked example 2: Galileo's odd numbers
From rest, u = 0, so sₙ = a(2n − 1)/2. The distances in successive seconds are in the ratio 1 : 3 : 5 : 7, the result NCERT's Example 2.5 proves. With g = 9.8 m/s², a dropped stone falls 4.9 m in the 1st second, 14.7 m in the 2nd and 24.5 m in the 3rd. Our profile of Galileo's inclined plane experiments shows how he found this pattern.
Worked example 3: distance in the last second
A stone dropped from rest falls 45 m in the last second of its fall. Take g = 10 m/s². Find the height.
sₙ = 5(2n − 1) = 45, so 2n − 1 = 9 and n = 5. The fall lasts 5 s.
Height = ½ × 10 × 5² = 125 m.
Check: x(5) − x(4) = 125 − 80 = 45 m.
Here sₙ works neatly because the total time comes out as a whole number. If it does not, solve with x = ½gt² for the full fall and for one second less.
Worked example 4: when the formula misleads
A ball is thrown up at 25 m/s. Take g = 10 m/s², with up positive. Find the distance it travels in the 3rd second.
The formula: s₃ = 25 − 10 × 2.5 = 0 m. That is the displacement.
But the ball stops at t = 25 ÷ 10 = 2.5 s, in the middle of the 3rd second.
At t = 2 s, v = 5 m/s. From 2 s to 2.5 s it rises 5² ÷ (2 × 10) = 1.25 m.
From 2.5 s to 3 s it falls 1.25 m.
Distance = 1.25 + 1.25 = 2.5 m.
sₙ is a displacement. When the velocity changes sign inside the nth second, find the turning point and add the two parts. Our equations of motion guide handles turning points over longer intervals.
Which mistakes cost marks?
Confusing sₙ with x(n). "In the 5th second" means one second of motion; "in 5 seconds" means all five.
Using sₙ as a distance through a turning point. See example 4.
Writing n instead of (2n − 1)/2. The middle of the 3rd second is t = 2.5 s, not 3 s.
Mixing signs. Choose a positive direction before writing u and a.
Using it when acceleration is not constant. The formula assumes uniform acceleration.
A practice routine
Derive sₙ yourself from x = ut + ½at² once, without looking.
Make three two-second problems by choosing u and a, computing two sₙ values, then solving backwards.
For a ball thrown up at 15 m/s, find the distance in the 2nd second, checking for a turning point first.
In past papers, mark every "nth second" question and solve each in two ways.
If a step in your working stops making sense, the Fermi Problem Library has worked problems with a step-by-step solver that explains the reasoning rather than printing an answer.
Start with step 3 above now: find the turning time before anything else.
