MixedJEE Physics · Original learning card10 original chapter questions

Distance in the Nth Second

For constant acceleration, displacement during the nth second of unit duration is s_n = u + (a/2)(2n-1). More generally, use x(n)-x(n-1), and distinguish signed displacement from distance if reversal occurs.

Why this shows up in the exam

Successive-second comparisons · Finding acceleration from interval data · Free-fall distance ratios

Learn the idea

Subtract cumulative displacements to isolate the displacement during one numbered second. The nth second is an interval, not the instant t = n. Its displacement is what the particle has accumulated by n seconds minus what it had accumulated by n-1 seconds.

🧠 Memory hook: Nth second means total at n minus total at n minus one.

Get this one clearly and it pays off every single time it shows up in the paper. 🎯

Formulas & facts to keep ready

  • s_n = x(n) - x(n-1) — displacement during the nth one-second interval
  • s_n = u + (a/2)(2n-1) — constant-acceleration nth-second displacement

How to approach it

  1. 1Write cumulative x(t)
  2. 2Evaluate at the two interval ends
  3. 3Check the velocity sign inside the interval

Common slip-ups that cost marks

  • •Using x(n) as the nth-second distance
  • •Forgetting the interval starts at n-1
  • •Using signed displacement as distance across a reversal

🌟 That's the whole idea — you've got this. Try the practice set below; every question you attempt makes it stick a little harder.

Original chapter practice

Original questions for this chapter, not past-paper questions or an exact mapping to this individual concept.

Question 1 of 10

A particle has initial speed 2 m/s and constant acceleration 3 m/s^2 for 4 s. What distance does it cover?

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