A body moves on a frictionless plane starting from rest. If Sₙ is distance moved between t=n-1 and t=n and Sₙ₋₁ is distance moved between t=n-2 and t=n-1, then the ratio Sₙ₋₁/Sₙ is (1-2/x) for n=10. The value of x is __________.

Solution

Physics sir · 4 steps

Starting from rest, distances in successive equal seconds go as 1, 3, 5, 7, the odd numbers.

1. Uniform acceleration. On a frictionless plane a constant force gives a constant acceleration a, and the body starts with u = 0. So the distance covered in the nth second follows the nth-second formula.

   Sₙ = u + a/2(2n-1) = a/2(2n-1)

2. Take the ratio. The factor a/2 cancels, so Sₙ₋₁/Sₙ = (2(n-1)-1)/(2n-1) = (2n-3)/(2n-1) = 1 - 2/(2n-1).
3. Put n = 10. S₉/S₁₀ = 17/19 = 1 - 2/19. Compare this with 1 - 2/x.
4. So. x = 2n - 1 = 19, so the answer is 19.

The common slip: 21. Writing the nth-second distance as a/2(2n+1) makes the ratio 19/21 and gives 21, not 19.

The idea to keep. From rest with constant acceleration, the distance in the nth second is proportional to (2n-1). Ratios of single-second distances are just ratios of odd numbers.

Answer from the NTA final answer key. Solution by PrepWiser.