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 i s ______.
Physics · Kinematics Part-1 (one dimension) · Motion under constant acceleration
JEE Mains 2024 05 Apr Morning Shift

Question

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 i s ______.

Physics · Kinematics Part-1 (one dimension) · Motion under constant acceleration

JEE Mains 2024 05 Apr Morning Shift

G1 Physics · Accepted Answer

For motion from rest with constant acceleration, distance travelled in nth second is: Sₙ = (a/2)(2n - 1) For n = 10: Sₙ = (a/2)(19) For the previous second: Sₙ₋₁ = (a/2)(17) Therefore: Sₙ₋₁/Sₙ = 17/19 Given: 17/19 = 1 - 2/x Therefore: x = 19