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JEE Main 2024 (4 Apr, Shift 1) · Equations of Kinematics

Equations of Kinematics question from JEE Main 2024 (4 Apr, Shift 1)

Q9JEE Main 2024 · 4 APR · SHIFT 1Easy

Equations of Kinematics

A body travels 102.5 m102.5\text{ m} in nnth second and 115.0 m115.0\text{ m} in (n+2)(n+2)th second. The acceleration is:

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Answer: (a)

  1. Sn=u+a2(2n−1)S_n = u + \frac{a}{2}(2n - 1), so Sn+2−Sn=a2(4)=2aS_{n+2} - S_n = \frac{a}{2}(4) = 2a.
  2. 2a=115.0−102.5=12.52a = 115.0 - 102.5 = 12.5
  3. a=6.25 m/s2a = 6.25\text{ m/s}^2
  4. So the answer is (a) 6.25 m/s26.25\text{ m/s}^2.
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