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We are given the distances covered in the 1st, 2nd, and 3rd seconds of motion: s₁ = 5 m (1st second) s₂ = 9 m (2nd second) s₃ = 13 m (3rd second) For uniformly accelerated motion, the distance covered in the n-th second is given by: sₙ = u + a/2 (2n - 1) Let’s apply this for each second: First second: s₁ = u + (a/2)(2×1 – 1) = u + (a/2) Second second: s₂ = u + (a/2)(3) = u + (3a/2) Third second: s₃ = u + (a/2)(5) = u + (5a/2) Now subtract: s₂ – s₁ = (u + 3a/2) – (u + a/2) = a ⇒ 9 – 5 = a = 4 m/s²
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