Solution:
Using \(S_n = u + \frac{a}{2}(2n - 1)\), we get \(u + 3.5a = 20\) and \(u + 4.5a = 25\). Solving gives \(a = 5\text{ m/s}^2\) and \(u = 2.5\text{ m/s}\). Thus, \(S_3 = 2.5 + 2.5(5) = 15\text{ m}\).
With a constant acceleration along x-axis, a car covers a distance of \(20\text{ m}\) during fourth second of its motion and \(25\text{ m}\) during fifth second of its motion. Distance covered by it during third second of motion was
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