The distance travelled by a particle starting from rest and moving with an acceleration \(\frac{4}{3} \text{ m s}^{-2}\), in the third second is
(2008)
Solution:
The distance in the \(n^{\text{th}}\) second is given by \(S_n = u + \frac{a}{2}(2n - 1)\). Here (u=0), \(a = \frac{4}{3} \text{ m s}^{-2}\) and (n=3). So \(S_3 = 0 + \frac{4/3}{2}(2 times 3 - 1) = \frac{2}{3}(5) = \frac{10}{3} \text{ m}\).
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