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The distance travelled by a particle starting from rest and moving with an acceleration $4/3 ms^{-2}$, in the third second is:
Explanation
Distance travelled in nth second is given by $t_{n} = u + a/2(2n - 1)$ put $u = 0$, $a = 4/3 ms^{-2}$, $n = 3$ $\therefore d = 0 + 4/(3 \times 2)(2 \times 3 - 1) = 4/6 \times 5 = 10/3 m$