The displacement of a particle moving with a constant acceleration in the n^thsecond is given by $S_{n} = 2 + 0.4 n$ . Then initial velocity is __________units

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answer is 2.2.

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Detailed Solution

$\begin{array}{rcl} S_{n} & = & 2 + 0.4 n \\ S_{n} & = & u + a (n + \frac{1}{2}) = u + a n - \frac{a}{2} \\ S_{n} & = & a n + (u - \frac{a}{2}) \\ ∴ a n & = & 0.4 n \\ a & = & 0.4 \\ ∴ u - \frac{a}{2} & = & 2 \end{array}$
u=2.2

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