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The velocity time relation of an electron starting from rest is given by $υ = kt$ , where $k = 2 m / s^{2}$ The distance traversed in 3 sec is

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9 m

16 m

27 m

36 m

answer is A.

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Detailed Solution

$υ = kt, k = 2 m / s$
$\begin{array}{l} \frac{dx}{dt} = kt \\ \int dx = \int ktdt \\ x = \frac{{kt}^{2}}{2} \\ if t = 3, k = 2 \\ x = \frac{2}{2} \times (3)^{2} = 9 m \end{array}$

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