An electron starting from rest has a velocity that increases linearly with time, i.e., $v = k t$ , where $k = 2 m / s^{2}$ . The distance covered in the first $3 s$ is

9 m

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16 m

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27 m

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36 m

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Solution

The correct option is A $9 m$
$\frac{d s}{d t} = k t$
Using integration $\Rightarrow s = \frac{1}{2} k t^{2} = \frac{1}{2} \times 2 \times 3 \times 3 = 9 m$

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