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A point particle starting from rest has a velocity that increases linearly with time such that $v = p t,$ where $p = 4 m s^{- 2} .$ The distance covered in the first 2 s will be

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4 m

10 m

8 m

6 m

answer is C.

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Detailed Solution

Given, $v = p t$

$\Rightarrow \int_{0}^{x} d x = p \int_{0}^{2} t d t = \frac{p t^{2}}{2} = \frac{4 \times 4}{2} = 8 m$

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