An electron starting from rest has a velocity that increases linearly with time that is $v = k t$ where $k = 2 m / s^{2}$ . Distance covered in first $3 s$ will be:

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 $9 m$
Acceleration of the electron is given by:
$a = Δ v / Δ t = k$

Since, acceleration is constant, we can use second equation of motion.
$s = u t + \frac{1}{2} a t^{2}$
Given: $u = 0, t = 3$ and from above $a = k = 2$
Substituting values,
$s = 0 \times 3 + \frac{1}{2} \times 2 \times 3^{2}$
$s = 9 m$

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An electron starting from rest has a velocity that increases linearly with time that is v=kt where k=2m/s2. Distance covered in first 3 s will be:

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