An electron starting from rest has a velocity that increases linearly with the time that is v = kt, where k = 2 m/sec² . The distance travelled in the first 3 seconds will be

The correct option is 9 m.

Step-by-step Explanation:

We are given that the velocity of an electron increases linearly with time according to the equation:
$v=kt$
where:
$k=2{\text{ m/s}}^2$

We know that velocity is the rate of change of displacement (distance) with respect to time:
$v=\frac{ds}{dt}$

Rearranging the equation to solve for distance $ds$:
$ds=vdt$

Substitute the expression for $v$ into the equation:
$ds=ktdt$

To find the total distance travelled in the first 3 seconds (from $t=0$ to $t=3$ seconds), we integrate both sides:
$s={\int }_0^{3}ktdt$

Since $k$ is a constant, we can pull it out of the integral:
$s=k{\int }_0^{3}tdt$

Integrating $t$ gives:
$s=k{\left[\frac{t^2}{2}\right]}_0^3$

Substitute the limits and the value of $k=2{\text{ m/s}}^2$:
$s=2\times (\frac{3^2}{2}-\frac{0^2}{2})$
$s=2\times \frac{9}{2}$
$s=9\text{ m}$

Therefore, the distance travelled in the first 3 seconds is 9 m.