I hope it is k equal to 2m per second square.
sol-
V=kt
v÷t=k,here k works as acceleration
so,a=2 m per second's square
S=ut+1/2 at2
=0×3+1/2×2×9
=9m ( ans )

Introduction:
In this problem, we are given that the velocity of an electron increases linearly with time. The relationship between velocity and time is given by v = kt, where k is a constant equal to 2 m/s. We need to determine the distance covered by the electron in the first three seconds.

Formula:
To solve this problem, we can use the formula for distance covered (s) as a function of time (t) and initial velocity (u):

s = ut + (1/2)at^2

However, since the electron starts from rest (initial velocity, u = 0) and the acceleration (a) is not given, we need to find an alternate approach to solve the problem.

Approach:
Since the velocity of the electron is increasing linearly with time, we can visualize this as a straight line on a velocity-time graph. The area under this line represents the distance covered by the electron.

Explanation:
To find the distance covered by the electron in the first three seconds, we need to calculate the area under the line on the velocity-time graph.

Step 1: Calculate the velocity at the end of the first three seconds.
Using the formula v = kt, we substitute t = 3 seconds and k = 2 m/s:
v = 2 * 3 = 6 m/s

Step 2: Calculate the average velocity during the first three seconds.
The average velocity is given by the formula:
Average velocity = (initial velocity + final velocity) / 2

Since the initial velocity is 0, we can substitute the final velocity:
Average velocity = (0 + 6) / 2 = 3 m/s

Step 3: Calculate the distance covered using the average velocity.
Using the formula s = vt, we substitute the average velocity and time:
Distance covered = 3 * 3 = 9 meters

Conclusion:
Therefore, the distance covered by the electron in the first three seconds is 9 meters.