Given:
Initial velocity (u) = 5.5 m/s due east
Acceleration (a) = -1 m/s^2 due west (negative sign indicates acceleration in the opposite direction of the initial velocity)
Time (t) = 6 seconds

To find:
The distance covered by the particle in the 6th second of its motion.

Explanation:
The particle is initially moving due east with a velocity of 5.5 m/s. However, it is subjected to a constant acceleration of -1 m/s^2 due west, which means its velocity is decreasing at a rate of 1 m/s every second. The negative sign indicates that the acceleration is in the opposite direction of the initial velocity.

Calculating the final velocity (v):
Using the formula v = u + at, where
v = final velocity
u = initial velocity
a = acceleration
t = time

Substituting the values, we get
v = 5.5 m/s + (-1 m/s^2) * 6 s
v = 5.5 m/s - 6 m/s
v = -0.5 m/s

The negative sign indicates that the particle is now moving in the opposite direction of its initial velocity, i.e., westward.

Calculating the distance covered in the 6th second:
To find the distance covered, we need to calculate the displacement of the particle in the 6th second. Displacement is defined as the change in position.

Using the formula s = ut + (1/2)at^2, where
s = displacement
u = initial velocity
t = time
a = acceleration

In this case, since the acceleration is constant, we can use the formula s = ut + (1/2)at^2 to calculate displacement.

Substituting the values, we get
s = 5.5 m/s * 6 s + (1/2) * (-1 m/s^2) * (6 s)^2
s = 33 m - 18 m
s = 15 m

Therefore, the particle covers a distance of 15 meters in the 6th second of its motion.