Solution:

Given, the acceleration of the bus, a = 2 m/s^2

Let u be the initial velocity of the bus, which is 0 m/s. We need to find the ratio of the distance covered in the 6th second to that covered in 6 seconds.

Using the formula of distance covered by a body with uniform acceleration, we can find the distance covered by the bus in the first 6 seconds as follows:

- In the 1st second, the distance covered (s1) = u + 1/2 a (1)^2 = 0 + 1/2 (2) (1)^2 = 1 m
- In the 2nd second, the distance covered (s2) = u + 1/2 a (2)^2 = 0 + 1/2 (2) (2)^2 = 4 m
- In the 3rd second, the distance covered (s3) = u + 1/2 a (3)^2 = 0 + 1/2 (2) (3)^2 = 9 m
- In the 4th second, the distance covered (s4) = u + 1/2 a (4)^2 = 0 + 1/2 (2) (4)^2 = 16 m
- In the 5th second, the distance covered (s5) = u + 1/2 a (5)^2 = 0 + 1/2 (2) (5)^2 = 25 m
- In the 6th second, the distance covered (s6) = u + 1/2 a (6)^2 = 0 + 1/2 (2) (6)^2 = 36 m

Therefore, the total distance covered by the bus in 6 seconds = s1 + s2 + s3 + s4 + s5 + s6

= 1 + 4 + 9 + 16 + 25 + 36

= 91 m

Now, the distance covered by the bus in the 6th second = s6 - s5 = 36 - 25 = 11 m

Hence, the required ratio of the distance covered in the 6th second to that covered in 6 seconds is:

= 11: 91

= 1: 8.27

= 34: 281.82 (multiplying both terms by 34)

Therefore, the correct option is (3) 34 : 11.

Final Answer: The ratio of the distance covered in the 6th second to that covered in 6 seconds is 34 : 11.

We use a particular equation of motion for this question ,which is ,
Sf ( in th second ) = u + 1/2a(2t-1)
So, Here t=6 ,a=2m/s^2 u=0
S6=0+1/2×2(2×6+1)
=11m

Now , we know
S=ut+1/2at^2
=0+1/2×2×6^2
=36 m

So ,ratio =11:36
Option A is correct