Question

a particle at rest starts moving in a horizontal straight line with uniform acceleration.find the ratio of the distance covered during the first and the third second.

Answer 1

Distance covered in 1st sec. = 0 + a/2 (2-1) = a/2
Distance traveled in 3rd sec. = 0 + a/2(6-1) = 5a/2
Hence ratio = 1:5

Answer 2

let s be distance covered in t=1 second and r be the distance covered in T=3 second
s=ut+(1/2)at^2
or, s=0+(1/2)a*(1)^2
or, s=(a/2) -------------(1)
now,
       r=uT+(1/2)aT^2
       r=0+(1/2)a*(3)^2
  or, r=9a/2 ------------(2)
distance covered in two sec(l). =o+(1/2)a*2^2
                                                l=4a/2

 distance covered in 3rd sec. =r-l
                                                  = 9a/2-4a/2
                                                  = 5a/2 

distance covered in 1st sec/distance covered in 3rd sec=(a/2)÷(5a/2)
                                                                             = 1:5