# 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.

2
by kishore8898

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by kishore8898

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Distance traveled in 3rd sec. = 0 + a/2(6-1) = 5a/2

Hence ratio = 1:5

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