Question

A particle moves with constant acceleration for 6 seconds after starting from rest. The distance travelled
during the consecutive 2 seconds interval are in the ratio
(1) 1:1:1
(2) 1:2:3
(3) 1:3:5
(4) 1:5:9​

Answer 1

Answer: answer is 3)1:3:5

Explanation:S=1/2×a×t^

At t=2

S=1/2×a×2^ =0.5a4=2a

At t=4

S=1/2×a×4^=0.5a16=8a

8a-2a=6a

At t=6s

S=1/2×a×6^=0.5a36=18a

18a-8a=10a

2a:6a:10a= 1:3:5

Answer 2

option (3) - 1:3:5

Explanation:

Given,u=0

we know,

$s = ut \: + \: 0.5at {}^{2}$

$s = at {}^{2}$

as u=O

for 2 seconds ,

$s = 0.5a{2}^{2}$

s=2a

for next 2 seconds,i.e,t=4

$s = 0.5a4 {}^{2}$

s1= 8a

( note: its total distance travelled from 0 to 4 seconds

we have subtract above distance for getting travelled distance between 2 to 4 seconds)

8a-2a = 6a = s2

for next 2 seconds,i.e.,t=6

$s = 0.5a6 {}^{2}$

s = 18 a

distance travelled between t =4 to 6 seconds

18 a- 8a= 10a = s3

Ratio of s1 : s2: s3

2a :6a: 10a

1:3:5,

so the correct option is (3)