Question

A particle experiences constant acceleration for 6 seconds after starting from rest. if it travels a distance d1 in the first two seconds, d2 in the next two seconds and a distance d3 in the last 2 seconds then find their ratio

Answer 1

Let s1 be the distance travelled in 2 sec
s= ut + 1/2 at²
s= s1
u=0 (Body was initially at rest)
s1 =1/2 × a ×4
s1 = 2a 

Now distance travelled in 4 sec is
s= 1/2 × a × (16)
s= 8a
So,
distance traveled during 2 sec to 4 sec interval is
=8a- 2a
=6a
s2= 6a

Now distance travelled in 6 seconds
= 1/2 ×a ×t²
=1/2 × a × (36)
=18 a
 
So,
distance traveled during 4 sec to 6 sec interval is 
=18a - 8a 
=10a
So,
s3= 10a

Thus,
s1:s2:s3 = 2a :6a:10a
s1 : s2: s3 = 1 :3 :5
This the ratio of the distances travelled

Answer 2

Answer:

1:3:5

Explanation:

is the answer

hope it helps you