Question

a particle experience a constant acceleration for 6 seconds after starting from rest if it travels a distance d one in the first to second d 2 in the next two seconds and a distance three in the last two second then​

Answer 1

Answer:

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

Explanation:

the answer will be 1:3:5