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a particle experiences constant acceleration for 6s after starting from rest. if it travels distance s 1 in the first 2s, distanse s 2 in the next 2s and distance s 3 in the last 2s; then s 1 :s 2 :s 3 is ?

a particle experiences constant acceleration for 6s after starting from rest. if it travels distance s1 in the first 2s, distanse s2 in the next 2s and distance s3 in the last 2s;  then s1:s2:s3 is ?

Grade:12

4 Answers

Archita
132 Points
8 years ago
s(1)=S(2)= u +[n-(1/2)a
       =   [2-(1/2)] 6
         = 9
s(2)=S(4)= sme formula
        = [4-(1/2)]6
         =21
 
s(3)= S(6)= same formula
      =[6-(1/2)]6
       =33
 s(1):s(2):s(3)= 9:21:33
                          =3:7:11
 
Archita
132 Points
8 years ago
im sry i misintepreted the question
s=ut+(1/2)at^2
 =(1/2)at^2
v=u+at
v=at
v=u
s(2)=ut+(1/2)at^2
       =at*t+(1/2)at^2
      =3at^2/2
v=u+at
v=at+at
v=2at
s(3)=ut+(1/2)at^2
       2at^t+(1/2)at^2
      5/2 at^2
 
s1:s2:s3= 1:3:5
sry for the earlier mistake
koushik kashyap
19 Points
8 years ago
it is not clear archita...... kindly, please explain me in detail.
Archita
132 Points
8 years ago
 
use the equation s=ut+(1/2)at2
the particle starts from rest.
hence u=0
s1=(1/2)at2
 
now find the fibal velocity of the body after travelling a distance s1
v=u+at
v=at
 
now for the next two seconds.
v1=u2
 
s2= ut+ ½ at2
s2=at*t+1/2at2
s2=at2+1/2at2
s2= (3/2)at2
s2= 3s1
v2=u2+at
v2=at+at
v2=2at
 
now for the next two seconds.
u3=v2
s3=u3+1/2at2
    =2at+1/2at2
     =(5/2) at2
       =5s1
 
hence s1:s2:s3=s1:3s1:5s1
=1:3:5
 
if u have any further doubts ask in answer box
 
even if u substitute the values of a and t
and use the formula sn=u+(n-[1/2])a
you will get the correct answer
 
 
 

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