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Grade: 12
        
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 ?
3 years ago

Answers : (4)

Archita
132 Points
							
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
 
3 years ago
Archita
132 Points
							
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
3 years ago
koushik kashyap
19 Points
							
it is not clear archita...... kindly, please explain me in detail.
3 years ago
Archita
132 Points
							
 
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
 
 
 
3 years ago
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