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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         = 9s(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 questions=ut+(1/2)at^2 =(1/2)at^2v=u+atv=atv=us(2)=ut+(1/2)at^2       =at*t+(1/2)at^2      =3at^2/2v=u+atv=at+atv=2ats(3)=ut+(1/2)at^2       2at^t+(1/2)at^2      5/2 at^2 s1:s2:s3= 1:3:5sry 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)at2the particle starts from rest.hence u=0s1=(1/2)at2 now find the fibal velocity of the body after travelling a distance s1v=u+atv=at now for the next two seconds.v1=u2 s2= ut+ ½ at2s2=at*t+1/2at2s2=at2+1/2at2s2= (3/2)at2s2= 3s1v2=u2+atv2=at+atv2=2at now for the next two seconds.u3=v2s3=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 tand use the formula sn=u+(n-[1/2])ayou will get the correct answer
```
3 years ago
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