A body covers 12m in 2nd second and 20 m in the 4th second how much distance distance will it cover in 4 seconds after 5th second
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Given :Case I:let initial velocity = u m/sdistance=s=12mTime =t=2secS=u+a(2t-1)/212=u+(2x2-1)/212=u+3a/22u+3a=24--------------(1)
Case II:Time=t=4secdistance =s=20m20=u+a(2x4-1)/220=u+7a/22u+7a=40--------(2)from 1 and 2:2u+3a=242u+7a=40- - -***********-4a=-16
a=4m/s2
substitute the value of a in equation 1 we get :
u=6m/s Case iii:Now, distance covered in 4sec after 5second=distance covered in 9th sec-distance covered in 5 sec=(6x9+1/24x9x9)-(6x5+1/2*4*5*5)=54+162-(30+50)=136m
Case II:Time=t=4secdistance =s=20m20=u+a(2x4-1)/220=u+7a/22u+7a=40--------(2)from 1 and 2:2u+3a=242u+7a=40- - -***********-4a=-16
a=4m/s2
substitute the value of a in equation 1 we get :
u=6m/s Case iii:Now, distance covered in 4sec after 5second=distance covered in 9th sec-distance covered in 5 sec=(6x9+1/24x9x9)-(6x5+1/2*4*5*5)=54+162-(30+50)=136m
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