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A body starting from rest moves with constant acceleration, then the ratio of the distance covered in 5th second to 5 seconds is?
Asked by ananya vyas | 13 Apr, 2011, 02:18: PM
answered-by-expert Expert Answer
 
 
Let us assume that constant acceleraion to be A.
u = o since the body starts from rest.
and we know that Dispacement in nth sec = snth = supto n - supto(n-1)
[un+(1/2)an2]-[u(n-1)+(1/2)a(n-1)2] 
Snth=u+(1/2) (2n-1)a
Now,
Distance covered in 5th second is 
S = 0+1/2 (2 x5-1)a =(9/2)a
 
Distance covered in 5 seconds is
S=ut+(1/2)at2
S=0+(1/2) (ax5x5) = (25/2) xa
Therefore the ratio of the distance covered in 5th second to 5 second is =
Answered by | 13 Apr, 2011, 07:52: PM

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