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A body starting from rest moves with constant accleration. The ratio of distance covered by the body during the 5th second to that covered in 5 second is

A body starting from rest moves with constant accleration. The ratio of distance covered by the body during the 5th second to that covered in 5 second is

Grade:11

2 Answers

Adarsh
733 Points
7 years ago
Distance covered in nth   second is given by
s = u + a/2(2n-1) 
 where u – intial velocity and a – acceleration
Distance covered in t seconds is given by
s = ut + a/2(t2)
But, here u = 0
So,s =  a/2(2n-1) and s = a/2(t2
So ratio of distance covered in 5th second and 5 seconds = {a/2(2(5)-1)}/{a/2(52)}
                                                                                            = 9/25
Hope you understand my solution.
 
Please approve only if my answer is correct.
Gosula Madhukar
84 Points
7 years ago
HERE ‘U’=0,
HENCE THE DISTANCE COVERED UPTO 5th second is=1/2at^2=(25a)/2.
and theTHE DISTANCE COVERED UPTO 4th second is=1/2a(42)=8a.
and ths distance travelled in 5th second=(25a)/2-8a=(9/2)a.
therefore the ratio is=(9a/2)/(25a/2)=9/25.
 

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