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Grade: 11
        
A body starting from rest moves with constant acceleration.the ratio of distance covered by the body during the 5th sec to that covered in 5 sec
 
4 months ago

Answers : (2)

chetan jangir
64 Points
							
IT is very simple to deal with such problems. We start with basics we can write distance travelled in 5th second is equal to total covered distance in 5 seconds minus total covered distance in 4 seconds.we get 1/2a5^2 -1/2 a4^2( we have let acceleration  a  it will also be cancelled out in taking ratio) We get 4.5a .And as we know distance covered in first 5 seconds is 1/2a5^2 that is 12.5a. Taking ratio we get 9/25.Which is required  answer .
Note .We can generalise a formula for distance covered in nth second it comes u/2 +a/2(2n-1)  where u=initial  velocity a = constant acceleration. 
4 months ago
Saurabh Koranglekar
askIITians Faculty
3156 Points
							Dear student

S5 = 0.5a(25)

S5th = 0.5a( 25-16) = 0.5a(9)

Ratio s5th/ s5 = 9 /25

Regards
4 months ago
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