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Grade 11Mechanics

A body starting from rest moves with constant acceleration.the ratio of distance covered by the body during the 5^th sec to that covered in 5 sec

prasann Bhende

6 Years agoGrade 11

2 Answers

chetan jangir

6 Years ago

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 1/2a $5^2$ 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.

Saurabh Koranglekar

6 Years ago

Dear student

S5 = 0.5a(25)

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

Ratio s5th/ s5 = 9 /25

Regards

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