Question

the ratio of distances travelled by a body starting from rest with constant acceleration in 9th and 8th second is?

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Answer 1

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Answer 2

Answer:

S9 / S8 = 17 / 15

Explanation:

Because body start from a = constant

We know that the distance at a fixed time use then S=(t)- St-1)-1 St ----- 1

after solving eq" 1 we get get S = u + 1 / 2a ( 2t -1 )

For 9 th sec

S9 = 0 + 1 / 2 a ( 18 - 1 )

S9 = 17 a / 2

For 8 th sec

S8 = 0 +1 /2 a ( 16 - 1 )

S8 = 15 / 2 a

Ratio of distance at 9th & 8th  Sec

S9 / S8 = 170 / 150

S9 / S8 = 17 / 15

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