Question

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48. A body starting from rest moves with constant acceleration. The ratio of distance covered by the body during

the 5th second to that covered in 5 seconds is​

Answer 1

Step-by-step explanation:

Using the second kinematical equation, we find

s = u t + ½ a t²

Since the body starts from rest, u=0.

Since the acceleration is constant, and the body started from rest, we may ignore the vector signs.

Thus, we get,

s(5) = ½a(5)²

s(4) = ½a(4)²

Thus, the ratio of distance travelled in 5th second to that travelled in 5 seconds, is :

{s(5)-s(4)} : s(5)

≈ {½a(25) - ½a(16)} : {½a(25)}

≈ (25–16):(25)

≈ 9:25