Question

A body starting from rst moves with constant acceleration the ratio of distance covered by the body during the 5th sec tp that covered in 5s is

Answer 1

Distance covered in nth second

S = u + a(n - 0.5)

In 5th second
S₁ = 0 + a(5 - 0.5)
S₁ = 4.5a


Distance covered in t seconds

S = ut + 0.5at²

In 5 seconds
S₂ = 0 + (0.5 × a × (5 s)²)
S₂ = 12.5a

S₁ / S₂ = 4.5a / (12.5a)
= 9 / 25

Ratio of distance travelled in 5th second to 5 seconds is 9 : 25