Answer:
9/25
Explanation:
Distance covered in the first 5 seconds:
Δx = v₀ t + ½ at²
Δx₀₋₅ = (0) (5) + ½ a (5)²
Δx₀₋₅ = 25a/2
Distance covered in the first 4 seconds:
Δx = v₀ t + ½ at²
Δx₀₋₄ = (0) (4) + ½ a (4)²
Δx₀₋₄ = 8a
So the distance covered during the 5th second is:
Δx₅ = 25a/2 − 8a
Δx₅ = 9a/2
So the ratio of the distance covered during the 5th second to the distance covered in the first 5 seconds is:
Δx₅ / Δx₀₋₅
(9a/2) / (25a/2)
9/25
