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A BODY STARTING FROM REST MOVES WITH CONSTANT ACCELERATON. what is the ratio of distance covered by the body during the fifth second of time to that covered in the first 5 s

Respuesta :

MathPhys

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

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