The body, moving uniformly accelerated, covers a distance of 2.5 m in the third second.

The body, moving uniformly accelerated, covers a distance of 2.5 m in the third second. determinant of the movement of the body in the fifth second.

Data: S (3) – the path covered in the 3rd second (S (3) = 2.5 m).

1) Find the constant acceleration of a given body: S (3) = S3 – S2 = a * t3 ^ 2/2 – a * t2 ^ 2/2 = a / 2 * (t3 ^ 2 – * t2 ^ 2), whence we express : a = S (3) * 2 / (t3 ^ 2 – * t2 ^ 2) = 2.5 * 2 / (3 ^ 2 – 2 ^ 2) = 1 m / s2.

2) Determine the path in the 5th second: S (5) = S (4) – S (3) = a * (t5 ^ 2 – t4 ^ 2) / 2 = 1 * (5 ^ 2 – 4 ^ 2) / 2 = 4.5 m.

Answer: For the 5th second, the movement of a given body is 4.5 m.