Moving uniformly accelerated from a state of rest, the body passed 18 m in the fifth second of the movement.

Moving uniformly accelerated from a state of rest, the body passed 18 m in the fifth second of the movement. What path did it cover in the second second?

Initial data: S (5) (the path that the body covered in the fifth second of its movement) = 18 m; uniformly accelerated motion (a = const).

1) Let’s calculate the acceleration with which the given body was moving:

S (5) = S5 – S4 = a * t5 ^ 2/2 – a * t4 ^ 2/2 = 0.5a * (5 ^ 2 – 4 ^ 2) = 0.5a * (25 – 16) = 4 , 5a.

a = S (5) / 4.5 = 18 / 4.5 = 4 m / s ^ 2.

2) S (2) = S2 – S1 = a * t2 ^ 2/2 – a * t1 ^ 2/2 = 0.5 * 4 * (2 ^ 2 – 1 ^ 2) = 2 * 3 = 6 m.

Answer: This body has passed 6 meters in the second second of its movement.