With uniformly accelerated movement from a state of rest, the body passes 90 cm in 5 seconds

With uniformly accelerated movement from a state of rest, the body passes 90 cm in 5 seconds, determine the movement of the body in 7 seconds.

Data: uniformly accelerated motion; V0 (initial speed) = 0 m / s; S (5) (the distance covered in the fifth second) = 90 cm (in SI S (5) = 0.9 m).

1) Constant acceleration: S (5) = S5 – S4 = a * t5 ^ 2/2 – a * t4 ^ 2/2 = a * (t5 ^ 2/2 – t4 ^ 2/2) and a = S ( 5) / (t5 ^ 2/2 – t4 ^ 2/2) = 0.9 / (5 ^ 2/2 – 4 ^ 2/2) = 0.9 / 4.5 = 0.2 m / s2.

2) Moving in the seventh second: S (7) = S7 – S6 = a * t7 ^ 2/2 – a * t6 ^ 2/2 = 0.2 * 7 ^ 2/2 – 0.2 * 6 ^ 2 / 2 = 4.9 – 3.6 = 1.3 m.

Answer: The movement in the seventh second is 1.3 m.