A fast train, leaving the station, moves uniformly with an acceleration of 0.5 m / s ^ 2.

A fast train, leaving the station, moves uniformly with an acceleration of 0.5 m / s ^ 2. What path will he cover in the ninth second of his movement?

To find the distance traveled by the train in the ninth second after leaving the station, we use the formula: S (9) = S9 – S8 = a * t9 ^ 2/2 – a * t8 ^ 2/2 = a / 2 * (t9 ^ 2 – t8 ^ 2).

Values of variables: a – constant train acceleration (a = 0.5 m / s2); t9 – the first time of movement (t9 = 9 s); t8 – three times the time of movement (t8 = 8 s).

Let’s perform the calculation: S (9) = a / 2 * (t9 ^ 2 – t8 ^ 2) = 0.5 / 2 * (9 ^ 2 – 8 ^ 2) = 4.25 m.

Answer: In the ninth second after leaving the station, the train must travel a distance of 4.25 m.