Find the speed of the passenger elevator acquired in 3.7 s and the distance traveled during this time.

Find the speed of the passenger elevator acquired in 3.7 s and the distance traveled during this time. The lift moves with an acceleration of 0.62 m / s. The initial speed is considered equal to zero.

Given:
a = 0.62 meters per second squared – passenger elevator acceleration;
t = 3.7 seconds – time span.
It is required to determine v (meter per second) – the speed of the elevator during the time interval t and S (meter) – the distance traveled during this time.
To simplify the solution, all results obtained will be rounded to one decimal place.
By the condition of the problem, the initial speed of the elevator is zero (v0 = 0). Then:
v = a * t = 0.62 * 3.7 = 2.3 meters per second.
S = a * t ^ 2/2 = 0.62 * 3.7 ^ 2/2 = 0.62 * 13.7 / 2 = 8.5 / 2 = 4.3 meters.
Answer: the speed is 2.3 m / s, the path is 4.3 meters.