Determine the lifting time of the elevator in a high-rise building, considering its movement during acceleration

Determine the lifting time of the elevator in a high-rise building, considering its movement during acceleration and deceleration to be equal to alternating with an acceleration equal in absolute value to 1m / s2, and in the middle section – uniform at a speed υ = 2m / s. Elevator lifting height h = 60m.

From the formula for acceleration, we find the acceleration time t₁.

a = (v-v₀) / t₁;

t₁ = (v-v₀) / a = 2 m / s / 1m / s² = 2 s,

where v is the final speed, v₀ is the initial speed.

Acceleration path:

h₁ = at² / 2 = (1m / s² * 4 s²) / 2 = 2 m.

When braking, the initial speed is v, the final speed is 0.

v-at₂ = 0;

Deceleration time t₂:

t₂ = V / a = 2 c;

Braking distance:

h₂ = vt-at² / 2 = 2 m.

Distance when braking and accelerating:

h₁ + h₂ = 2 m + 2 m = 4 m.

Path traveled evenly:

h₃ = h – (h₁ + h₂) = 60 – 4 = 56 m.

Steady motion time:

t₃ = 56 m / 2 m / s = 28 s.

Total time

t = t₁ + t₂ + t₃ = 2 + 2 + 28 = 32 s.

Answer: 32 p.