Question

The elevator starts from rest at the first floor of the building and comes to a complete stop at the 6th floor. It can accelerate at 6 ft/s² and then decelerate at 2 ft/s² Determine the shortest time is takes to reach the 6th floor, which is 60 ft above the ground. Draw the v?t and s?t graphs for the motion of the elevator.

Answer 1

the elevator starts at rest, the initial velocity of the elevator is zero. U = 0.

maximum acceleration =a_max = 6 ft/sec².

maximum velocity = V = U+a_max t.

Maximum velocity = V_max = 0 + 6t = 6t₁

Minimum acceleration = a_min = 2 ft/sec²

When elevator come to rest, velocity is zero, V₃ = 0.

V₃ = V_max - a_min t₂

0 = V_max - 2t₂

V_max = 2t₂

we know Maximum velocity = V_max = 0 + 6t = 6t₁

we need a compromise between the maximum and minimum velocities.

6 t₁ = 2 t₂

t₁ = 0.33 t₂.

distance travelled through maximum acceleration = h.

Distance travelled through minimum acceleration = 60-h.

From S= Ut+1/2 at² relation, for maximum aceleration,

h = 0 + (1/2) * 6 *t₁².

h = 3 t₁².

For minimum acceleration,

60-h = 0+ V_max t₂ - (1/2)*2*t₂².

From relation V² - U² = 2as,

V²_max = 0+2(6)(h-0)

V²_max = 12 h.

final velocity should be zero,

0 = V²_max +2*(-2)*(60-h)

V²max = 240-4h.

we already know

V²_max = 12 h.

By equating above two equations we get,

240-4h = 12 h,

Height h = 15 ft.

Distance covered with high acceleration = h = 15 ft.

We know V²_max = 12 h.

So V_max = (12*15)^0.5 = 13.41 ft/sec.

From V_max = 2t_2,

t₂ = 13.41 / 2 = 6.705 sec.

from t₁ = 0.33 t₂.

time t1 = 2.21 sec.

Shortest time = t1 + t2 = 2.21 + 6.705 = 8.91 sec.