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Determine- the rates of acceleration and deceleration, max velocity achieved, length of the lift shaft?

by alexkess

Question by Alex Rimmington: Determine- the rates of acceleration and deceleration, max velocity achieved, length of the lift shaft?
A direct drive hoist is initially required to accelerate a load of 1.5 tonne through a vertical distance of 140m in a time of 12 seconds. It then maintains a period of constant velocity for 10.6 seconds before uniform deceleration brings the load to reat at the top of the lift in a time of 15 seconds.
need to find average velocity

Best answer:

Answer by wingstwo
Where v = final velocity, vi = initial velocity, x = distance

Phase 1: Accelerating
x1 = xi + ½(v+vi)·t
140m = 0 + ½(v+0)·12s
v1 = 23 1/3 m/s
also,
v1 = vi + at
23 1/3 m/s = 0 + a1(12)
a1 = 1 17/18 m/s²

Phase 2: Constant velocity
x2 = rate * time
x2 = vt = 23 1/3 m/s²*10.6sec
x2 = 247 1/3 m

Phase 3: Deceleration
v3 = vi + a3·t
0 = 23 1/3 m/s + a3·(15s)
a3= -1 5/9 m/s

x3 = xi + ½(v+vi)·t
x3 = 0 + ½(0+23 1/3 m/s)·15s
x3 = 175m

ANSWERS
1. Max acceleration:
a = 1 17/18 m/s²

2. Max Deceleration:
a= -1 5/9 m/s

3. Max Velocity:
v = 23 1/3 m/s

4. Length of Lift Shaft:
x = 140m + 247 1/3 m + 175m
x = 562 1/3m

5. Average speed = dist/time
= 562 1/3m / (12 + 10.6 + 15) = 14.96 m/s

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