A certain elevator cab has a total run of 190 m and a maximum speed of 305 m/min, and it accelerates from rest and then back to rest at 1.22 m/s2. (a) How far does the cab move while accelerating to full speed from rest? (b) How long does it take to make the nonstop 190 m run, starting and ending at rest?

Question

.sam. · Answer

Vo = 0

10.6m = 1/2at^2
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or Vx = Vox + at

5.0833 = 0 + 1.22 t

t = 4.17 s
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2t  to find the time to both accelerate and decelerate.

Distance traveled at 5.0833 m/s will be

d = 190m -2(10.6m) = 168.8 m

Time to travel 168.8 m at 5.0833 m/s

T = d/v = 168.8m / 5.0833 m/s = 33.21 s

Total time is
Total time = 33.21 +2( 4.17 ) = 41.5 s