Question

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. How long does it take to make the nonstop 190 m run, starting and ending at rest?

Answer 1

a)
V-speed
a-acceleration
V=a*t if you take Vmax, you ill get time needed to reach max speed.
t=V/a=305/(60*1.22)=4.16666666s (the 60 in the fraction is for converting from m/min to m/s)

the distance s=(a*t^2)/2=(1.22*4.1666^2)/2=10.6m

b) as already shown in a, to gain max speed it takes 4.16666s, and since the acceleration is the same for slowing down it also takes 4.16666s to stop the elevator from max speed, so the only unknown value is how long does the elevator travel by max speed.
s=(a*t^2)/2= 1.22*4.1666^2/2=10,6m is the distance needed to be passed to gain max speed, and to stop from max speed.
190m-2*10.6m=168.8m
t2- time traveling with the max speed is t2=168.8m*60/305=33.206s
t=2*t+t2=2*4.16666+33.206s=41.5s

Answer 2

maximum speed in m/s =5.08 m/s; accelaration=1.22 m/sec^2 time taken to reach that speed = 5.08/1.22=4.16 sec distance traveled = 1/2 at^2 =1/2 * 1.22*4.16*4.16=10.55 m distance traveled by the lift =190-10.55*2= 168.9 meter time taken to cover this distance=168.9/5.08=33.24 seconds net time taken to cover this distance= 33.24+4.16*2 =41.56 seconds; ans=41.56 sec;

Answer 3

41.5 sec