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 . (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?

Answer 1

Answer:

Explanation:

Given

length of run=190 m

maximum speed$=305 m/min\approx 5.08 m/s$

$a=1.22 m/s^2$

time require to reach max speed

v=u+at

$5.08=0+1.22\times t$

$t=\frac{5.08}{1.22}=4.16 s$

distance traveled in t=4.16 s

$s=ut+\frac{at^2}{2}$

$s=0+\frac{1.22\times 4.16^2}{2}=10.55 m$

Now time taken to stop from max speed to zero

v=u+at

$0=5.08-1.22\times t$

t=4.16 s

so the distance travel by car during the deceleration is 10.55 m

total time time taken=4.16+4.16=8.32 s