Question

A certain elevator cab has a total run of 198 m and a maximum speed is 317 m/min, and it accelerates from rest and then back to rest at 1.11 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 198 m run, starting and ending at rest?

Answer 1

Answer:

Explanation:

Given

Total run=198 m

maximum speed=317 m/min=5.28 m/s

acceleration$=1.11 m/s^2$

(a)Distance traveled while  accelerating

$v^2-u^2=2as$

$(5.28)^2-0=2\times 1.11\times s$

s=12.57 m

time taken

v=u+at

$5.28=0+1.11\times t$

t=4.75 s

(b)Time taken to cover 198 m

if car takes 12.57 m & 4.75 s to reach max speed so it also takes 12.57 m and 4.75 s to stop from max speed

Distance traveled with max speed=198-25.14=172.86 m

time taken$=\frac{172.86}{5.28}=32.73 s$

total time=4.75+32.73+4.75=42.25 s