Question

A certain elevator cab has a total run of 214 m and a maximum speed is 315 m/min, and it accelerates from rest and then back to rest at 1.21 m/s2. (a) How far does the cab move while accelerating to full speed from rest?

Answer 1

Answer:

11.4 m

Explanation:

We are given that  

Total distance traveled by cab,d=214 m

Maximum speed of cab=v=315 m/min=$\frac{315}{60}=5.25m/s$

1 min =60 s

Acceleration of cab=$1.21 m/s^2$

We have to find the distance traveled by the cab while accelerating to full speed from rest.

Initial speed o f cab=u=0

$v^2-u^2=2ax$

Substitute the values in the formula

$(5.25)^2-0=2(1.21)x$

$27.5625=2.42x$

$x=\frac{27.5625}{2.42}$

$x=11.4 m$

Hence, the cab travel 11.4 m from rest to full speed.