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

11.4 m

Step-by-step explanation:

We are given that

Total distance traveled by cab,d=214 m

Maximum speed of cab=v=315 m/min=
`(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=(27.5625)/(2.42)`

`x=11.4 m`

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