Originally Posted by

**ticbol** The acceleration being 1.39 m/sec both speeding and slowing down means the effect of gravity is ignored. Umm, the system is an elevaror. So the acceleration is controlled by the system. Okay.

The 1.39 m/sec/sec acceleration or deceleration is assumed constant.

*"How far does the cab move while accelerating to full speed from rest"*

full speed, V = 302 m/min.

In m/sec, V = (302m/min)(1min/60sec) = 5.03333m/sec

V = Vo +at

5.03333 = 0 +(1.39)t

t = 5.03333 / 1.39 = 3.6211 seconds.

distance travelled, s = Vo*t +(1/2)a*t^2

s = 0*3.6211 +(1/2)(1.39)(3.6211)^2 = 9.113 meterss -----answer.

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**"How long does it take to make the non-stop 165m run, starting and ending at rest."**

Aceelerating to max speed, from rest, is shown to take 3.6211 seconds, and the cab travells 9.113 meters.

The accleration and deceleration being the same in magnitude, then the sopping time and stopping distance will be the same as in the acceleration.

So, from rest to max speed, and from max speed to rest, the cab will spend:

time = 2(3.6211) = 7.2422 seconds

distance = 2(9.113) = 18.226 meters

Hence for the interval when the cab is at max speed,

remaining distance, d = 165 -16.226= 146.774 meters.

V = 5.03333 m/sec ------no more acceleration.

d = V*t

146.774 = 5.03333*t

t = 146.774 / 5.0333 = 29.1604 sec

Therefore, for the nonstop trip, the cab will take 7.2422 +29.1604 = 36.4026 seconds. -------answer.