# science

posted by
**Jen** on
.

An elevator cab in NY Marriott has a total run of 190 m . IT's max speed is 305m/min Its acceleration and deceleration both have a magnitude of 1.22m/s^2

a) how far does the cab move while accelerating to full speed from rest?

b) How long does it take to make the nonstop trip 190m start and end at rest?

What would the equation be for this problem?

There is no single equation to plug into. You need to do this problem, like many in enginering and physocs, one step at a time.

First calculate the time it takes to accelerate to maximum speed and the (equal) time that it takes to decelerate from that speed to zero. That time is

t = Vmax/a = (305 m/min)*(1/60min/s)/1.22 m/s^2 = 4.17 s

So the elevator spends 8.34 s accelerating and decelerating. Figure out how far in goes during each of those intervals. That distance is

X1 = X3 = (1/2) a t^2 = 10.6 m

That is the answer to part (a) of your question. For part (b), calculate how much of the total run is traversed at constant velocity. That would be

X2 = 190 - (2)(10.6) = 168.8 m

For the total transit time, add the time spent accelerating, decelerating, and the time spend traveling at maximum speed. The latter time is just

t2 = 168.8 m/305 m/min = 0.553 min = 33.2 s.

Add 2t to that for your answer.