# How to find the minimum time for an elevator going up and deccelerating in a building?

The problem is as follows:

The magnitude of the acceleration and decceleration of an elevator is $$4\frac{m}{s^2}$$ and its maximum vertical speed is $$6\frac{m}{s}$$. Find the minimum time (in seconds) such that the elevator goes up and gets to $$90\,m$$ of height departing from rest and arriving with zero speed.

The alternatives given are:

$$\begin{array}{ll} 1.&12.5\,s\\ 2.&13.5\,s\\ 3.&14.5\,s\\ 4.&15.5\,s\\ 5.&16.5\,s\\ \end{array}$$

What I thought doing here was to use the fact that the combined displacement for going up and deccelerating will add to $$90\,m.$$

This is summarized as follows:

$$y=y_{o}+v_ot+\frac{1}{2}at^2$$

The first part reduces to:

$$y_{h}=\frac{1}{2}(4)t^2=2t^2$$

Then for the second equation is:

$$90=y_{h}+6t-\frac{1}{2}4t^2$$

But adding the two expressions result into:

$$90=6t$$

$$t=\frac{90}{6}$$

Where exactly did I made an error?. Can somebody help me with this?.

Note first that the time to reach maximum speed from rest is $$6/4=1.5$$, so distance travelled in accelerating from rest to maximum speed is $$\frac{1}{2}4(1.5)^2=4.5$$ That is also the distance travelled in decelerating. So it must go 81 at max speed of 6, requiring a further 81/6=13.5$sec. So total time 16.5 s. Check: to travel the full 90m at full speed would take 15 sec. So this is longer, but not much. However, options 1-3 must be wrong, and option 4 would require even faster acceleration. Incidentally an acceleration of nearly $$g/2$$ is fast for a lift carrying people. • Interesting approach. I totally overlooked the fact that the deccelerating length was the same in accelerating. It has to be noted that during the$81\,m\$ the elevator goes with constant speed. This part I was initially confused but then it made sense. I didn't noticed your comment but it is true regarding half of the acceleration of gravity is fast for a lift. Jan 27, 2020 at 12:30