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

Question

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?.

Answer 1

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.