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

#### Chemist116

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?.

#### skeeter

Math Team
1.5 seconds to 6 m/s from a resting start and 1.5 seconds to come to a stop from 6 m/s = 3 seconds of acceleration

total distance = distance traveled while undergoing acceleration + distance traveled at the max constant speed

$90 = 2 \cdot \dfrac{1}{2} \cdot 4 \cdot 1.5^2 + 6t \implies 90 = 9 + 6t \implies t = 13.5$

total time = time accelerating + time at the constant speed = 3 + 13.5 = 16.5 sec