Answer to Question #152453 in Mechanical Engineering for omar

Question

The elevator starts from rest at the first floor of the building. It can accelerate at 1.5m/s2 and then decelerate at 0.6m/s2. Determine the shortest time it takes to reach a floor 12 m above the ground. The elevator starts from rest and stops.

Answer 1

Explanations & Calculations

• Consider the figures below

• A possible pattern of motion for the lift is, acceleration $\to$ uniform velocity$\to$ deceleration as given in the first figure.
• Then the total time can be written as,

$\qquad\qquad \begin{aligned} \small T=t_1+t_2+t_3\\ \end{aligned}$

• since acceleration & deceleration are a must, time taken for them is always positive.($\small t_1,t_2>0$ )
• For $\small T$ to be less $\small t_2 \to 0$
• Then $\small T_{min}=t'+t''$ (as shown in the figure to the left)
• Then applying $\small V=u+at$ for the periods of acceleration & deceleration,

$\qquad\qquad \begin{aligned} \small V_1&= \small 1.5t'=0.6t''\\ \small 1.5t'&=\small 0.6(T_{min}-t')\\ \small T_{min}&=\small 3.5t' \end{aligned}$

• It's about reducing $\small t'$ to minimize the total time but $\small t'$ is fixed due to the set value of acceleration meaning that there is only one shortest possible time for this lift to travel that distance.
• If we write the equation for the distance traveled,

$\qquad\qquad \begin{aligned} \small 12&= \small \frac{V_1 }{2}T_{min}\\ &= \small \frac{1.5t'}{2}T_{min}\\ & =\small \frac{3}{14}T^2_{min}\\ \small T_{min}&= \small \bold{7.48s} \end{aligned}$