Answer to Question #152453 in Mechanical Engineering for omar

Question #152453
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.
1
Expert's answer
2020-12-24T04:12:32-0500

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\n\\begin{aligned}\n\\small T=t_1+t_2+t_3\\\\\n\\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\n\\begin{aligned}\n\\small V_1&= \\small 1.5t'=0.6t''\\\\\n\\small 1.5t'&=\\small 0.6(T_{min}-t')\\\\\n\\small T_{min}&=\\small 3.5t'\n\\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\n\\begin{aligned}\n\\small 12&= \\small \\frac{V_1 }{2}T_{min}\\\\\n&= \\small \\frac{1.5t'}{2}T_{min}\\\\\n& =\\small \\frac{3}{14}T^2_{min}\\\\\n\\small T_{min}&= \\small \\bold{7.48s}\n\\end{aligned}"


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