The elevator starts from rest at the first floor of the building. It can accelerate at 5 ft/s^2 and then decelerate at 2 ft/s^2. Determine the shortest time it takes to reach a floor 40 ft above the ground. The elevator starts from rest and then stops. Draw the a–t, v–t, and s–t graphs for the motion

Question

The elevator starts from rest at the first floor of the building. It can accelerate at 5 ft/{ s }^{ 2 } and then decelerate at 2 ft/{ s }^{ 2 }. Determine the shortest time it takes to reach a floor 40 ft above the ground. The elevator starts from rest and then stops. Draw the a–t, v–t, and s–t graphs for the motion.

Accepted Answer

[latex]\begin{aligned} + \uparrow \quad &{ v }_{ 2 } = { v }_{ 1 } + { a }_{ c } { t }_{ 1 } \ &{ v }_{ max } = 0 + 5 { t }_{ t1 } \end{aligned} \ \begin{aligned} + \uparrow \quad &{ v }_{ 3 } = { v }_{ 2 } + { a }_{ c } { t } \ & 0 = { v }_{ max } - 2 { t }_{ 2 } \end{aligned}[/latex]

Thus

[latex]\quad\quad\space\space\space { t }_{ 1 } = 0.4 { t }_{ 2 } \ \begin{aligned} + \uparrow \quad & { s }_{ 2 } = { s }_{ 1 } + { v }_{ 1 } { t }_{ 1 } \frac { 1 } { 2 } { a }_{ c } { t }_{ 1 }^{ 2 } \ & h = 0 + 0 + \frac { 1 } { 2 } (5)({ t }_{ 1 }^{ 2 }) = 2.5 \space { t }_{ 1 }^{ 2 } \end{aligned} \ + \uparrow 40 - h = 0 + { { v }_{ max } }^{ t_2 } - \frac{ 1 } { 2 }(2) { t }_{ 2 }^{ 2 } \ + \uparrow { v }^{ 2 } = { v }_{ 1 }^{ 2 } + 2 { a }_{ c } ( { s } - { s }_{ 1 } ) \ { v }_{ max }^{ 2 } = 0 + 2(5)(h - 0) \ { v }_{ max }^{ 2 } = 10h \ 0 = { v }_{ max }^{ 2 } + 2(-2)(40 - h) \ { v }_{ max }^{ 2 } = 160 - 4h[/latex]

Thus,

[latex]10 h = 160 - 4h \ h = 11.429 ft \ { v }_{ max } = 10.69 ft/s \ { t }_{ 1 } = 2.138s \ { t }_{ 2 } = 5.345s \ t = { t }_{ 1 } + { t }_{ 2 } = 7.48s[/latex]

When t = 2.145, [latex]v = { v }_{ max } = 10.7 ft/s[/latex] and h = 11.4 ft.

Question : The elevator starts from rest at the first floor of the buil...

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