## 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 at, vt, and st graphs for the motion.

## Step-by-step

\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}

Thus

\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

Thus,

$10 h = 160 – 4h \\ h = 11.429 ft \\ { v }_{ max } = 10.69 ft/s \\ { t }_{ 1 } = 2.138s \\ { t }_{ 2 } = 5.345s \\ t = { t }_{ 1 } + { t }_{ 2 } = 7.48s$

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