91Ó°ÊÓ

$$\begin{gathered} {\mathrm{v}}_{\mathrm{f}}=0\mathrm{m}/\mathrm{s} \\ {\mathrm{v}}_0=305\mathrm{m}/\min \\ =5.08\mathrm{m}/\mathrm{s} \\ \mathrm{a}=1.22\mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$

The problem deals with the kinematic equation of motion in which the motion of an object is described at constant acceleration. Using the given acceleration and kinematic equations, the distance traveled by elevator to achieve full speed can be found. Similarly, using kinematic equations, the time it takes to start, run, and stop in the given distance can be calculated.

$$\begin{gathered} {\mathrm{v}}_{\mathrm{f}}^2={\mathrm{v}}_0^2+2\mathrm{a}×\mathrm{x}\left(\mathrm{i}\right) \\ {\mathrm{v}}_{\mathrm{f}}={\mathrm{v}}_0+\mathrm{a}×\mathrm{t}\left(\mathrm{ii}\right) \end{gathered}$$

Where,

$v_0$is the initial velocity,

x is the displacement,

a is the acceleration and

t is the time.

Conversion of m/min to m/s

$$\begin{gathered} {\mathrm{v}}_0=305\frac{\mathrm{m}}{\min }×\frac{1\mathrm{m}/\sec }{60\mathrm{m}/\min } \\ =5.08\mathrm{m}/\mathrm{s} \end{gathered}$$

Using equation (i), the displacement can be given by,

$$\begin{gathered} \mathrm{x}=\frac{{\mathrm{v}}_{\mathrm{f}}^2-{\mathrm{v}}_0^2}{2×\mathrm{a}} \\ =\frac{5.{08}^2-0}{2×1.22} \\ =10.6\mathrm{m} \end{gathered}$$

Using equation (ii),

$$\begin{gathered} {\mathrm{v}}_{\mathrm{f}}={\mathrm{v}}_0+\mathrm{a}×\mathrm{t} \\ 0=5.0833-1.22×\mathrm{t} \\ \mathrm{t}=4.166\sec \\ ≈4.17\mathrm{s} \end{gathered}$$

The elevator can accelerate and decelerate at the same rate. So, the time required for both would be the same. Also, the distance traveled during acceleration and deceleration would be the same.

So, the time for acceleration and deceleration

$$\begin{gathered} {\mathrm{t}}^{\text{'}}=2×\mathrm{t} \\ =2×4.166 \\ =8.33\sec \end{gathered}$$

The distance traveled during acceleration and deceleration

$$\begin{gathered} {\mathrm{d}}^{\text{'}}=2\mathrm{d} \\ =2\left\{10.6\right\} \\ =21.2\mathrm{m} \end{gathered}$$

So, elevator would travel for 190-21.2=168.8 m of distance with the maximum velocity.

We can find the time taken to travel this distance as

$$\begin{gathered} {\mathrm{t}}^{"}=\frac{168.8}{5.0833} \\ =33.21\sec \end{gathered}$$

The total time

$$\begin{gathered} \mathrm{T}={\mathrm{t}}^{\text{'}}+{\mathrm{t}}^{"} \\ =8.33+33.21 \\ =41.54\sec \\ ≈41.5\sec \end{gathered}$$

So, an elevator would take 41.5 sec to start, run, and stop.