Q29P - Fundamentals Of Physics

Question

A certain elevator cab has a horizontal run of 190 m and a maximum speed of  305 m/min , and it accelerates from rest and then back to rest at $\mathbf{1}\mathbf{.}\mathbf{22}\mathbf{ }\mathbf{m}\mathbf{/}{\mathbf{s}}^{\mathbf{2}}$. (a) How far does the cab move while accelerating to full speed from rest?  (b)How long does it take to make the nonstop 190 m run, starting and ending at rest?

Step-by-Step Solution

Answer

(a) Distance is 10.6 m.

(b) Time is  41.5 s

Step 1: Given information

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

Step 2: Concept and formula used in the given question

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.

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

Where,

             $v_0$is the initial velocity,  

              x is the displacement, 

              a is the acceleration and 

               t is the time.

Step 3: (a) Calculation for the how far does the cab move while accelerating to full speed from rest

Conversion of m/min to m/s 

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

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

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

Step 4: (b) Calculation for the how long does it take to make the nonstop run, starting and ending at rest

Using equation (ii),

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

 

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

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

 

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 

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

 

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