A certain elevator cab has a total run 2

A certain elevator cab has a total run of 190 m and a maximum speed of 305 m/min, and it accelerates from rest and then back to rest at 1.22 m/s^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?

Solution:

Let a_1= +1.22m/s^2 and a_2= -1.22 m/s^2 representing acceleration and deceleration.

Let us also convert 305 m/min to m/s as these would be the conventional SI units.

305 m/min=5.08m/s.

a) To solve this problem, we will use the following equation:

v^2=v_0^2+2a_{1}\triangle x where v is final velocity, v_0 is initial velocity, a is acceleration, and x is displacement.

Isolating for \triangle x gives us:

\triangle x=\frac{(5.08m/s)^2}{2(1.22m/s^2)}

\triangle x=10.59m

b) To do this part of the problem, we will use the following equation:

v=v_0+at

Substituting our values and isolating for t gives us:

t_1=\frac{v-v_0}{a_1}

t=\frac{5.08m/s}{1.22m/s^2}

t=4.17s

Because the deceleration is the same as the acceleration, the time it takes is also same, thus, the total time is 4.17s \times 2=8.33s. Because the time is the same, the total distance can also be calculated by simply multiplying 10.59m \times 2= 21.18m. It is important to realize that the distance we found only represents the acceleration and deceleration distance, which means 190m-21.18m gives us the distance at which the elevator is in constant velocity. This distance is equal to 168.82m. To find the time taken for the elevator to travel this distance, we can write:

t=\frac{168.82m}{5.08 m/s}

t=33.21s

Therefore, the total time is 8.33s+33.21s=41.54s

This question can be found in Fundamentals of Physics, 10th edition, chapter 2, question 29.