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 ?

Answer

To solve the problem step-by-step, we will break it down into parts (a) and (b) as requested.

### Part (a): How far does the cab move while accelerating to full speed from rest?

1. **Identify Given Values:**
- Maximum speed (V_max) = 305 m/min
- Convert V_max to m/s:
\[ V_{max} = \frac{305 \text{ m/min}}{60 \text{ s/min}} = 5.083 \text{ m/s} \]
- Initial speed (u) = 0 m/s (since it starts from rest)
- Acceleration (a) = 1.22 m/s²

2. **Use the Third Equation of Motion:**
The third equation of motion is given by:
\[ V^2 = U^2 + 2aS \]
Here, we want to find the distance (S) while accelerating to maximum speed (V):
\[ V_{max}^2 = 0^2 + 2 \cdot a \cdot S_1 \]
Substituting the known values:
\[ (5.083)^2 = 2 \cdot 1.22 \cdot S_1 \]
\[ 25.83 = 2.44 \cdot S_1 \]
\[ S_1 = \frac{25.83}{2.44} \approx 10.59 \text{ m} \]
Rounding off, we get:
\[ S_1 \approx 10.6 \text{ m} \]

### Part (b): How long does it take to make the nonstop 190 m run, starting and ending at rest?

1. **Determine Total Distance:**
The total distance covered is 190 m. The distance during acceleration (S_1) and deceleration (S_3) is the same:
\[ S_3 = S_1 = 10.6 \text{ m} \]
Let S_2 be the distance covered at constant speed:
\[ S_2 = 190 - S_1 - S_3 = 190 - 10.6 - 10.6 = 168.8 \text{ m} \]

2. **Calculate Time at Constant Speed (T_2):**
Using the formula:
\[ T_2 = \frac{S_2}{V_{max}} = \frac{168.8}{5.083} \approx 33.2 \text{ s} \]

3. **Calculate Time During Acceleration (T_1):**
Using the formula:
\[ S_1 = ut + \frac{1}{2} a t_1^2 \]
Since u = 0:
\[ 10.6 = 0 + \frac{1}{2} \cdot 1.22 \cdot t_1^2 \]
\[ 10.6 = 0.61 \cdot t_1^2 \]
\[ t_1^2 = \frac{10.6}{0.61} \approx 17.38 \]
\[ t_1 \approx 4.16 \text{ s} \]

4. **Time During Deceleration (T_3):**
Since the time taken to decelerate is the same as the time taken to accelerate:
\[ T_3 = T_1 \approx 4.16 \text{ s} \]

5. **Total Time (T):**
\[ T = T_1 + T_2 + T_3 = 4.16 + 33.2 + 4.16 \approx 41.52 \text{ s} \]
Rounding off, we get:
\[ T \approx 41.5 \text{ s} \]

### Final Answers:
- (a) The cab moves approximately **10.6 m** while accelerating to full speed from rest.
- (b) The total time taken to make the nonstop 190 m run, starting and ending at rest, is approximately **41.5 seconds**.