A certain elevator cab has a total run of 204 m and a maximum speed is 321 m/min, and it accelerates from rest and then back to rest at 1.21 m/s2. (a) How far does the cab move while accelerating to full speed from rest? (b) How long does it take to make the nonstop 204 m run, starting and ending at rest?
To solve this problem, we need to use the equations of motion for uniformly accelerated linear motion.
(a) How far does the cab move while accelerating to full speed from rest?
To find the distance covered while accelerating to full speed, we can use the equation:
\(v^2 = u^2 + 2as\),
where:
- v = final velocity (321 m/min),
- u = initial velocity (0 m/min, as the cab is at rest initially),
- a = acceleration (1.21 m/s²), and
- s = distance covered while accelerating.
Rearranging the equation, we get:
\(s = \frac{{v^2 - u^2}}{{2a}}\).
Plugging in the given values, we have:
\(s = \frac{{(321\,m/min)^2 - (0\,m/min)^2}}{{2 \times 1.21\,m/s^2}}\).
Converting the final velocity into m/s:
\(321\,m/min = \frac{{321\,m}}{{60\,s}} \approx 5.35\,m/s\).
Substituting the values:
\(s = \frac{{(5.35\,m/s)^2 - (0\,m/s)^2}}{{2 \times 1.21\,m/s^2}}\).
Simplifying:
\(s = \frac{{28.6225\,m^2/s^2}}{{2 \times 1.21\,m/s^2}}\),
\(s \approx 11.88\,m\).
Therefore, the cab moves approximately 11.88 meters while accelerating to full speed from rest.
(b) How long does it take to make the nonstop 204 m run, starting and ending at rest?
To find the time taken, we can use the equation:
\(v = u + at\),
where:
- v = final velocity (321 m/min),
- u = initial velocity (0 m/min),
- a = acceleration (1.21 m/s²), and
- t = time taken.
Rearranging the equation, we have:
\(t = \frac{{v - u}}{{a}}\).
Plugging in the values:
\(t = \frac{{321\,m/min - 0\,m/min}}{{1.21\,m/s^2}}\).
Converting the final velocity into m/s:
\(321\,m/min = \frac{{321\,m}}{{60\,s}} \approx 5.35\,m/s\).
Substituting the values:
\(t = \frac{{5.35\,m/s - 0\,m/s}}{{1.21\,m/s^2}}\).
Simplifying:
\(t = \frac{{5.35\,m/s}}{{1.21\,m/s^2}}\),
\(t \approx 4.42\,s\).
Therefore, it takes approximately 4.42 seconds to make the nonstop 204 m run, starting and ending at rest.
To calculate the distance and time taken for the elevator cab to accelerate and decelerate, we can use the following equations of motion:
1. Distance (d) = Initial velocity (u) * time (t) + 0.5 * acceleration (a) * time (t)^2
2. Final velocity (v) = Initial velocity (u) + acceleration (a) * time (t)
Let's start with part (a), finding the distance traveled while accelerating to full speed from rest.
(a) To find the distance traveled during the acceleration phase, we need to determine the time it takes for the cab to reach its maximum speed.
Given:
Total run = 204 m
Maximum speed = 321 m/min
Acceleration = 1.21 m/s^2
First, let's convert the maximum speed from m/min to m/s:
Maximum speed = 321 m/min * (1 min/60 s) = 5.35 m/s
Using equation 2, we can find the time taken to reach maximum speed:
v = u + at
5.35 m/s = 0 + 1.21 m/s^2 * t
t = 5.35 m/s / 1.21 m/s^2
t ≈ 4.42 s
Now, we can calculate the distance traveled during acceleration using equation 1:
d = ut + (0.5) * a * t^2
d = 0 m (since the cab starts from rest) + (0.5) * 1.21 m/s^2 * (4.42 s)^2
d ≈ 5.95 m
Therefore, the cab moves approximately 5.95 meters while accelerating to full speed from rest.
Moving on to part (b), finding the time taken for the nonstop 204 m run.
For the nonstop 204 m run, the cab first accelerates to full speed, then maintains that speed for the remaining distance, and finally decelerates to rest.
The distance traveled during the constant speed phase can be calculated by subtracting twice the distance traveled during acceleration from the total run:
Distance during constant speed phase = Total run - 2 * Distance during acceleration
Distance during constant speed phase = 204 m - 2 * 5.95 m
Distance during constant speed phase ≈ 192.1 m
To find the time taken during the constant speed phase, we can use the equation:
Distance (d) = Speed (v) * Time (t)
192.1 m = 5.35 m/s * t
t = 192.1 m / 5.35 m/s
t ≈ 35.93 s
Since the cab accelerates to full speed and then decelerates to rest, the total time will be twice the time taken during the constant speed phase (as the acceleration and deceleration times are equal):
Total time = 2 * Time during constant speed phase
Total time = 2 * 35.93 s
Total time ≈ 71.86 s
Therefore, it takes approximately 71.86 seconds to make the nonstop 204 m run, starting and ending at rest.