An elevator cab in NY Marriott has a total run of 190 m . IT's max speed is 305m/min Its acceleration and deceleration both have a magnitude of 1.22m/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 trip 190m start and end at rest?
What would the equation be for this problem?
There is no single equation to plug into. You need to do this problem, like many in enginering and physocs, one step at a time.
First calculate the time it takes to accelerate to maximum speed and the (equal) time that it takes to decelerate from that speed to zero. That time is
t = Vmax/a = (305 m/min)*(1/60min/s)/1.22 m/s^2 = 4.17 s
So the elevator spends 8.34 s accelerating and decelerating. Figure out how far in goes during each of those intervals. That distance is
X1 = X3 = (1/2) a t^2 = 10.6 m
That is the answer to part (a) of your question. For part (b), calculate how much of the total run is traversed at constant velocity. That would be
X2 = 190 - (2)(10.6) = 168.8 m
For the total transit time, add the time spent accelerating, decelerating, and the time spend traveling at maximum speed. The latter time is just
t2 = 168.8 m/305 m/min = 0.553 min = 33.2 s.
Add 2t to that for your answer.
To solve this problem, you need to follow a step-by-step approach. Here's how to do it:
a) To find out how far the cab moves while accelerating to full speed from rest, you need to calculate the distance covered during the acceleration phase.
First, find the time it takes to accelerate to maximum speed using the equation:
t = Vmax / a
Where Vmax is the maximum speed (305 m/min) and a is the acceleration (1.22 m/s^2).
t = 305 m/min / (1.22 m/s^2) = 4.17 s
Now, use the equation for distance covered during acceleration:
X1 = (1/2) * a * t^2
Where X1 is the distance covered during acceleration and t is the time calculated above.
X1 = (1/2) * 1.22 m/s^2 * (4.17 s)^2 = 10.6 m
Therefore, the cab moves a distance of 10.6 meters while accelerating to full speed from rest.
b) To calculate the time it takes to make the nonstop trip of 190 meters starting and ending at rest, you need to consider the time spent accelerating, decelerating, and traveling at constant velocity.
The time spent accelerating and decelerating is calculated by taking 2 times the time calculated in part (a). Therefore, it is:
2t = 2 * 4.17 s = 8.34 s
Next, calculate the distance covered while traveling at constant velocity, which is the remaining distance after subtracting the distances covered during acceleration and deceleration from the total run.
X2 = Total run - 2 * X1 = 190 m - 2 * 10.6 m = 168.8 m
Finally, calculate the time spent traveling at constant velocity by dividing X2 by the maximum speed:
t2 = X2 / Vmax
t2 = 168.8 m / 305 m/min = 0.553 min = 33.2 s
To find the total time it takes for the nonstop trip, add the times spent accelerating/decelerating and traveling at constant velocity:
Total time = 2t + t2 = 8.34 s + 33.2 s = 41.54 s
So, the elevator takes approximately 41.54 seconds to make the nonstop trip of 190 meters starting and ending at rest.
The equations used in this problem are:
- t = Vmax / a (to find the time spent accelerating)
- X1 = (1/2) * a * t^2 (to find the distance covered during acceleration)
- X2 = Total run - 2 * X1 (to find the distance covered while traveling at constant velocity)
- t2 = X2 / Vmax (to find the time spent traveling at constant velocity)
- Total time = 2t + t2 (to find the total time for the nonstop trip)
a) The cab moves a distance of 10.6 m while accelerating to full speed from rest.
b) The total time it takes to make the nonstop trip, starting and ending at rest, is 41.74 seconds.