The slevator starts at 1st floor and accelerated 1m/s^2 for 6 seconds and continues at a constant velocity for 12 seconds more and is then stopped in 4 seconds with constant deceleration. If the floors are 3 meters apart at what floor does the elevator stops?
with what constant deacceleration value?
Ok, I see it now, the total displacement is the sum of
d1=1/2 *1*6^2
d2= 1*6*12
d3=1/2 (-1*6*12)/4 *4^2
check those...
then floor=displacement/3 + 1
assuming ground floor is called floor 1
displacement= d1+d2+d3
Hmmm. I get
d3 = 6*4 + 1/2 (-6/4)*4^2
since v goes from 6 to 0 in 4 seconds.
bob's the physicist, but ...
To find the floor at which the elevator stops, we need to calculate the distance traveled during each phase of its motion and add them all up.
Step 1: Calculating distance during acceleration phase
During the acceleration phase, the elevator starts from rest (0 m/s) and accelerates at a rate of 1 m/s^2 for 6 seconds. We can use the formula:
distance = initial velocity * time + 1/2 * acceleration * time^2
The initial velocity is 0 m/s, the acceleration is 1 m/s^2, and the time is 6 seconds. Plugging these values into the formula:
distance = 0 * 6 + 1/2 * 1 * (6^2)
distance = 0 + 1/2 * 1 * 36
distance = 0 + 1/2 * 36
distance = 0 + 18
distance = 18 meters
So, during the acceleration phase, the elevator travels 18 meters.
Step 2: Calculating distance during constant velocity phase
During the constant velocity phase, the elevator is moving at a constant speed, which means there is no acceleration. The velocity remains constant for 12 seconds. We can calculate the distance traveled using the formula:
distance = velocity * time
Since the velocity is constant, we only need to multiply the velocity by the time. However, we don't know the velocity yet, so let's find that next.
We know that acceleration is the change in velocity divided by the time taken. In this case, the elevator accelerated for 6 seconds with an acceleration of 1 m/s^2. So:
acceleration = (final velocity - initial velocity) / time
1 m/s^2 = (final velocity - 0) / 6s
final velocity = 1 m/s^2 * 6s
final velocity = 6 m/s
Now we can calculate the distance during the constant velocity phase:
distance = 6 m/s * 12s
distance = 72 meters
So, during the constant velocity phase, the elevator travels 72 meters.
Step 3: Calculating distance during deceleration phase
During the deceleration phase, the elevator is stopped in 4 seconds with a constant deceleration. We can calculate the final velocity first using the formula:
final velocity = initial velocity + acceleration * time
Since the elevator comes to a stop, the final velocity is 0 m/s. The initial velocity is the velocity achieved during the constant velocity phase, which is 6 m/s. The time is 4 seconds. Plugging these values into the formula:
0 m/s = 6 m/s + acceleration * 4s
acceleration * 4s = -6 m/s
acceleration = -6 m/s / 4s
acceleration = -1.5 m/s^2
The negative sign indicates that the elevator is decelerating. Now we can calculate the distance using the formula:
distance = initial velocity * time + 1/2 * acceleration * time^2
The initial velocity is 6 m/s, the time is 4 seconds, and the acceleration is -1.5 m/s^2. Plugging these values into the formula:
distance = 6 m/s * 4s + 1/2 * (-1.5 m/s^2) * (4^2)
distance = 24 m + 1/2 * (-1.5 m/s^2) * 16
distance = 24 m - 0.5 m/s^2 * 16
distance = 24 m - 8 m
distance = 16 meters
So, during the deceleration phase, the elevator travels 16 meters.
Step 4: Adding up the distances
To find the final floor at which the elevator stops, we add up the distances traveled during each phase:
18 meters + 72 meters + 16 meters = 106 meters
Since each floor is 3 meters apart, we divide the total distance by 3 to find the number of floors:
106 meters / 3 meters = 35.333 floors
Since we can't have a fraction of a floor, we round down to the nearest whole number. Therefore, the elevator stops at the 35th floor.