An elevator in a high rise building accelerates and decelerates at the rate of 1 foot per second squared. Its maximum speed is 8 feet per second. It starts from rest, accelerates to its maximum speed and stays at that maximum speed until it approaches its destination where it decelerates to rest.
How far does the elevator travel if it accelerates from rest until it reaches its maximum speed?
How far has the elevator traveled after starting from rest and traveling for 10 seconds without decelerating?
a. vf^2=vi^2+2ad
64=0+2*1*d solve for d
b. find the time for the above first,
vf=at solve for t, I get about 8 seconds.
which means 2 seconds left at speed, so distance total=distance in a+2*8
To find the distance traveled by the elevator during acceleration, we can use the formula:
distance = (initial velocity^2) / (2 * acceleration)
In this case, the elevator starts from rest, so its initial velocity is 0 feet per second, and the acceleration is 1 foot per second squared. The maximum speed is given as 8 feet per second.
Plugging in the values into the formula, we get:
distance = (8^2) / (2 * 1)
distance = 64 / 2
distance = 32 feet
Therefore, the elevator travels a distance of 32 feet during its acceleration phase.
Now, to calculate how far the elevator has traveled after starting from rest and traveling for 10 seconds without decelerating, we need to determine the distance covered during the constant speed phase.
Since the elevator stays at the maximum speed of 8 feet per second during this phase, we can calculate the distance using the formula:
distance = speed * time
Plugging in the values, we get:
distance = 8 * 10
distance = 80 feet
Therefore, the elevator has traveled a distance of 80 feet after starting from rest and traveling for 10 seconds without decelerating.