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?
velocity v = at
distance s = 1/2 at^2
You have a, so plug it in.
To find the distance the elevator travels when it accelerates from rest until it reaches its maximum speed, we can use the formula for displacement:
\[ d = \frac{1}{2} \cdot a \cdot t^2 \]
where \( d \) is the displacement, \( a \) is the acceleration, and \( t \) is the time taken. In this case, the acceleration is 1 foot per second squared.
Plugging in the values, we get:
\[ d = \frac{1}{2} \cdot (1 \, \text{ft/s}^2) \cdot (t=0 \, \text{s})^2 \]
Since \( t = 0 \) at the start, the displacement is 0 feet. Hence, the elevator does not travel any distance while accelerating from rest until it reaches its maximum speed.
To find how far the elevator has traveled after starting from rest and traveling for 10 seconds without decelerating, we first need to determine the distance covered during acceleration. We know the maximum velocity achieved by the elevator is 8 feet per second, and it accelerates at a rate of 1 foot per second squared.
The formula for the final velocity after acceleration is given by:
\[ v = u + a \cdot t \]
where \( v \) is the final velocity, \( u \) is the initial velocity, \( a \) is the acceleration, and \( t \) is the time taken. Since the elevator starts from rest, the initial velocity is 0 feet per second. Plugging in the values, we get:
\[ v = 0 + (1 \, \text{ft/s}^2) \cdot (t = 10 \, \text{s}) \]
\[ v = 10 \, \text{ft/s} \]
So, the elevator achieves a velocity of 10 feet per second after 10 seconds of acceleration.
Now, to find the distance covered during this time, we use the formula:
\[ d = u \cdot t + \frac{1}{2} \cdot a \cdot t^2 \]
where \( d \) is the displacement, \( u \) is the initial velocity, \( a \) is the acceleration, and \( t \) is the time taken. Since the initial velocity \( u \) is 0 feet per second, the equation simplifies to:
\[ d = \frac{1}{2} \cdot a \cdot t^2 \]
Plugging in the values, we get:
\[ d = \frac{1}{2} \cdot (1 \, \text{ft/s}^2) \cdot (t = 10 \, \text{s})^2 \]
\[ d = \frac{1}{2} \cdot (1 \, \text{ft/s}^2) \cdot 100 \, \text{s}^2 \]
\[ d = 50 \, \text{ft} \]
So, the distance traveled by the elevator after starting from rest and traveling for 10 seconds without decelerating is 50 feet.