An elevator is stopped at the ground floor. It starts moving upwards at constant acceleration a>0 for 5 seconds. It then keeps a constant speed for 35 seconds. Finally, it slows down with an acceleration of the same magnitude (but opposite direction) −a , until it comes to a halt at the top floor. The top floor is 320 meters above the ground floor.
(a) What is the maximal speed v of the elevator ? (in m/s)
(b) What is the acceleration a ? (in m/s2)
Please post the answer fast
anon did you happen to solve the second last question? the one with the inclined roof
To find the maximal speed of the elevator, we need to calculate the acceleration during the first phase of motion and then use it to determine the maximum speed.
(a) Calculation of maximal speed:
1. Determine the displacement of the elevator during the first phase of motion:
The elevator starts from rest and accelerates with a constant acceleration a for 5 seconds. The formula to calculate displacement during constant acceleration is given by:
\[ S = ut + \frac{1}{2}at^2 \]
Where:
S = Displacement
u = Initial velocity (0 since the elevator starts from rest)
t = Time (5 seconds)
a = Acceleration
Plugging in the values, we get:
\[ S = (0)(5) + \frac{1}{2}a(5)^2 \]
\[ S = \frac{25a}{2} \]
2. Determine the displacement of the elevator during the third phase of motion:
The elevator comes to a halt at the top floor, so its final velocity is 0. It decelerates with a constant acceleration -a until it stops. The formula to calculate displacement during constant deceleration is the same as during constant acceleration, but with the negative value of acceleration:
\[ S = ut + \frac{1}{2}at^2 \]
Where:
S = Displacement
u = Initial velocity (the final velocity from the second phase of motion)
t = Time (5 seconds, as the elevator started decelerating at this time)
a = Decceleration (-a, since it is in the opposite direction)
Plugging in the values, we get:
\[ S = v(5) - \frac{1}{2}a(5)^2 \]
\[ S = 5v - \frac{25a}{2} \]
3. Total displacement during the entire motion:
The total displacement is given by the vertical distance between the ground floor and the top floor, which is 320 meters.
To calculate the total displacement during the entire motion:
\[ S_{total} = \frac{25a}{2} + 5v - \frac{25a}{2} \]
Since the \(\frac{25a}{2}\) terms cancel each other out, the equation simplifies to:
\[ 320 = 5v \]
\[ v = \frac{320}{5} = 64 m/s \]
Therefore, the maximal speed of the elevator is 64 m/s.
(b) Calculation of acceleration:
The given problem states that the magnitude of the acceleration during the first and third phases of motion is the same and denoted as a.
From the second phase of motion, we know that the elevator moves at a constant speed for 35 seconds. Since the speed is constant, the acceleration during this phase is zero (0 m/s^2).
Hence, the value of acceleration a is equal to zero (a = 0 m/s^2).