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)
38
can someone post the ecuations?
thanks
To solve this problem, we'll use kinematic equations, which describe the motion of an object in terms of its position, velocity, acceleration, and time.
Let's break down the elevator's motion into three parts:
1. Accelerating upwards for 5 seconds.
2. Moving at a constant speed for 35 seconds.
3. Decelerating until it comes to a halt at the top floor.
For Part 1:
We can use the equation of motion for uniform acceleration: v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time.
In this case, the elevator starts from rest (u = 0) and accelerates upwards for 5 seconds. So, the equation becomes v = 0 + a * 5.
For Part 2:
The elevator moves at a constant speed for 35 seconds, which means its acceleration is zero. Therefore, the velocity remains constant.
For Part 3:
The elevator decelerates with an acceleration of -a. We can use the same equation of motion: v = u + at, where u is the initial velocity and a is the acceleration. In this case, the elevator slows down until it comes to a halt, so the final velocity is 0. The equation becomes 0 = u + (-a)t.
Now, we can solve the questions:
(a) What is the maximal speed v of the elevator?
To find the maximum speed, we need to determine the velocity at the end of Part 2, just before deceleration begins. Since the velocity remains constant during Part 2, the velocity at the end of Part 2 is the same as the velocity at the beginning of Part 3. Therefore, we can write:
v = u + at, where u is the velocity at the end of Part 1, a is the deceleration (-a), and t is the time taken for Part 3.
We need to find the value of u, which is the velocity at the end of Part 1. We can substitute the equation from Part 1 into this equation to get:
v = a*5 + (-a)*t.
To find t, we can use the equation of motion for uniform acceleration again:
320 = 0 + (a/2)*t^2, where 320 is the distance covered during deceleration.
Now, we have two equations:
v = a*5 - a*t
320 = (a/2)*t^2
By solving these simultaneous equations, we can find the value of a, and then substitute it back into the first equation to find v.
(b) What is the acceleration a?
By solving the second equation, we can find the value of t.
Once we have t, we can substitute it into the first equation and solve for a.
Please note that the given information is not sufficient to provide numerical values for both a and v. You would need more information in order to get specific values.