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.

did you got all the answers?

yes

to solve its the distance traveled/ 40(this is velocity)
for acceleration its the velocity/5
so v=d/40
a=v/5

thnx dude...........and what other questions did you got?

How did you get the velocity 40 ?

I mean how did you get it d/40* ?

speed = distance /time

and time is 35+5=40.

To solve this problem, we can use the equations of motion under constant acceleration. Let's break down the problem into three parts: the initial acceleration, the constant speed, and the deceleration.

Part 1: Initial Acceleration
We know that the elevator starts from rest and accelerates for 5 seconds. We can use the equation:
v = u + at,
where v is the final velocity, u is the initial velocity (which is 0 in this case), a is the acceleration, and t is the time.

Since u = 0, the equation simplifies to:
v = at,

We're given that the elevator accelerates at a constant rate for 5 seconds, so we can substitute these values into the equation:
v = a * 5.

Part 2: Constant Speed
During this period, the elevator maintains a constant speed. Therefore, the acceleration is zero, and we can conclude that the velocity remains constant throughout this duration. Let's say the constant speed is represented by the variable "v_max."

Part 3: Deceleration
The elevator slows down with the same magnitude of acceleration (but in the opposite direction). This means that the acceleration is -a (negative because it is in the opposite direction). Again, we can use the equation:
v = u + at,

Since the final velocity is 0 (the elevator comes to a halt), we can rewrite the equation as:
0 = v_max + (-a)t,

But we know the duration of this deceleration period is 5 seconds, so we can substitute the values:
0 = v_max - 5a.

Lastly, we're given that the top floor is 320 meters above the ground floor, which means that the elevator travels a total distance of 320 meters. We know that the distance covered during constant speed is given by the equation:
distance = speed * time.

Since the elevator travels at a constant speed for 35 seconds, we can rewrite this equation as:
320 = v_max * 35.

Now, we have a system of equations that we can solve to find the answers to both parts (a) and (b).

Equation 1: v = a * 5,
Equation 2: 0 = v_max - 5a,
Equation 3: 320 = v_max * 35.

Let's solve this system of equations to find the maximal speed (v_max) and the acceleration (a).