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)

The average speed during the acceleration, and deacceleration is V/2. Theconstant velocity is V.

distance=V(t-10)+V/2 * 10

solve for V.

the distance during accelerating is V/2*5

Vf^2=vo^2+2ad

where d=avg velocity*time=V/2 * 5
Vf^2=Vo^2+2ad
V^2/(v*5)=a=V/5
check that.

Hello

Can you explain the first equation?

distance=V(t-10)+V/2 * 10

Where is this distance coming from for me to be able to solve for V?

Also where is the 10 coming from in V(t-10) and V/2*10?

My final equation solving for V is :

V= 10d / 4(t-10)

To solve this problem, we can break it down into three phases:

Phase 1: The elevator accelerates upward for 5 seconds.
Phase 2: The elevator maintains a constant speed for 35 seconds.
Phase 3: The elevator decelerates until it comes to a halt at the top floor.

Let's calculate the values for each phase and then combine them to find the answers.

Phase 1:
In this phase, the elevator starts from rest and accelerates upward for 5 seconds. We can use the equation of motion:

s = ut + (1/2)at^2
where:
s = distance covered,
u = initial velocity (0 m/s in this case),
a = acceleration,
t = time (5 seconds in this case).

Since we are given that the elevator moves upwards, the acceleration is positive, and we need to find the distance covered in this phase.

s = 0 + (1/2)at^2
320 = 0 + (1/2)(a)(5^2)
320 = (25/2)a
a = (2/25)(320)
a = 12.8 m/s^2

Phase 2:
In this phase, the elevator maintains a constant speed. Therefore, the acceleration is zero, and the elevator covers a distance of 320 meters in 35 seconds since it moves at a constant speed for this duration.

Phase 3:
In this phase, the elevator decelerates until it comes to a halt at the top floor. The distance covered is the same as in Phase 1, 320 meters, and the time taken is 5 seconds. However, the acceleration is now in the opposite direction, so it is negative.

Now, let's find the answers:

(a) Maximal speed (v):
In Phase 2, the elevator maintains a constant speed. Therefore, the maximal speed is the same as the speed during Phase 2, which is:

v = distance/time
v = 320/35
v = 9.14 m/s

(b) Acceleration (a):
Since we are given that the magnitude of the deceleration in Phase 3 is the same as the acceleration in Phase 1, the acceleration is:

a = -12.8 m/s^2

Therefore, the maximal speed of the elevator is 9.14 m/s, and the acceleration is -12.8 m/s^2.