How Do You Calculate Time for a Nonstop 190m Elevator Run?

[blinkbubble]

I solved part a of this problem but I'm having trouble figuring out the different steps for part b...could someone break it down clearly?

A certain elevator cab has a total run of 190m and a maximum speed of 305m/min, and it accelerates from rest and then back to rest at 1.22m/s^2
a. how far does the cab move while accelerating to full speed from rest?
b. how long does it take to make the nonstop 190m run, starting and ending at rest?

thanks :)

 

[rootX]

Draw a v-t graph,
you know that v is contant when it reaches 350 m/min

and, what would be the area under graph?

P.S. Is this from Halliday, I solved this two days ago lol

 

[m2003]

I understand your struggle with part b of this problem. Let me break it down for you. In order to solve this problem, we need to use the equation v^2 = u^2 + 2as, where v is the final velocity, u is the initial velocity, a is the acceleration, and s is the distance traveled.

For part a, we can use the information given to solve for the distance traveled while accelerating to full speed. We know that the initial velocity is 0 (since the cab starts from rest) and the final velocity is 305m/min. We also know that the acceleration is 1.22m/s^2. Plugging these values into the equation, we get:

305^2 = 0^2 + 2(1.22)s
93025 = 2.44s
s = 93025/2.44 = 38093.85m

Therefore, the cab moves 38093.85m while accelerating to full speed.

For part b, we need to find the time it takes for the cab to make the nonstop 190m run. We can use the equation v = u + at, where t is the time taken. Again, we know that the initial velocity is 0 and the final velocity is 305m/min. We also know that the distance traveled is 190m. Plugging these values into the equation, we get:

305 = 0 + (1.22)t
t = 305/1.22 = 250 seconds

Therefore, it takes 250 seconds for the cab to make the nonstop 190m run, starting and ending at rest. I hope this breakdown helps you understand the problem better. Keep up the good work!