# What rate does the lift accelerates in the first 5 sec?

Hi,

I'm trying to solve this problem and it's driving me a little crazy. Any help greatly appreciated.

Q) A lift travels to the top of a tower through a vertical displacement of 48 m. The total journey takes 17 s. The lift accelerates from rest at a constant rate for the first 5 seconds. Then it moves at constant speed and then decelerates to rest at a constant rate for the last 5 seconds.

What rate does the lift accelerate during the first 5 seconds?

My Attempt:

I have the initial and final velocity which are both zero (elevator starts and stops), displacement (48m) and time (overall time 17sec and three time intervals of acceleration, constant velocity and deceleration). Clearly the acceleration and deceleration will be same rates as they occur during the same time intervals. However with the information I have I feel I'm unable to use the usual constant acceleration equations.

The average velocity is 48/17 = 2.8 m/s. However I'm not sure how that helps.

To find the acceleration during the first 5 seconds, I have the time and initial velocity but I need the velocity
it reaches at 5 seconds to obtain the acceleration and I just can't work out how to get it.

Related Introductory Physics Homework Help News on Phys.org
PeroK
Homework Helper
Gold Member
2020 Award
There are several approaches. Since you've been thinking about average velocity, you might like to try to solve the problem using the average velocity for each stage of the motion. Then you don't need any of the "usual" equations.

• henrco
gneill
Mentor
Hi Conal Henry, Welcome to Physics Forums.

Please retain the formatting headers (provided in the edit window) when you post a problem here.

Start with a graphic approach to gain insight. Have you tried making a sketch of velocity versus time? What does the area under a v vs t graph give you?

• henrco
There are several approaches. Since you've been thinking about average velocity, you might like to try to solve the problem using the average velocity for each stage of the motion. Then you don't need any of the "usual" equations.
Hi Perok,

1) For stage 1, the first 5 seconds, the acceleration is constant, so I've taken the angle the acceleration makes to be 45 degrees. (This is intuitive and if it's correct,
would you mind explaining why?).
The velocity for this 5 x Sin (45) = 3.5m/s.
With initial velocity is 0, the acceleration a = (3.5-0)/5 = 0.7m/s2

2) Find the average velocity for each stage. Starting with Stage 1, the first 5 seconds. I obviously have the time but not the displacement.
Avg Vel = Displacement / 5.
Work out area of each stage. Stage 1 = ( 5 x v/2 ) Stage 2 = ( 7 x v) Stage 3 = ( 5 x v/2). This all comes to 12 v
We know Area = displacement, therefore: 12v = 48. So v = 4.
Acceleration = (4-0)/5 = .8 m/s2

My second attempt seems to be correct? As I then worked out the displacement for each stage to be (S1 = 10, S2 = 28 and S3 = 10).
Could you please let know if this is correct?

Also you mentioned that there are several approaches, if you have time would you mind briefly outlining another approach.
I'd like to try to understand this problem from different approaches.

Hi Conal Henry, Welcome to Physics Forums.

Please retain the formatting headers (provided in the edit window) when you post a problem here.

Start with a graphic approach to gain insight. Have you tried making a sketch of velocity versus time? What does the area under a v vs t graph give you?
Hi gneill,

Point noted, will post future problems using the template. Thank you for your advice regarding the problem.
I replied to another helpful suggestion above and I think the answer below was the direction you were sending me in?

Find the average velocity for each stage. Starting with Stage 1, the first 5 seconds. I obviously have the time but not the displacement.
Avg Vel = Displacement / 5.
Work out area of each stage. Stage 1 = ( 5 x v/2 ) Stage 2 = ( 7 x v) Stage 3 = ( 5 x v/2). This all comes to 12 v
We know Area = displacement, therefore: 12v = 48. So v = 4.
Acceleration = (4-0)/5 = .8 m/s2

Conal

gneill
Mentor
I replied to another helpful suggestion above and I think the answer below was the direction you were sending me in?
Yes, you get to the same result. A graphical depiction to begin can sometimes help. For example: You know the area must be your displacement, you have the time periods, the only thing you don't have is the maximum velocity Vm. But Vm is easily found given the other information. Acceleration is just the slope of a line in the figure.

• henrco
PeroK
Homework Helper
Gold Member
2020 Award
Hi Perok,

1) For stage 1, the first 5 seconds, the acceleration is constant, so I've taken the angle the acceleration makes to be 45 degrees. (This is intuitive and if it's correct,
would you mind explaining why?).
The velocity for this 5 x Sin (45) = 3.5m/s.
With initial velocity is 0, the acceleration a = (3.5-0)/5 = 0.7m/s2

2) Find the average velocity for each stage. Starting with Stage 1, the first 5 seconds. I obviously have the time but not the displacement.
Avg Vel = Displacement / 5.
Work out area of each stage. Stage 1 = ( 5 x v/2 ) Stage 2 = ( 7 x v) Stage 3 = ( 5 x v/2). This all comes to 12 v
We know Area = displacement, therefore: 12v = 48. So v = 4.
Acceleration = (4-0)/5 = .8 m/s2

My second attempt seems to be correct? As I then worked out the displacement for each stage to be (S1 = 10, S2 = 28 and S3 = 10).
Could you please let know if this is correct?

Also you mentioned that there are several approaches, if you have time would you mind briefly outlining another approach.
I'd like to try to understand this problem from different approaches.

You seem to have understood the graphical approach. I'd say that's one thing never to forget: displacement = area under the velocity vs time graph.

Regarding average velocity. For constant acceleration from or to rest, the average velocity is half the final or initial velocity respectively. More generally, if you accelerate from ##u## to ##v## at a constant acceleration, then the average velocity is ##(u+v)/2##. You can check that graphically.

Another approach was to note the symmetry of the motion. Although, in this case, that just makes things a little easier. Always look out for symmetry in a problem. As problems get harder, using the symmetry of a problem can makes things a lot easier. And I mean a lot!

• henrco
Thank you very much, very helpful.

Yes, you get to the same result. A graphical depiction to begin can sometimes help. For example:
View attachment 92896
You know the area must be your displacement, you have the time periods, the only thing you don't have is the maximum velocity Vm. But Vm is easily found given the other information. Acceleration is just the slope of a line in the figure.
Thank you very much, this was very helpful.