How to Calculate Distance in a Two-Stage Lift Journey Using SUVAT Equations?

  • Thread starter Thread starter NinaL
  • Start date Start date
  • Tags Tags
    Exam Mechanics
Click For Summary

Homework Help Overview

The problem involves calculating the distance traveled by a lift during a two-stage journey, where the lift accelerates and then decelerates. The context is within the subject area of mechanics, specifically using SUVAT equations related to uniformly accelerated motion.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • The original poster attempts to understand how to apply SUVAT equations given two different accelerations and the total time for the journey. Some participants suggest treating the motion in two parts and using the formula V_f = V_i + aT to find the time intervals for each stage of the lift's journey.

Discussion Status

Participants are exploring the problem by breaking it down into two segments of motion and discussing the application of kinematic equations. There is an indication of helpful guidance being provided, particularly in how to set up the equations based on the initial and final velocities and the respective accelerations.

Contextual Notes

The original poster expresses uncertainty about how to determine the time spent during each phase of acceleration and deceleration, which is a critical aspect of solving the problem. The total journey time is fixed at 10 seconds, which adds a constraint to the discussion.

NinaL
Messages
5
Reaction score
0
I have a mechanics exam coming up, and I'm going through a textbook answering questions. I'm really stuck on the following:

A lift ascends from rest with an accceleration of 0.5ms before slowing with an acceleration of -0.75ms for the next stop. If the total journey time is 10 secs, what is the distance between the two stops?


The thing is, I know that you have to use the SUVAT equations (displacement, initial velocity, final velocity, time) but I have no idea what to do with this. Why are they giving me two accelerations?

I've tried splitting the question into two parts, but i don't know the time traveled at each acceleration.

Please help! This is for my AS maths exam, on Jan the 12th, and with my luck, this kind of thing will come up, just because I'm not prepared for it!. :)
Thank you!
 
Last edited:
Physics news on Phys.org
Treat the motion in two parts. In part one, the lift goes from a speed of 0 to V in [itex]T_1[/itex] seconds; in part two, it goes from V to 0 in [itex]T_2[/itex] seconds. Figure out [itex]T_1[/itex] and [itex]T_2[/itex]. (Hint: use [itex]V_f = V_i + aT[/itex].) Then use the times to figure out the distance traveled.
 
Doc Al said:
(Hint: use [itex]V_f = V_i + aT[/itex].) .

Thank you.

I'm sorry, but I'm not feeling particularly intelligent today, or perhaps I've just called it something different, but could you explain the above formula?

Thank you ever so much!
 
Since it started from rest V initial = 0m/s
So first Velocity V= 0m/s + (0.5m/s^2)(t1)
Second final velocity since it stoped
0m/s = V(initial) + (-0.75m/s^2)(t2 = 10s)
and then you plug first Velocity in the second equation.

0m/s = [0m/s + (0.5m/s^2)(t1)] + (-0.75m/s^2)(t2 = 10s)
 
Last edited:
[itex]V_f = V_i + aT[/itex] is one of the basic kinematic formulas describing uniformly accelerated motion. It tells you how to calculate the final speed ([itex]V_f[/itex]) a uniformly accelerated object will attain after T seconds given the initial speed ([itex]V_i[/itex]).

Hints: In part one, the initial speed is 0, call the final speed V. In part two, the initial speed is V, the final speed is zero. Now apply that equation for each part, and make use of the fact that [itex]T_1 + T_2 = 10[/itex] seconds. You should be able to solve for the two times.
 
Thanks a lot, I finally figured out how to do it! :)

On the subject of Mechanics though, does anyone know any good websites that feature revision material, explanations etc?

Thank you!
 

Similar threads

  • · Replies 2 ·
Replies
2
Views
2K
Replies
7
Views
3K
  • · Replies 1 ·
Replies
1
Views
4K
Replies
4
Views
4K
Replies
7
Views
2K
Replies
3
Views
6K
Replies
12
Views
15K
  • · Replies 7 ·
Replies
7
Views
10K
  • · Replies 13 ·
Replies
13
Views
21K
Replies
8
Views
9K