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

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The discussion focuses on calculating the distance traveled by a lift during a two-stage journey using SUVAT equations. The lift accelerates at 0.5 m/s² from rest and then decelerates at -0.75 m/s², with a total journey time of 10 seconds. Participants emphasize the need to split the motion into two parts, applying the formula V_f = V_i + aT to determine the time intervals T_1 and T_2 for each stage. Ultimately, the correct application of these equations allows for the calculation of the total distance between the two stops.

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  • Understanding of SUVAT equations in kinematics
  • Basic knowledge of acceleration and deceleration concepts
  • Ability to manipulate algebraic equations
  • Familiarity with initial and final velocity definitions
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  • Study the application of SUVAT equations in various motion scenarios
  • Learn how to derive time intervals from acceleration and velocity
  • Explore additional kinematic equations for uniformly accelerated motion
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Students preparing for AS-level mathematics exams, particularly those focusing on mechanics and kinematics, as well as educators seeking to enhance their teaching materials on motion equations.

NinaL
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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:
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Treat the motion in two parts. In part one, the lift goes from a speed of 0 to V in T_1 seconds; in part two, it goes from V to 0 in T_2 seconds. Figure out T_1 and T_2. (Hint: use V_f = V_i + aT.) Then use the times to figure out the distance traveled.
 
Doc Al said:
(Hint: use V_f = V_i + aT.) .

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:
V_f = V_i + aT is one of the basic kinematic formulas describing uniformly accelerated motion. It tells you how to calculate the final speed (V_f) a uniformly accelerated object will attain after T seconds given the initial speed (V_i).

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 T_1 + T_2 = 10 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!
 

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