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

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To calculate the distance in a two-stage lift journey using SUVAT equations, split the motion into two parts: acceleration and deceleration. For the first part, the lift accelerates from rest at 0.5 m/s², while in the second part, it decelerates at -0.75 m/s². The total journey time is 10 seconds, and the equations V_f = V_i + aT can be applied to find the time for each stage. By determining the final velocity after the first stage and using it as the initial velocity for the second stage, the distances for both parts can be calculated. Understanding these concepts is crucial for exam preparation in mechanics.
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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!
 
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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)
 
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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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