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SUMMARY

The discussion centers on analyzing the motion of an object described by the equation -4(t²) + 5t - 1, where the acceleration is -4 m/s², the initial velocity is 5 m/s, and the initial distance is -1 m. Participants confirm that the object first moves upwards before descending, reaching a maximum height before falling below the reference point. It is recommended to plot both velocity versus time and position versus time graphs to visualize the motion accurately.

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A dummy progression
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Homework Statement
A mobile (M) is animated by a uniformly varied rectilinear motion.
At time=0, it is on abscissa x= -1 meter
At time = 1s, it is on abscissa X= +2 meters
At time = 3 s , it is on abscissa x= -4 meters
Relevant Equations
Find the time equation of the movement
IMG_20220123_111059_884.jpg


This is where I'm stucked i don't even know if my reasoning is correct.
 
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When it says at time ##t = 0##, ##x = -1## that means you can substitute ##t=0## into your equation and get ##x = -1##. That may help!
 
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PeroK said:
When it says at time ##t = 0##, ##x = -1## that means you can substitute ##t=0## into your equation and get ##x = -1##. That may help!
I fond that the accelaration is -4
The initial velocity is 5
And the initial distance is -1

The time equation is : -4(t^2) + 5t -1

Is it correct?

But i don't understand something the accelaration is -4 and how is it possible that a time=1, The abscissa is positive and after time=3, it becomes negative. Shouldn't the mobile stop?
 
A dummy progression said:
I fond that the accelaration is -4
The initial velocity is 5
And the initial distance is -1

The time equation is : -4(t^2) + 5t -1

Is it correct?
Yes, although you should include the relevant units in metres and seconds.
A dummy progression said:
But i don't understand something the accelaration is -4 and how is it possible that a time=1, The abscissa is positive and after time=3, it becomes negative. Shouldn't the mobile stop?
This motion is similar to a ball being thrown up and falling back down: it starts ##1m## below some reference point (perhaps ##1m## below the garage roof); it has an initial velocity of ##5m/s## upwards; and, the acceleration due to gravity is ##-4m/s^2##. That might be nearer the figure for Mars than Earth, I guess.

At time ##t = 1## it has moved above the reference point and is still moving up. By time ##t = 3## it has reached its maxium height and fallen all the way back to ##4m## below the reference point (##3m## below where it started.

It might be worth plotting a velocity against time and a position against time graph for your equation.
 
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PeroK said:
Yes, although you should include the relevant units in metres and seconds.

This motion is similar to a ball being thrown up and falling back down: it starts ##1m## below some reference point (perhaps ##1m## below the garage roof); it has an initial velocity of ##5m/s## upwards; and, the acceleration due to gravity is ##-4m/s^2##. That might be nearer the figure for Mars than Earth, I guess.

At time ##t = 1## it has moved above the reference point and is still moving up. By time ##t = 3## it has reached its maxium height and fallen all the way back to ##4m## below the reference point (##3m## below where it started.

It might be worth plotting a velocity against time and a position against time graph for your equation.
Should i do two different graphs? One for the velocity and the other for the position?
 
A dummy progression said:
Should i do two different graphs? One for the velocity and the other for the position?
Yes.
 
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PeroK said:
Yes.
Oh, thank you Perok 👌👍
 
A dummy progression said:
Oh, thank you Perok 👌👍
I'll show you the graphs
 

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