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In summary, at time 1 the mobile was below the garage roof and had an initial velocity of 5m/s upwards. It reached its maximum height at time 3 and then fell all the way back to 4m below the garage roof.
  • #1
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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  • #2
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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  • #3
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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?
 
  • #4
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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  • #5
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?
 
  • #6
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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  • #7
PeroK said:
Yes.
Oh, thank you Perok 👌👍
 
  • #8
A dummy progression said:
Oh, thank you Perok 👌👍
I'll show you the graphs
 

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