Mechanics, Distance and speed question

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The relationship between the distance y traveled by an object falling freely from rest and the time x it has been falling is that y is proportional to x². This is derived from the kinematic equation for distance under constant acceleration, specifically s = ut + (1/2)at², where initial velocity u is zero for free fall. By substituting the variables to match the problem's notation, the equation simplifies to y = (1/2)at². Other equations are not applicable here as they relate different variables that do not directly link distance to time. Thus, the conclusion that distance is proportional to the square of time is confirmed.
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What is the relationship between the distance y traveled by an object falling freely from rest and the time x the object has been falling?

The answer to this question is: y is proportional to x2

Can someone tell me how this answer is correct.
 
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Are you familiar with the kinematics of falling bodies? (An example of constant acceleration motion.)
 
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Doc Al said:
Are you familiar with the kinematics of falling bodies? (An example of constant acceleration motion.)
Yes, I understand that but I still do not understand why it is X^2. Can you please explain and thank you.
 
Gajan1234 said:
Yes, I understand that but I still do not understand why it is X^2.
They are just using, for some strange reason, "X" to stand for the time. Is that the concern?

Can you write the kinematic equation for distance as a function of time?
 
Doc Al said:
They are just using, for some strange reason, "X" to stand for the time. Is that the concern?

Can you write the kinematic equation for distance as a function of time?
s=ut+at^2/2 and s=(u+v)/2 x t there is a lot more
 
Gajan1234 said:
s=ut+at^2/2
That's the one you need. Since the object is falling from rest, what is u? And change the standard variable names to what they want in the problem, Y and X.
 
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Doc Al said:
That's the one you need. Since the object is falling from rest, what is u? And change the standard variable names to what they want in the problem, Y and X.
Thank you indeed but last question: why do we reject the other equations.
and also even if I make the initial velocity zero, then I would end up with 2s=at^2, t
 
Gajan1234 said:
Thank you indeed but last question: why do we reject the other equations.
Only because we want an equation that links distance to time. The other equations link other variables, so they won't help us here.

Gajan1234 said:
and also even if I make the initial velocity zero, then I would end up with 2s=at^2,
Sure, you end up with s=at^2/2 = (a/2)t^2.

Which tells you that the distance fallen is proportional to the square of the time. (You can replace s with y, and t with x, to make it match the variable names used in the problem statement.)
 
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Thank you very much, great mentor!
 
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