How does halving the time affect the distance traveled by a falling object?

AI Thread Summary
The discussion centers on calculating the distance a falling object covers in the first three seconds after falling 72 meters in six seconds. Using the equation of motion, the acceleration is determined to be 4 m/s², leading to a calculated distance of 18 meters in the first three seconds. Participants debate the implications of air resistance and the translation of the problem statement, which uses the term "overcome" instead of "fall." The confusion arises as many incorrectly assume uniform velocity, while the correct interpretation involves understanding that under uniform acceleration, the distance covered in half the time is a quarter of the total distance. Ultimately, the correct answer is confirmed to be 18 meters, highlighting the importance of recognizing the effects of acceleration in such problems.
M. M. Fahad Joy
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Homework Statement



If a falling object overcome 72 metres in 6 second, how much distance it overcame in first 3 second.

Homework Equations


s = ut+(at^2)/2

The Attempt at a Solution


Here,
s = 72 m
u = 0
t = 6s
a = ?
We know,
s = ut+(at^2)/2
Or, 72 = 0*6 + (a*6^2)/2
Or, 72 = 36a/2
Or, 72 = 18a
Or, a = 4 m/s

In the second part,
u = 0
t = 3s
a = 4 m/s^2
s = ?
Again,
s = ut+(at^2)/2
= 0*t + (4*3^2)/2
= (4*9)/2
= 36/2
= 18 m (ans.)
 
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M M FAHAD JOY said:

Homework Statement



If a falling object overcome 72 metres in 6 second, how much distance it overcame in first 3 second.

Homework Equations


s = ut+(at^2)/2

The Attempt at a Solution


Here,
s = 72 m
u = 0
t = 6s
a = ?
We know,
s = ut+(at^2)/2
Or, 72 = 0*6 + (a*6^2)/2
Or, 72 = 36a/2
Or, 72 = 18a
Or, a = 4 m/s

In the second part,
u = 0
t = 3s
a = 4 m/s^2
s = ?
Again,
s = ut+(at^2)/2
= 0*t + (4*3^2)/2
= (4*9)/2
= 36/2
= 18 m (ans.)
You have apparently assumed the falling object is on another planet or somewhere in space above the Earth's surface where the acceleration of gravity is not the same as it is at Earth's surface. Since a object would fall more than 72m in 6s near Earth's surface, your assumption seems reasonable, and you have correctly worked out the result based on that assumption. But are your sure that was what was intended in the original question?
 
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tnich said:
You have apparently assumed the falling object is on another planet or somewhere in space above the Earth's surface where the acceleration of gravity is not the same as it is at Earth's surface. Since a object would fall more than 72m in 6s near Earth's surface, your assumption seems reasonable, and you have correctly worked out the result based on that assumption. But are your sure that was what was intended in the original question?
It's from my school exam question.
It's about falling object on Earth. But the gravity of Earth is 9.8 as I know.
We know that the gravity doesn't depend on mass. But for the friction of air reduces the acceleration of gravity. So it a problem me seeing the first law of falling bodies.
 
M M FAHAD JOY said:
It's from my school exam question.
It's about falling object on Earth. But the gravity of Earth is 9.8 as I know.
We know that the gravity doesn't depend on mass. But for the friction of air reduces the acceleration of gravity. So it a problem me seeing the first law of falling bodies.
The force that air resistance applies to a falling body (called drag) depends on the speed of the falling body. So you could not solve the problem by assuming drag is constant. You might look at the original problem statement and if there is anything you missed.
 
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tnich said:
The force that air resistance applies to a falling body (called drag) depends on the speed of the falling body. So you could not solve the problem by assuming drag is constant. You might look at the original problem statement and if there is anything you missed.
I have checked it again. It is ok.
 
M M FAHAD JOY said:
I have checked it again. It is ok.
In the problem statement you have written "overcome [a distance]". That is not a common expression. I assume that is your translation of the original problem. What do you think it means?
 
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tnich said:
In the problem statement you have written "overcome [a distance]". That is not a common expression. I assume that is your translation of the original problem. What do you think it means?
I am not so well in English. I have translated it from my own language. That's the problem.
 
M M FAHAD JOY said:
I am not so well in English. I have translated it from my own language. That's the problem.
Do you think the object could have started out going upward, reached a maximum height and then fallen?
 
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tnich said:
Do you think the object could have started out going upward, reached a maximum height and then fallen?
I didn't mean to criticize your English. I meant to ask if "overcome 72 m" could mean something different than "fall 72m".
 
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  • #10
tnich said:
I didn't mean to criticize your English. I meant to ask if "overcome 72 m" could mean something different than "fall 72m".
Yes, the translation will be fall 72 metres.
By the way, it was an MCQ.
a) 36m b) 24m c) 18m d) 8m

Most of my friends answered 36 metres. But I have answered 18 metres. That's my confusion.
 
  • #11
M M FAHAD JOY said:
Most of my friends answered 36 metres.
Then they fell into the trap.
Moving half the distance in half the time would be right for uniform velocity, but for uniform acceleration from rest it is a quadratic, so a quarter of the distance in half the time.
 
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  • #12
haruspex said:
Then they fell into the trap.
Moving half the distance in half the time would be right for uniform velocity, but for uniform acceleration from rest it is a quadratic, so a quarter of the distance in half the time.
Are you sure, quarter of the distance in half time?
 
  • #13
M M FAHAD JOY said:
Are you sure, quarter of the distance in half time?
½at2. What happens to that if you halve t? Note, this is only for falling from rest.
 
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  • #14
haruspex said:
½at2. What happens to that if you halve t? Note, this is only for falling from rest.
Ok, Thanks.
 
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