- #1

yhPscis

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I don't understand what the logic is behind the formula for the distance traveled by an object under the influence of a constant accelerating force, namely [(Vf+Vi)/2].t

How come it equals the distance?

I mean, I understand that you can derive the distance based on a constant velocity by solving for d = V.t

Velocity means that with each elapsed fraction of time, a certain distance is covered, so you naturally add up as many amounts of distance as there are fractions of time.

But when the body accelerates, the distance per fraction of time always changes. It doesn't make any sense to me that one could still derive the distance despite that.

My teacher said that since the velocity isn't constant, you take the average, hence the fact that "a" is substituted for "[(Vf+Vi)/2]" in the original formula "v.t equals distance". But by adding the initial velocity (which often equals 0 anyway) and the final one to have an averga, you may have a rough idea of the distance, but this shouldn't be accurate, isn't it?

To me it's like taking the age of the eldest (16) and the youngest (2) in a group of people where most are aged 10 and then use that outcome (16 years old) as though it represents the age in that group. Clearly, it doesn't.

You need to sum up all the elements of a group and divide by the amounts of elements to get an average and The velocity of an accelerating object changes every second, every 1/10th of a second, 1/100th of a second, 1/1000th... so it's a group an infinite number of values between which the difference is an infitesimal change. You can't possibly add that up and devide...

And yet this formula is widely used in physics... I'm no Einstein, so I won't contradict the world, but is the distance obtained with it just a rough approximation or is it accurate after all? How come it is accurate if it is? What's the logic behind it? What's the flaw in my logic?

I'm a person who absolutely needs to understand where formulas come from to memorize them, just studying them by heart is not an option if I don't want to fail my exam (and I don't), so this is really important to me.

I think my lack of understanding of this formula may be related to the fact that I have trouble conceptualizing acceleration. I will lay it out here so as to let you glimpse inside my mind when I try to imagine what "acceleration" means:

To me, an acceleration of, let's say, 3 m/s² means that, with each second passed, the velocity of the body increases with 3 m/s right.

So if I drop the ball in my hand and it falls in a straight line to the ground with an initial velocity of 0 under the influence of the accelerating force of gravity (which equals 10 m/s² in this exercise) during the first 4 seconds, the following happens:

After 0 second: distance traveled = 0, no increase in velocity

In this interval [0s , 1s]: velocity = 0 m/s + infinite amount of infinitesimal changes

After 1 second: distance traveled = ?m, velocity + 10 m/s

In this interval [1s , 2s]: velocity = 10 m/s + infinite amount of infinitesimal changes

After 2 seconds: distance traveled ?m + 10 m, velocity + 10 m/s

In this interval [2s , 3s]: velocity = 20 m/s + infinite amount of infinitesimal changes

After 3 seconds: distance traveled = ?m + 20 m , velocity + 10 m/s

In this interval [3s , 4s]: velocity = 30 m/s + infinite amount of infinitesimal changes

After 4 seconds: distance traveled = ?m + 30 m, velocity + 10 m/s

"?" is a number that I can't figure out because acceleration only tells me what happens each second, so I can only calculate starting from the first second, right? I just don't see how one can possibly work with something like that... Maybe I sound really stupid right now, but it's worth succeeding my exam, I guess... :(

Thanks a lot for reading and (hopefully helping)!

How come it equals the distance?

I mean, I understand that you can derive the distance based on a constant velocity by solving for d = V.t

Velocity means that with each elapsed fraction of time, a certain distance is covered, so you naturally add up as many amounts of distance as there are fractions of time.

But when the body accelerates, the distance per fraction of time always changes. It doesn't make any sense to me that one could still derive the distance despite that.

My teacher said that since the velocity isn't constant, you take the average, hence the fact that "a" is substituted for "[(Vf+Vi)/2]" in the original formula "v.t equals distance". But by adding the initial velocity (which often equals 0 anyway) and the final one to have an averga, you may have a rough idea of the distance, but this shouldn't be accurate, isn't it?

To me it's like taking the age of the eldest (16) and the youngest (2) in a group of people where most are aged 10 and then use that outcome (16 years old) as though it represents the age in that group. Clearly, it doesn't.

You need to sum up all the elements of a group and divide by the amounts of elements to get an average and The velocity of an accelerating object changes every second, every 1/10th of a second, 1/100th of a second, 1/1000th... so it's a group an infinite number of values between which the difference is an infitesimal change. You can't possibly add that up and devide...

And yet this formula is widely used in physics... I'm no Einstein, so I won't contradict the world, but is the distance obtained with it just a rough approximation or is it accurate after all? How come it is accurate if it is? What's the logic behind it? What's the flaw in my logic?

I'm a person who absolutely needs to understand where formulas come from to memorize them, just studying them by heart is not an option if I don't want to fail my exam (and I don't), so this is really important to me.

**You can stop reading here and help me with an answer (please) if you don't have much time**, but here are some more details that might help you understand my problem a little more easily if you have.I think my lack of understanding of this formula may be related to the fact that I have trouble conceptualizing acceleration. I will lay it out here so as to let you glimpse inside my mind when I try to imagine what "acceleration" means:

To me, an acceleration of, let's say, 3 m/s² means that, with each second passed, the velocity of the body increases with 3 m/s right.

So if I drop the ball in my hand and it falls in a straight line to the ground with an initial velocity of 0 under the influence of the accelerating force of gravity (which equals 10 m/s² in this exercise) during the first 4 seconds, the following happens:

After 0 second: distance traveled = 0, no increase in velocity

In this interval [0s , 1s]: velocity = 0 m/s + infinite amount of infinitesimal changes

After 1 second: distance traveled = ?m, velocity + 10 m/s

In this interval [1s , 2s]: velocity = 10 m/s + infinite amount of infinitesimal changes

After 2 seconds: distance traveled ?m + 10 m, velocity + 10 m/s

In this interval [2s , 3s]: velocity = 20 m/s + infinite amount of infinitesimal changes

After 3 seconds: distance traveled = ?m + 20 m , velocity + 10 m/s

In this interval [3s , 4s]: velocity = 30 m/s + infinite amount of infinitesimal changes

After 4 seconds: distance traveled = ?m + 30 m, velocity + 10 m/s

"?" is a number that I can't figure out because acceleration only tells me what happens each second, so I can only calculate starting from the first second, right? I just don't see how one can possibly work with something like that... Maybe I sound really stupid right now, but it's worth succeeding my exam, I guess... :(

Thanks a lot for reading and (hopefully helping)!

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