Velocity divided by Acceleration gives distance?

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Homework Help Overview

The discussion revolves around the relationship between velocity, acceleration, and distance in the context of kinematic equations. Participants explore the implications of dividing velocity by acceleration and the correct interpretation of symbols and units in physics.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants examine the reasoning behind the division of velocity by acceleration, questioning the validity of the original poster's conclusion that this operation yields distance. There are attempts to clarify the meanings of symbols and units involved in the equations.

Discussion Status

The discussion has seen some participants acknowledging misunderstandings regarding the symbols used for distance and time. There is a mix of clarifications and corrections being offered, with some participants expressing realization of their errors while others continue to question the implications of the original statement.

Contextual Notes

Some participants highlight confusion between the symbol 's' representing distance and its use as a unit for seconds. The conversation reflects ongoing exploration of these concepts without reaching a definitive consensus.

Mongster
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See I figured that since Velocity = m/s
Acceleration = m/s^2

If I have velocity divided by Acceleration
----> m/s ÷ m/s^2 = s
Relevant equations

Velocity --> s/t
Acceleration --> (v-u)/t
The attempt at a solution
My idea seems reasonable to me but somehow I couldn't apply this logic to related questions. Based on my understanding, velocity divided by acceleration gives distance as 's' but it don't seems applicable when I attempted questions with this approach.
 
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Mongster said:
. Based on my understanding, velocity divided by acceleration gives distance as 's' .

No it doesn't. You're mixing symbols with units, s is short for seconds. eg velocity is metres per second (m/s)
In the kinematic equations (SUVAT) 's' is used to represent distance (which has units of metres)

s is distance in metres (m)
u is initial velocity in metres per second (m/s)
v is final velocity in metres per second (m/s)
a is acceleration in metres per second squared (m/s2)
t is time in seconds (s)
 
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Oh wait... I see the mistake now oh my, hahaha! It is really stupid... *cringing*
But thanks a lot for the detailed explanation there, appreciate it really!

Cheers!
 
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Mongster said:
See I figured that since Velocity = m/s
Acceleration = m/s^2

If I have velocity divided by Acceleration
----> m/s ÷ m/s^2 = s
Relevant equations

Velocity --> s/t
Acceleration --> (v-u)/t
The attempt at a solution
My idea seems reasonable to me but somehow I couldn't apply this logic to related questions. Based on my understanding, velocity divided by acceleration gives distance as 's' but it don't seems applicable when I attempted questions with this approach.
That's only because the masses (m) canceled out. :rolleyes: o_O :confused:
 
Yes. s is correct. How long it takes to reach the velocity. "Long" being the "distance". It is something like a period vs frequency.

Don't confuse velocity x time = distance.
 
Dumisa Ngwenya said:
Yes. s is correct. How long it takes to reach the velocity. "Long" being the "distance". It is something like a period vs frequency.

Don't confuse velocity x time = distance.
Ummmm... What? How long it takes to reach the velocity... from what starting point?

Mongster even admitted that they confused the s (distance) with the unit s (seconds.) The problem was already solved 7 years ago, no need to add to it.

-Dan
 
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