Physics dimensions of acceleration question

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Homework Help Overview

The discussion revolves around the dimensions of acceleration, specifically how to express acceleration in terms of length, mass, and time. The original poster references a question from an online homework platform regarding the dimensional analysis of acceleration.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants explore the relationship between the dimensions of acceleration and its components, questioning the dimensions of velocity and time. There is a focus on ensuring the answer adheres to the specified dimensional format.

Discussion Status

The discussion includes attempts to clarify the dimensions of velocity and time, with some participants confirming the dimensional expression of acceleration as LT-2. However, the conversation reflects a mix of understanding and verification rather than a definitive conclusion.

Contextual Notes

The problem is constrained to expressing acceleration solely in terms of length, mass, and time, which influences the nature of the discussion and the reasoning applied by participants.

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i have a question from mastering physics(online site that i do my homework on) that i don't get.

Question: Find the dimensions [a] of acceleration.

Express your answer as powers of length (l), mass (m), and time (t).

and here's a small hint they give you for the equation for acceleration:

In physics, acceleration a is defined as the change in velocity in a certain time. This is shown by the [tex]a = \triangle v/\triangle t[/tex].. The [tex]\triangle[/tex] is a symbol that means "the change in."

[a] = _______ (answer can only be expressed in l,m and/or t)
 
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the change in v has same demension of v... same rule applied to t,too
 
vincentchan said:
the change in v has same demension of v... same rule applied to t,too


yea, but the answer can only be in length (l), mass (m), and time (t).
 
what is the demension of v, then
 
got the answer, it's [tex]LT^{-2}[/tex]
 
yup......
 

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