Why Does Stopping Distance Increase with Speed?

  • Context: High School 
  • Thread starter Thread starter Bashyboy
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    Stopping distance
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Discussion Overview

The discussion centers around the relationship between stopping distance and initial speed in the context of physics, specifically exploring the kinematic equations that describe motion under constant acceleration. Participants seek to understand why stopping distance increases with the square of the initial speed.

Discussion Character

  • Exploratory, Technical explanation, Mathematical reasoning

Main Points Raised

  • One participant references a physics textbook stating that stopping distance increases with the square of the initial speed and requests an example to illustrate this concept.
  • Another participant prompts a discussion on how distance relates to speed under constant acceleration, suggesting the need to identify a relevant kinematic formula.
  • A participant proposes the kinematic equation v^2 = vi^2 + 2a(x - xi) as a means to understand the relationship, indicating that it can be solved for the change in distance (delta x).
  • The same participant reiterates the use of the kinematic equation, clarifying that the initial speed (Vi) is the starting speed and the final speed (V) is zero when considering stopping distance.

Areas of Agreement / Disagreement

The discussion does not reach a consensus, as participants are exploring the relationship and deriving equations without concluding on specific examples or applications.

Contextual Notes

Participants have not fully resolved the implications of the kinematic equations, and there may be assumptions regarding constant acceleration that are not explicitly stated.

Bashyboy
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In my physics textbook, it states that the stopping distance increases with the square of the intial speed. Could someone provide me with an example of why this is true?

Thank you
 
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For constant acceleration, how does distance relate to speed? (Look for a connecting kinematic formula.)
 
Would it be the kinematic equation v^2 = vi^2 + 2a(x - xi) solved for delta x?
 
Bashyboy said:
Would it be the kinematic equation v^2 = vi^2 + 2a(x - xi) solved for delta x?
Exactly. Vi would be the initial speed; V = 0, the final speed.
 

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