Distance and acceleration problem

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Discussion Overview

The discussion revolves around a physics problem involving distance and acceleration, specifically focusing on the conditions under which a quadratic equation has real solutions. Participants explore how to express the minimum speed in terms of acceleration and distance.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant notes the importance of the discriminant being positive for the time of catching, t_catch, to have a real value.
  • Another participant suggests that the minimum speed, c_min, will yield a discriminant of zero and proposes solving for c_min and substituting it back into the expression for t_catch.
  • A participant asks how to solve for c_min, indicating a need for clarification.
  • Further clarification is provided regarding the quadratic formula and its application to the problem, leading to the condition c^2 - 2ab > 0 for a positive root.
  • One participant expresses confusion about the connection between the provided solution and the original question.
  • Another participant reiterates that the question seeks to express the minimum speed in terms of a and b, emphasizing that c^2 - 2ab = 0 defines the minimum value.

Areas of Agreement / Disagreement

Participants exhibit some agreement on the importance of the discriminant and the condition for real solutions, but there is disagreement regarding the clarity of the connection to the original question and the interpretation of the minimum speed.

Contextual Notes

Some participants express uncertainty about the steps involved in solving for c_min and how it relates to the original problem, indicating potential gaps in understanding or assumptions that may not have been fully articulated.

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this picture shows the problem:
http://s93755476.onlinehome.us/phys.jpg

i realize the discriminant must be positive for t_catch to have a real value. i just can't figure out how to write an expression with a and b. thanks in advance.
 
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Realizing that the discriminant must be greater than or equal to 0 is really the whole problem. The minimum speed, c, will give a discriminant of 0 (i.e., cmin2 - 2ab = 0). Just solve for cmin and plug back into your expression for tcatch from the quadratic formula.
 
how do i solve for c_min?
 
cmin2 - 2ab = 0
 
sorry I am retarded, i didn't read your response correctly. thanks for your help!
 
if you have a quadratic formula:
[tex]ax^{2} + bx + c = 0[/tex]
the solution for its roots is:

[tex]x = \frac {-b {+-} \sqrt{b^{2} - 4ac}}{2a}[/tex]

now your equation is{
[tex]\frac{1}{2}at^{2} - ct + b = 0[/tex]

so the solution to have a positive x walue would be easy to find, just sub in your constants in the equation above:

[tex]x = \frac {-(-c) {+-} \sqrt{(-c)^{2} - 4((\frac{1}{2}a)(b)}}{2(\frac{1}{2}a)}[/tex]

[tex]x = \frac {c {+-} \sqrt{c^{2} - 2ab}}{a}[/tex]

so in order to have a positive x value, there must be a positive root,
[tex]c^2 - 2ab > 0[/tex]

there is your answer. Do you understand now?
 
Hmm, I don't see the connection to the original question.
 
what don't you see about it, the question is asking to express the minimum values of the mans speed in terms of a and b. As long as

[tex]c^2 - 2ab > 0[/tex]

there is your answer. The min value would be:

[tex]c^2 - 2ab = 0[/tex]
 

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