Solving Quadratic Equation: Find the Time

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Homework Help Overview

The discussion revolves around solving a quadratic equation related to motion, specifically in the context of determining time in a physics problem involving acceleration and position. The original poster attempts to find which of the two calculated times is relevant for their scenario.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the implications of obtaining two positive time values from the quadratic formula and question how to interpret these in the context of motion. They explore the significance of negative acceleration and the physical meaning of the two times in relation to the object's trajectory.

Discussion Status

Participants are actively engaging with the problem, offering insights into the nature of the solutions derived from the quadratic equation. There is a recognition that both times could be valid under certain conditions, and some guidance is provided regarding the interpretation of the results in the context of projectile motion.

Contextual Notes

There is a noted absence of the original question details, which may affect the clarity of the discussion. Participants are also considering the implications of negative acceleration and the physical context of the motion described.

intenzxboi
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k so i used
x-xo= Vo t + 1/2 at^2

plugged in all my values
then moved everything to one side and used the quadratic formula

here is my question.

when i solve for it i get
t=40.28 and t= 40.58

So which t is my time?
 
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Er.. well, you seem to not have included the actual question?
 
lets just use imaginary numbers

if u solve for t and get one + and one - answer u know time can't be negative if you are traveling forward.. so the answer is the + one

but what if your t comes out to be both + and you are going foward with -acceleration
 
This isn't something that has a general rule - sometimes both of them can be the correct time and other times there's some condition that rules the other one out.

Looking at your equation now it's the one used for 2D motion.. Taking that as an example, if you're considering a projectile and get two times it should be because the object is actually at the same horizontal position twice - once when going up and once when going down. It depends on what you want to calculate, in that case.

Probably this is no help, but I honestly can't do much more. Sorry ;/
 
I actually got it... had to think about it for a while..
my guess is that its the lower time because that's when the object gets there first...
the second time is when v=0 and the -acc. is making it go backwards so it goes back past the same position

right?
 
Yeah, if you're considering something that's thrown straight up into the air that would be the case, I'd say.
 

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