Solving for t in a Physics Problem for Students

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Homework Help Overview

The discussion revolves around solving a quadratic equation derived from a physics problem involving motion, specifically relating to the variables of time and acceleration. The original poster seeks assistance in determining the value of time, t, from the equation provided.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss the transformation of the original equation into a standard quadratic form and mention the quadratic formula as a method for finding solutions. There are suggestions to utilize graphing calculators for efficiency. Additionally, some participants reflect on the distinction between mathematical and physics-related aspects of the problem.

Discussion Status

There is an ongoing exchange of ideas, with some participants providing mathematical insights while others emphasize the conceptual understanding of the physics involved. Guidance has been offered regarding the mathematical approach, but there is no explicit consensus on the next steps or the interpretation of the solutions.

Contextual Notes

Participants are navigating the challenge of distinguishing between the physics concepts and the mathematical techniques required to solve the problem. There is an acknowledgment of the potential confusion that can arise when students conflate physics with mathematical difficulties.

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1.5 = (0.25 m/s)t +(0.48m/s2 )t 2

How do I find what t is?

If you know can you please show me how to do this!

This question is relating to a Physics problem that my lab group is stump on

My Aim is Atarikid4000, Thanks. :smile:
 
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You have 0.48 t2+ 0.25 t- 1.5= 0.

In general, at2+ bt+ c= 0 has solutions
[tex]t= \frac{-b+\sqrt{b^2-4ac}}{2a}[/tex] and
[tex]t= \frac{-b-\sqrt{b^2-4ac}}{2a}[/tex].

In your problem, a= 0.48, b= 0.25, c= -1.5.
 
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I recommend writing a program in your graphing calculator that automatically will solves using the quadratic formula. Saves time.

My $.02

Paden Roder
 
Thanks for the help! :smile:
 
One more thing:

You probably already "got most of the physics" when you arrived at that equation (assuming it's correct)! [That's great!] Given your equation, "solving for t" is a math problem... not a physics problem. The only physics is deciding which of the two roots to accept for your answer.

I tell my students to try to distinguish
"problems with physics" (e.g., what is going on?, how do I set up the problem?, am I using the appropriate physical laws and definitions?, how do I interpret my answer?)
from
"problems with math" (e.g., how do I solve for x?, what is the volume of a cylinder?, should I use sin or cos?, what is the x-component of this vector?).
This helps the student (and me) zero-in on where the real problems lie.

(It's a peeve of mine that "physics" is often mistakenly blamed for problems in math.)
 

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