How Do You Linearize the Equation d=vit + 1/2at^2?

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SUMMARY

The discussion focuses on linearizing the equation d = vit + 1/2at² by transforming it into different variable relationships: d/t² vs 1/t, t/d vs t²/d, and 1/d vs 1/t. The key approach involves rearranging the equation into the linear form y = mx + b. Specifically, for the first transformation, the equation becomes d/t² = vi/t + 1/2a, where y is defined as d/t² and x as 1/t, resulting in a linear graph with a slope of vi. The participant expresses confusion about manipulating the original equation but ultimately identifies the correct substitutions and transformations.

PREREQUISITES
  • Understanding of kinematic equations, specifically d = vit + 1/2at²
  • Knowledge of linear equations in the form y = mx + b
  • Familiarity with graphing techniques for linear and non-linear relationships
  • Basic algebraic manipulation skills, including substitution and division
NEXT STEPS
  • Study the process of linearizing quadratic equations in physics
  • Learn about graphing techniques for analyzing linear vs non-linear relationships
  • Explore the use of the quadratic formula in solving for variables in kinematic equations
  • Practice transforming other kinematic equations into linear forms for graphing
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Students studying physics, particularly those focusing on kinematics, as well as educators looking for methods to teach linearization of equations effectively.

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Homework Statement



basically I am suppose to linearize d=vit + 1/2at^2 into d/t^2 vs 1/t, t/d vs t^2/d and 1/d vs 1/t
and its asking which graphs would be linear and non-linear.

Homework Equations



d=vit + 1/2at^2

The Attempt at a Solution



all i know is that in order to linearize this i need to make it into the format y=mx+b from there i can see which variables alternate and which remain constant.
now I am confused on how I am suppose to get d/t^2 vs 1/t for the first one so i started to solve for t but i ended up needing to use the quadratic equation since its ax^2 + bx + c=0 but that really didnt help me since i coudlnt isolate "d" again. So basically I am kinda lost on how to manipulate this formula.
 
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For the first part, I'd let y = d/t^2, let x = 1/t. Can you get the d and t equation into an equation with y and x... substitute appropriately and manipulate the formulas.
 
If you divide the equation by t^2 you get

\frac{d}{t^2} = \frac{v_i}{t} + \frac{1}{2}a

If you wanted that against 1/t, basically make x a substitution of 1/t, you would have a graph that looked pretty linear with a slope of vi.

Do you see how it works?
 

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