Kinematic Equations: V_0, V, t, a, x

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SUMMARY

The discussion focuses on the kinematic equations used to describe motion in physics, specifically detailing the relationships between initial velocity (V_0), final velocity (V), time (t), acceleration (a), and distance (x). Key equations highlighted include x = V_0t + (1/2)at², V = V_0 + at, and V² = V_0² + 2ax. The clarification of the correct formulation for distance traveled over time is emphasized, correcting misconceptions about the equation structure.

PREREQUISITES
  • Understanding of basic physics concepts, particularly motion.
  • Familiarity with algebraic manipulation of equations.
  • Knowledge of constant acceleration scenarios.
  • Ability to interpret and apply kinematic equations in problem-solving.
NEXT STEPS
  • Study the derivation of kinematic equations in physics textbooks.
  • Practice solving problems involving constant acceleration using the kinematic equations.
  • Explore graphical representations of motion and their relation to kinematic equations.
  • Learn about the applications of kinematic equations in real-world scenarios, such as projectile motion.
USEFUL FOR

Students of physics, educators teaching motion concepts, and anyone interested in understanding the principles of kinematics and their applications in real-world problems.

Tom McCurdy
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Since a good amount of the posts refer to problems involving kinematics I figured this may be helpful

[tex]V_0 =[/tex] Inital Velocity
[tex]V =[/tex] Final Velocity
[tex]t=[/tex] time
[tex]a=[/tex] acceleration
[tex]x=[/tex] distance

[tex]x=V_0t+\frac{1}{2}at^2[/tex]

[tex]V=V_0+at[/tex]

[tex]V^2=V_0^2+2ax[/tex]

[tex]x= \frac{1}{2}(V_0+V)*t[/tex]
 
Last edited:
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Assuming that you mean "x is the distance traveled in time t" and that a is a constant, then
[tex]x= V_0t+ \frac{1}{2}t^2[/tex]
NOT
[tex]x= V_0t*\frac{1}{2}t^2[/tex]
 
Very True... it does equal plus
 

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