Can You Derive Equations Using UAM Principles?

  • Thread starter Thread starter jensgt
  • Start date Start date
  • Tags Tags
    deriving
Click For Summary
SUMMARY

The discussion focuses on deriving equations 5-1 and 5-2 using Uniform Acceleration Motion (UAM) principles. Participants emphasize the importance of utilizing standard constant acceleration equations for both the x and y directions separately, rather than relying on memorized range or height equations. Key equations mentioned include Vx = Vox + Axt and X = Xo + Voxt + 1/2axt². The conversation highlights the necessity of understanding the foundational equations of motion to successfully derive the desired equations.

PREREQUISITES
  • Understanding of Uniform Acceleration Motion (UAM) principles
  • Familiarity with basic kinematic equations
  • Knowledge of vector components in physics
  • Ability to manipulate algebraic equations
NEXT STEPS
  • Study the derivation of kinematic equations in physics
  • Learn how to apply UAM principles to solve projectile motion problems
  • Explore the relationship between horizontal and vertical motion in projectile dynamics
  • Practice deriving equations from first principles in physics
USEFUL FOR

This discussion is beneficial for physics students, educators, and anyone seeking to deepen their understanding of kinematic equations and their applications in motion analysis.

jensgt
Messages
7
Reaction score
0

Homework Statement


Can you come up with (i.e. derive) equations 5-1 and 5-2 on your own using UAM equations? Try it


Homework Equations



This is the equation I need to end up at...

R=Vox/g(Voy+(Voy^+2gh)^1/2


The Attempt at a Solution



I am not really sure where to start. I was trying to use the equations for range and height but it does not seem to be getting me anywhere. This is my first physics course.
 
Physics news on Phys.org
welcome to pf!

hi jensgt! welcome to pf! :smile:

(try using the X2 button just above the Reply box :wink:)

don't try to memorise (or use) the range or height equations :wink:, go back to the standard constant acceleration equations, for the x and y directions (separately)

they'll use the same t …

show us what you get :smile:
 
The equations I know are

Vx = Vox + Axt

X = Xo + Voxt + 1/2axt^2

X-Xo = [(Vox + Vx)/2]t

Vx^2 = Vox^2 +2ax(X-Xo)

I am just confused as to where to start. I derived the other one he asked us about this one has me stuck.
 
hi jensgt! :smile:

(just got up :zzz:)
jensgt said:
I am just confused as to where to start.

use x = xo + Vxot + 1/2axt2

and y = yo + Vyot + 1/2ayt2

(obviously, ax = 0) …

show us what you get :smile:

(btw, never use "x" for times, it's too confusing … use "*" instead, or nothing at all! :wink:)
 

Similar threads

Replies
3
Views
2K
Replies
13
Views
2K
Replies
1
Views
1K
  • · Replies 4 ·
Replies
4
Views
1K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 14 ·
Replies
14
Views
3K
  • · Replies 4 ·
Replies
4
Views
1K
  • · Replies 2 ·
Replies
2
Views
3K