Understanding Kinematics: Uniformly Accelerated Motion Equation Explained

  • Thread starter Thread starter LuGoBi
  • Start date Start date
  • Tags Tags
    Kinematics
Click For Summary
SUMMARY

The discussion focuses on the equations governing uniformly accelerated motion, specifically addressing the common misconception in applying the equations incorrectly. The correct relationship for distance traveled under uniform acceleration is derived from the equations S = So + Vot + (1/2)at^2. The error arises when substituting final velocity (V) into the distance equation, which only applies to constant velocity scenarios. The average velocity must be used instead, leading to the correct formulation of S = So + (Vo + Vf)/2 * t.

PREREQUISITES
  • Understanding of basic kinematics concepts
  • Familiarity with the equations of motion
  • Knowledge of calculus, specifically derivatives
  • Ability to manipulate algebraic equations
NEXT STEPS
  • Study the derivation of the equations of motion in detail
  • Learn about the concept of average velocity in uniformly accelerated motion
  • Explore the application of calculus in physics, particularly in motion analysis
  • Practice solving problems involving uniformly accelerated motion using real-world examples
USEFUL FOR

Students of physics, educators teaching kinematics, and anyone interested in mastering the principles of motion under uniform acceleration.

LuGoBi
Well, I think this one is pretty simple, but still, I don't know how to solve it.

We all know that for uniform velocity in a straight line the following equation sets the relationship between time and distance traveled: S = So + Vt (Eq. 1)

When it comes to uniformly accelerated motion, the only difference is that the velocity is changing constantly, according to the following equation: V = Vo + at (Eq. 2)

Now, if you insert Eq. 2 in Eq. 1 you get: S = So + Vot + at^2 (Eq. 3)

But we all know the correct equation is S = So + Vot + at^2/2! Besides, the second derivative of Eq. 3 is 2a, when the correct one is, by definition, a, obviously. So it's clearly wrong. What's the deal with this?
 
Physics news on Phys.org
In Eq. 2, V stands for the final speed after some time. To use it in Eq. 1, you'd have to replace V by the average speed, since that equation only applies for constant speed (or average speed). Since the acceleration is uniform, the average speed is just (Vo + Vf)/2 = (Vo + Vo + at)/2 = Vo + at/2. Plug that into Eq. 1 and see what happens.
 
Damn it, that's beautiful. Thank you very much.
 
The reason for the wrong answer is that ;
v = u + at is equation to find final velocity in constant acceleration, and you are putting this final velocity in 2nd equation (BUT VELOCITY IS CHANGING AT EVERY POINT)

SO,YOUR EQUATION MODIFIES AS:-

S = So + v1t1 + v2t2 + ...(where t1 + t2 +...= t)

so,in general,u MUST USE,

dS = v.dt
 

Similar threads

  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 49 ·
2
Replies
49
Views
4K
  • · Replies 10 ·
Replies
10
Views
3K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 2 ·
Replies
2
Views
2K