Integrating V = V(initial) + at

  • Thread starter Alem2000
  • Start date
  • Tags
    Integrating
In summary, the conversation is about integrating the equation v=v(initial)+at to get x=x(initial)+v(initial)t+1/2at^2 and the confusion around using the uv-integral theorem to get the second equation for constant a. The conversation also touches on the integration process and the role of x(initial) and v(initial).
  • #1
Alem2000
117
0
sorry i didnt use latex..I tried...its annoying


V=V(initial) + at...when you integrate both sides you get x=x(initial) + ..?
I had a little trouble trying to integrate the (at) do you use the uv-integral(v DV) theorm...well i used that and it and it didnt give me the second equation for constant a...near to it but off by two terms. Would anyone be kind enough to show me?
 
Last edited:
Physics news on Phys.org
  • #2
Somethings messed up with your formatting. Is your question how to get from this:
[tex]v = v_0 + at[/tex]
to this:
[tex]x = x_0 + v_0 t + 1/2 a t^2[/tex]
If so, it's straightforward integration, since v = dx/dt.
 
  • #3
I understand the first two terms must give you x and x initial. But I can't seem to grasp where the rest..are you integrating from time initial to time t?
 
  • #4
Yes, you are integrating both terms, [itex]v_0[/itex] and [itex]a t[/itex], from the initial time (t = 0) to t. [itex]x_0[/itex] is just the integration constant = initial position.
 

1. What is the meaning of V = V(initial) + at?

The equation V = V(initial) + at represents the final velocity of an object (V) after a given amount of time (t) starting from an initial velocity (V(initial)) and accelerating at a constant rate (a).

2. How is this equation derived?

This equation can be derived from the basic kinematic equation V = u + at, where V is the final velocity, u is the initial velocity, a is the acceleration, and t is the time. By rearranging this equation, we get V = V(initial) + at.

3. What units are used for each variable in this equation?

V is typically measured in meters per second (m/s), V(initial) in meters per second (m/s), a in meters per second squared (m/s^2), and t in seconds (s).

4. Can this equation be used for objects with changing acceleration?

No, this equation can only be used for objects with a constant acceleration. For objects with changing acceleration, the equation becomes more complex and involves calculus.

5. How is this equation useful in real-world applications?

This equation is commonly used in physics and engineering to calculate the final velocity of an object under constant acceleration. It can be applied to various scenarios such as calculating the speed of a falling object or the velocity of a car accelerating from rest.

Similar threads

I How did Hamilton derive the characteristic function V in his essay?
    • Classical Physics
Replies
3
Views
582
I Impulsive force and simple integration
    • Classical Physics
Replies
12
Views
792
I Time derivative of the moment of inertia tensor
    • Classical Physics
Replies
5
Views
2K
I What is the Order of Integration in a Double Integral?
    • Classical Physics
Replies
7
Views
1K
I Is There a Better Model for the Relationship Between Friction and Velocity?
    • Classical Physics
2
Replies
63
Views
3K
B Integration by Parts, an introduction I get confused with
    • Calculus
Replies
2
Views
931
I Pushing a rod in a magnetic field
    • Classical Physics
Replies
15
Views
536
A The equations of variable mass systems
    • Classical Physics
Replies
4
Views
724
I Integrating Force: Is F=(m/t)*v Correct?
    • Classical Physics
Replies
14
Views
1K
A Dx in an integral vs. differential forms
    • Classical Physics
Replies
6
Views
1K

Hot Threads

Recent Insights

Back
Top