Integrating V = V(initial) + at

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To derive the equation x = x_0 + v_0 t + 1/2 a t^2 from v = v_0 + at, one must integrate both sides with respect to time. The integration of v, which is the velocity, gives the displacement x, while the initial velocity v_0 contributes to the first term. The acceleration term at is integrated to yield 1/2 a t^2, accounting for the change in velocity over time. The integration should be performed from the initial time (t = 0) to time t, with x_0 representing the initial position as the integration constant. This process clarifies the relationship between velocity, acceleration, and displacement in kinematics.
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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?
 
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Somethings messed up with your formatting. Is your question how to get from this:
v = v_0 + at
to this:
x = x_0 + v_0 t + 1/2 a t^2
If so, it's straightforward integration, since v = dx/dt.
 
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?
 
Yes, you are integrating both terms, v_0 and a t, from the initial time (t = 0) to t. x_0 is just the integration constant = initial position.
 
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