How were these linear motion equations derived?

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SUMMARY

The linear motion equations, specifically x = x0 + v0t + 1/2at^2 and v^2 = v0^2 + 2a(xf - xi), are derived from the principles of calculus applied to constant acceleration. The first equation is obtained by integrating the acceleration function x''(t) = a twice, leading to the position function that includes initial velocity v0 and initial position x0. The second equation is derived from the velocity function, showing the relationship between final velocity, initial velocity, and displacement under constant acceleration.

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jshpark
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how were these equations derived or found?

x = xo + vot + 1/2at2I tried to use v=(xf-xi)/(tf-ti) and a=(vf-vi)/(tf-ti)

but I don't know where the 1/2 comes from

also this one:
v2 = vo2 + 2a(xf-xi)
 
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The first one is what you get if you integrate the equation that describes constant acceleration twice. If the acceleration is constant (=a), we have
$$x''(t)=a$$ This implies that there's a constant ##v_0## such that $$x'(t)=at+v_0.$$ The constant is denoted by ##v_0## because ##x'(t)## is the velocity at time t, and the result we just found implies that ##x'(0)=v_0##, so ##v_0## is the velocity at time 0. The result we found also implies that there's a constant ##x_0## such that
$$x(t)=\frac 1 2 at^2+v_0t+x_0.$$ The constant is denoted by ##x_0## because it's equal to ##x(0)##, the position at time 0.

The other one follows from the result for the velocity above, and
$$v(t)-v_0=\int_0^t x''(t)dt=at.$$ These results imply that
$$v(t)^2=(v_0+at)^2=v_0{}^2+a^2t^2+2v_0at=v_0{}^2+2a\bigg(\frac 1 2 at^2+v_0t\bigg)=(v_0)^2+2a(x(t)-x_0).$$

I think the formula you wrote down only holds if ##v(t_i)=v_0##, so you may want to check your book to see if it says exactly what you wrote, or something slightly different.
 
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