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Deriving kinematic equations

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For a = constant
\[v=\int a dt=at+c\]c = initial velocity So, \[v=u+at\]
good job

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Other answers:

if a constan, a =0 v = u + at v = u + 0 v = u
\[x_f = \int v(t) dt=\int (\int a dt) dt = \int(at+u)dt = \frac{1}{2}at^2 + ut +c\], c = initial position So, \[x_f = ut +\frac{1}{2}at^2 + x_i\]\[x_f-x_i = ut +\frac{1}{2}at^2\]\[s = ut+\frac{1}{2}at^2\]s=displacement.
for a = 0; s = ut + 0.5 at^2 s = ut + 0 s = ut
From the third post, \[s=ut+\frac{1}{2}at^2\]\[at^2+2ut -2s = 0\]\[t^2 + \frac{2u}{a}t - \frac{2s}{a}=0\]\[(t+\frac{u}{a})^2 - \frac{u^2}{a^2}-\frac{2s}{a}=0\]\[(t+\frac{u}{a})^2 = \frac{u^2-2as}{a^2}\]\[t+\frac{u}{a} = \frac{\sqrt{u^2-2as}}{a}\]\[t = \frac{\sqrt{u^2-2as}-u}{a} -(1)\] Put (1) into v=u+at \[v =u+a(\frac{\sqrt{u^2-2as}-u}{a})\]\[v = u +\sqrt{u^2-2as}-u\]\[v^2 = u^2 +2as\]
\[x_f = \int v(t) dt = v_{ave}t+C\]c = initial position \[x_f = v_{ave}t +x_i\]\[x_f = \frac{1}{2}(u+v)t +x_i \]\[s = \frac{1}{2}(u+v)t\]displacement = s = \(x_f - x_i\)
Probably something wrong with the last post, which is s=(1/2) (u+v)t.
that's correct ... it assumes constant acceleration. that;s all.
Seriously?! I did it!?!
let me see how can i put it up logically.
now you just need to show that for constant accn V_av = (u+v)/2 ... wanna try it?
Huh!? I thought it's some maths.. Oh..How to start??
try using MVT
MVT.... again.../_\
well, you can do this without MVT ,,, i was wondering if i could improve my skills with MVT. try expanding v(t) in the expression of average.
or, V_av = s/t
expanding v(t)?
v(t) = u + at
or simply put ,,, average velocity = distance/time
Fail ._.
\[v_f= v_i+\int_0^{t_f} a dt=v_i + at\] So, \[v_f=v_i + at\]
\[x_f = x_i+\int_0^{t_f} v(t) dt= x_i+\int_0^{t_f} (v_i+at) dt = x_i+v_it+\frac{1}{2}at^2\] So, \[x_f = x_i+v_it+\frac{1}{2}at^2\] That is \[x_f - x_i = v_it+\frac{1}{2}at^2\]\[s= ut+\frac{1}{2}at^2\]

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