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Callisto
 3 years ago
Deriving kinematic equations
Callisto
 3 years ago
Deriving kinematic equations

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Callisto
 3 years ago
Best ResponseYou've already chosen the best response.3\[v=\int a dt=at+c\]c = initial velocity So, \[v=u+at\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0if a constan, a =0 v = u + at v = u + 0 v = u

Callisto
 3 years ago
Best ResponseYou've already chosen the best response.3\[x_f = \int v(t) dt=\int (\int a dt) dt = \int(at+u)dt = \frac{1}{2}at^2 + ut +c\], c = initial position So, \[x_f = ut +\frac{1}{2}at^2 + x_i\]\[x_fx_i = ut +\frac{1}{2}at^2\]\[s = ut+\frac{1}{2}at^2\]s=displacement.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0for a = 0; s = ut + 0.5 at^2 s = ut + 0 s = ut

Callisto
 3 years ago
Best ResponseYou've already chosen the best response.3From the third post, \[s=ut+\frac{1}{2}at^2\]\[at^2+2ut 2s = 0\]\[t^2 + \frac{2u}{a}t  \frac{2s}{a}=0\]\[(t+\frac{u}{a})^2  \frac{u^2}{a^2}\frac{2s}{a}=0\]\[(t+\frac{u}{a})^2 = \frac{u^22as}{a^2}\]\[t+\frac{u}{a} = \frac{\sqrt{u^22as}}{a}\]\[t = \frac{\sqrt{u^22as}u}{a} (1)\] Put (1) into v=u+at \[v =u+a(\frac{\sqrt{u^22as}u}{a})\]\[v = u +\sqrt{u^22as}u\]\[v^2 = u^2 +2as\]

Callisto
 3 years ago
Best ResponseYou've already chosen the best response.3\[x_f = \int v(t) dt = v_{ave}t+C\]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 = \(x_f  x_i\)

Callisto
 3 years ago
Best ResponseYou've already chosen the best response.3Probably something wrong with the last post, which is s=(1/2) (u+v)t.

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.0that's correct ... it assumes constant acceleration. that;s all.

Callisto
 3 years ago
Best ResponseYou've already chosen the best response.3Seriously?! I did it!?!

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.0let me see how can i put it up logically.

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1351833416633:dw

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.0now you just need to show that for constant accn V_av = (u+v)/2 ... wanna try it?

Callisto
 3 years ago
Best ResponseYou've already chosen the best response.3Huh!? I thought it's some maths.. Oh..How to start??

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.0well, 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.

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.0or simply put ,,, average velocity = distance/time

Callisto
 3 years ago
Best ResponseYou've already chosen the best response.3\[v_f= v_i+\int_0^{t_f} a dt=v_i + at\] So, \[v_f=v_i + at\]

Callisto
 3 years ago
Best ResponseYou've already chosen the best response.3\[x_f = x_i+\int_0^{t_f} v(t) dt= x_i+\int_0^{t_f} (v_i+at) dt = x_i+v_it+\frac{1}{2}at^2\] So, \[x_f = x_i+v_it+\frac{1}{2}at^2\] That is \[x_f  x_i = v_it+\frac{1}{2}at^2\]\[s= ut+\frac{1}{2}at^2\]

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1351853269560:dw

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1351853406639:dw
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