Question

Deriving kinematic equations

Answer 1

For a = constant

Answer 2

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

Answer 3

good job

Answer 4

if a constan, a =0

v = u + at
v = u + 0
v = u

Answer 5

\[x_f = \int v(t) dt=\int (\int a dt) dt = \int(at+u)dt = \frac{1}{2}at^2 + ut +c\], 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.

Answer 6

for a = 0;

s = ut + 0.5 at^2
s = ut + 0
s = ut

Answer 7

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
\[v =u+a(\frac{\sqrt{u^2-2as}-u}{a})\]\[v = u +\sqrt{u^2-2as}-u\]\[v^2 = u^2 +2as\]

Answer 8

\[x_f = \int v(t) dt = v_{ave}t+C\]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 = \(x_f - x_i\)

Answer 9

Probably something wrong with the last post, which is s=(1/2) (u+v)t.

Answer 10

that's correct ... it assumes constant acceleration. that;s all.

Answer 11

Seriously?! I did it!?!

Answer 12

let me see how can i put it up logically.