8Ckinematicsformula


 * Deriving the Kinematics Formulae ** (not in course)

Assume that acceleration is constant.

The four kinematics formulae that are widely used in Physics and Maths can be obtained using some simple calculus and then algebra.

We start with the definition of acceleration math \\ . \qquad \dfrac{dv}{dt} = a \\. \\ . \qquad v = \displaystyle{ \int \, a \; dt} \\. \\ . \qquad v = at + c math

in kinematics we use u = v(0), so math \\ . \qquad u = a(0) + c \\. \\ . \qquad c = u math

**hence** math . \qquad v = u + at \qquad \textbf {Equation 1} math

math \\ . \qquad v = \dfrac{dx}{dt} \\ .\\ . \qquad \dfrac{dx}{dt} = u + at \\. \\ . \qquad x = \displaystyle{ \int \, u+at \; dt} \\. \\ . \qquad x = ut + \frac{1}{2}at^2 + c math
 * Now continue with Equation 1 and apply the definition of velocity. **

If we assume that x is displacement from the position at t = 0, then we have that x(0) = 0, which gives c = 0

**hence** math . \qquad x = ut + \frac{1}{2}at^2 \qquad \textbf{Equation 2} math

math \\ . \qquad v=u+at \\. \\ . \qquad at=v-u \\. \\ . \qquad a=\dfrac{v-u}{t} math
 * Now return to Equation 1 and make __a__ the subject: **

Substitute into Equation 2: math \\ . \qquad x = ut + \frac{1}{2}at^2 \\. \\ . \qquad x = ut + \frac{1}{2} \Big( \dfrac{v-u}{t} \Big) \Big( t^2 \Big) \\ .\\ . \qquad x = ut + \frac{1}{2} \big( v-u \big) t \\ .\\ . \qquad x = ut + \frac{1}{2} vt - \frac{1}{2}ut \\ .\\ . \qquad x = \frac{1}{2}ut + \frac{1}{2}vt \\ .\\ math

**hence** math . \qquad x = \frac{1}{2} \big( u + v \big) t \qquad \textbf{Equation 3} math

math \\ . \qquad v = u + at \\ .\\ . \qquad at = v-u \\. \\ . \qquad t = \dfrac{v-u}{a} math
 * Now, return to Equation 1 and make __t__ the subject **

Substitute into Equation 2: math \\ . \qquad x=ut+\frac{1}{2}at^2 \\ .\\ . \qquad x = u \Big( \dfrac{v-u}{a} \Big) + \frac{1}{2} a \Big( \dfrac{v-u}{a} \Big)^2 \\ .\\ . \qquad x = \dfrac{uv-u^2}{a} + \dfrac{a(v-u)^2}{2a^2} math

{multiply the first fraction by 2/2 and cancel a from the second fraction} math \\ . \qquad x = \dfrac{2uv - 2u^2}{2a} + \dfrac{ u^2 - 2uv + v^2 }{2a} \\. \\ . \qquad 2ax = 2uv - 2u^2 + u^2 - 2uv + v^2 \\. \\ . \qquad 2ax = v^2 - u^2 math

**hence** math . \qquad v^2 = u^2 + 2ax \qquad \textbf{Equation 4} math

.