PX155 - B4 - equations of motion

constant acceleration

• constant $a$ in 1D, not a function of $t,x,v$
  $a=\frac{d{v}^{\prime }}{d{t}^{\prime }}$
  $a{\int }_0^{t}d{t}^{\prime }={\int }_u^{v}d{v}^{\prime }$

  $⟹at=v-u$

$v=\frac{d{x}^{\prime }}{d{t}^{\prime }}$
$vd{t}^{\prime }=d{x}^{\prime }$
$⟹{\int }_0^t(u+at)d{t}^{\prime }={\int }_0^{s}d{x}^{\prime }$
$⟹[u{t}^{\prime }+\frac{1}{2}a{t}^{\prime 2}{]}_0^t=s$

- from $(1)$: $t=\frac{v-u}{a}$ in $(2)$:
$s=u\frac{v-u}{a}+\frac{1}{2}a\frac{(v+u{)}^2}{a^2}$

• eg: in the UK driving theory test, the breaking distance at $70mph$ is $75m$. what is the acceleration?
  $1mile=1600m$
  $u=1600\times 70/3600=31m{s}^{-1}$
  • $a=\frac{v^2-u^2}{2s}=-6.4m{s}^{-2}$

time dependent acceleration

• suppose $a=a(t)=\frac{d{v}^{\prime }}{d{t}^{\prime }}$

- $LHS$ might be integrable analytically

position dependent acceleration

• if $a=a(x)=\frac{dv}{dt}$
  ${\int }_u^{v}dv={\int }_0^{s}a(x)dt$
  $a(x)=\frac{dx}{dt}\frac{dv}{dx}=v\frac{dv}{dx}$
  ${\int }_0^{s}a(x)dx={\int }_u^{v}v.dv$$$\frac{1}{2}(v^2-u^2)={\int }_0^{s}a(x)dx$$

velocity dependent acceleration

• if $a=a(v)=\frac{dv}{dt}$
  ${\int }_u^v\frac{1}{a(v)}dv={\int }_0^{t}dt$
• eg: starting from rest at $t=0$, an object falls through the air subject to air resistance force $F=-\frac{mv}{\tau }$ , $\tau$ is constant. find $v(t)$ and $x(t)$
  • PX155 - A2 - newton's second law : $mg-\frac{mv}{t}=ma$
    $a=g-\frac{v}{\tau }$
    ${\int }_0^v\frac{1}{g-\frac{v}{\tau }}dv=t$
    $t=[-\tau ln|g-\frac{v}{\tau }|{]}_0^v$
    $t=-\tau (ln|g-\frac{v}{\tau }|-lng)$
    $t=-\tau ln\frac{|g-\frac{v|}{\tau }}{g}$
    • rearranging for v: $e^{\frac{-t}{\tau }}=\frac{g-\frac{v}{\tau }}{g}$
      $v=g\tau (1-e^{\frac{-t}{r}})$
    • setting $v=\frac{dx}{dt}$, we can integrate again: $x={\int }_0^{t}g\tau (1-e^{\frac{-t}{r}})$
      $=[g\tau (t+\tau e^{-t/\tau }){]}_0^t$
      $=g\tau [t-\tau (1-e^{-t/\tau })]$