We assume [tex]a=const[/tex] (acceleration is constant. We apply the equation
[tex]v^2=v0^2+2as[/tex] where s is the distance to stop [tex]v=0(m/s)[/tex]. We find the acceleration from this equation
[tex]a=-v0^2/(2s)=-70^2/(2*3) =-816.7 (m/s^2)
[/tex]
We know the acceleration, thus we find the distance necesssary to stop when initial speed is [tex]v=140 (m/s)[/tex]
[tex]s=-v0^2/(2a) =140^2/(2*816.7)=12 (m)[/tex]
