Question 1195776

{{{D(t)=18t^2−130t}}}

given: {{{D(t)=150}}} feet

{{{150=18t^2−130t}}}

{{{0=18t^2−130t-150}}}.......use quadratic formula

 {{{t = (-(-130) +- sqrt( (-130)^2-4*18*(-150) ))/(2*18) }}}

{{{t = (130 +- sqrt( 27700 ))/36 }}}

{{{t = (130 +- 10sqrt(277))/36 }}}


exact solutions:

{{{t = (130 +10sqrt(277))/36 }}} -> {{{t=65/18 + (5sqrt(277))/18}}}

or
{{{t = (130 - 10sqrt(277))/36 }}}->{{{t=65/18 -(5sqrt(277))/18}}}


approximately:

{{{t=8.23 }}} seconds

or
{{{t=-1.012032493637}}}-> disregard negative solution, time cannot be negative


so, the time it would take for the car to skid {{{150}}} feet is approximately {{{8.23 }}} seconds