Question 332220
The braking distance (in feet) of a car going V mph is given by
 {{{d(v)=v^2/20+v}}} v is greater or equal to 0.
how fast would the car have been traveling for a braking distance of 150feet?
 round to nearest mile per hour.
:
Write it:
{{{v^2/20+v}}} = 150
multiply by 20, results
v^2 + 20v = 20(150)
:
v^2 + 20v = 3000
:
v^2 + 20v - 3000 = 0
Solve for v using the quadratic formula
{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}} 
in this equation; x=v; a=1; b=20; c= -3000
{{{v = (-20 +- sqrt(20^2-4*1*-3000 ))/(2*1) }}}
: 
{{{v = (-20 +- sqrt(400-(-12000) ))/2 }}}
:
{{{v = (-20 +- sqrt(12400 ))/2 }}}
Two solutions, we only want the positive solution
{{{v = (-20 + 111.355)/2 }}}
v = {{{91.355/2}}}
v = 45.68 mph for a stopping distance of 150 ft
:
:
See if that flies in the original equation
{{{d(v)=45.68^2/20+45.68}}}
{{{d(v)=2086.45/20+45.68}}}
{{{d(v)=104.32+45.68}}}
d(v) = 150.00, confirms our solution