Question 1075031
Tommy's head start in miles is:
{{{ 3*2 = 6 }}} mi
Let {{{ t }}} = the time that they both spend
running starting when Zach starts and ending
when they are {{{ 8 }}} mi apart
-------------------------------------------
Tommy's distance:
{{{ 6 + 3t }}}
Zach's distance:
{{{ 4t }}}
-------------------------
using Pythagorean theorem:
{{{ ( 6 + 3t )^2 + ( 4t )^2 = 8^2 }}}
{{{ 36 + 36t + 9t^2 + 16t^2 = 64 }}}
{{{ 25t^2 + 36t  - 28 = 0 }}}
Use quadratic formula
{{{ t = ( -b +- sqrt( b^2 - 4*a*c ))/(2*a) }}}
{{{ a = 25 }}}
{{{ b = 36 }}}
{{{ c = -28 }}}
{{{ t = ( -36 +- sqrt( 36^2 - 4*25*(-28) ))/(2*25) }}}
{{{ t = ( -36 +- sqrt( 1296 + 2800 ))/50 }}}
{{{ t = ( -36 +- sqrt( 4096 ))/50 }}}
{{{ t = ( -36 + 64)/50 }}}
{{{ t = 28/50 }}}
{{{ t = .56 }}} hrs
convert to min
{{{ t = .56*60 }}}
{{{ t = 33.6 }}}
They will be 8 mi apart in 33.6 min
-----------------------------------
check:
{{{ ( 6 + 3t )^2 + ( 4t )^2 = 8^2 }}}
{{{ ( 6 + 3*.56 )^2 + ( 4*.56 )^2 = 8^2 }}}
{{{ ( 6 + 1.68 )^2 + 2.24^2 = 64 }}}
{{{ 7.68^2 + 2.24^2 = 64 }}}
{{{ 58.9824 + 5.0176 = 64 }}}
{{{ 64 = 64 }}}
OK