Question 641310
Let {{{ s }}} = the speed of the slower train in km/hr
{{{ s + 10 }}} = the speed of the faster train in km/hr
Let {{{ t }}} = time in hrs for the slower train to go 160 km
{{{ t - .5 }}} = time in hrs for faster train to go 160 km
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Slower train:
(1) {{{ 160 = s*t }}}
Faster train:
(2) {{{ 160 = ( s + 10 )*( t - .5 ) }}}
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(2) {{{ 160 = s*t + 10t - .5s - 5 }}}
From (1)
(1) {{{ t = 160/s }}}
and
(2) {{{ 160 = s*( 160/s) + 10*( 160/s ) - .5s - 5 }}}
(2) {{{ 160s = 160s + 1600 - .5s^2 - 5s }}}
(2) {{{ .5s^2 + 5s - 1600 = 0 }}}
(2) {{{ 5s^2 + 50s - 16000 = 0 }}}
(2) {{{ s^2 + 10s - 3200 = 0 }}}
{{{ s = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}
{{{ a = 1 }}}
{{{ b = 10 }}}
{{{ c = -3200 }}}
{{{ s = (-10 +- sqrt( 10^2 - 4*1*(-3200) ))/(2*1) }}}
{{{ s = (-10 +- sqrt( 100 + 12800 )) / 2 }}}
{{{ s = (-10 +- sqrt( 12900 )) / 2 }}}
Check my work and then check the results
for {{{ s }}}, the slower train and {{{ s+10 }}},
the faster train