Question 429287
Let {{{d}}} = distance front car has to go
Let {{{t}}} = time for them to meet
equation for 40 mi/hr car
(1) {{{d = 40t}}}
equation for 53 mi/hr car
(2) {{{d + 300/5280 = 53t }}}
Substitute (1) into (2)
(2) {{{40t + 300/5280 = 53t }}}
{{{ 13t = 300/5280 }}}
{{{ t = .0568/13 }}}
{{{ t = .00437 }}} hrs
{{{ t = .00437 * 60*60 = 15.73 }}} hrs x min/hr x sec/min
They hit in 15.73 sec
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If the front car goes 10 mi/hr
(1) {{{ d = 10t }}}
(2) {{{d + 300/5280 = 53t }}}
(2) {{{10t + 300/5280 = 53t }}}
{{{ 43t = .0568 }}}
{{{ t = .00132 }}} hrs
{{{ t = .00132 * 60 * 60 = 4.757 }}} sec
They hit in 4.757 sec