Question 696510
Equation for normal trip from A to B is
(1) {{{ d = r*t }}}
In {{{ 1 }}} hr, the train covers {{{ d[1] = r*1 }}}
The train is stopped for 30 minutes
{{{ 1.5 }}} hrs have now passed
Now it's rate is {{{ (4/5)*r }}} for the remaining time of
{{{ t + 2 - 1.5  = t + .5 }}}
The distance covered is {{{ d[2] = (4/5)*r*( t + .5 ) }}}
Now I have {{{ d = d[1] + d[2] }}}
(2) {{{ d = r + (4/5)*r*( t + .5 ) }}}
So far I have 2 equations with 3 unknowns,  so
I need 1 more equation
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If the train covered {{{ 80 }}} mi at rate = {{{ r }}}, 
it traveled {{{ 80 = r*t[1] }}}
{{{ t[1] = 80/r }}}
Now the trip takes {{{ t + 1 }}} hrs, so the rest of the trip takes
{{{ t + 1 - 80/r }}}
The train is stopped for 30 minutes, so
There is {{{ t + 1 - .5 - 80/r }}} hrs left
The distance covered after the accident is then
{{{ d[3] = (4/5)*r*( t + .5 - 80/r ) }}}, and
(3) {{{ d = 80 + (4/5)*r*( t + .5 - 80/r ) }}}
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Substitute (1) into (2)
(1) {{{ d = r*t }}}
(2) {{{ d = r + (4/5)*r*( t + .5 ) }}}
(2) {{{ r*t = r + (4/5)*r*( t + .5 ) }}}
(2) {{{ 5r*t = 5r + 4*r*( t + .5 ) }}}
(2) {{{ 5r*t = 5r + 4r*t + 2r }}}
(2) {{{ r*t = 7r }}}
(2) {{{ t = 7 }}}
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(3) {{{ d = 80 + (4/5)*r*( t + .5 - 80/r ) }}}
(3) {{{ d = 80 + (4/5)*r*( 7 + .5 - 80/r ) }}}
(3) {{{ d = 80 + (4/5)*( 7.5r - 80 ) }}}
(3) {{{ d = 80 + 6r - 64 }}}
(3) {{{ d = 16 + 6r }}}
and since
{{{ d = r*t }}}
{{{ d = 7r }}}
(3) {{{ 7r = 16 + 6r }}}
(3) {{{ r = 16 }}}
The usual rate of the train is 16 mi/hr
check answer:
(1) {{{ d = r*t }}}
(1) {{{ d = 16*7 }}}
(1) {{{ d = 112 }}} mi
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(2) {{{ d = r + (4/5)*r*( t + .5 ) }}}
(2) {{{ d = 16 + (4/5)*16*( 7 + .5 ) }}} 
(2) {{{ d = 16 + (4/5)*16*7.5 }}} 
(2) {{{ d = 16 + .8*120 }}} 
(2) {{{ d = 16 + 96 }}}
(2) {{{ d = 112 }}} mi
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(3) {{{ d = 80 + (4/5)*r*( t + .5 - 80/r ) }}}
(3) {{{ d = 80 + .8*16*( 7 + .5 - 80/16 ) }}}
(3) {{{ d = 80 + .8*16*( 7 + .5 - 5 ) }}}
(3) {{{ d = 80 + .8*16*2.5 }}}
(3) {{{ d = 80 + 32 }}}
(3) {{{ d = 112 }}} mi
OK