Question 297107
Let {{{d}}} = the dog's speed running
Let {{{t}}} = the time for the train to just reach the bridge
Let {{{b}}} = the length of the bridge in miles
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In words:
(dog's speed) x (time for train to reach bridge + time for train to cross bridge) = (3/4)x(the length of the bridge)
As an equation:
(1) {{{d*(t + b/30) = (3/4)*b}}}
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In words:
(dog's speed) x (time for train to reach bridge) = (1/4)*(length of bridge)
As an equation:
(2) {{{d*t = (1/4)*b}}}
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In words:
(train's speed plus dog's speed) x (time for train to reach bridge) = (distance train travels to reach bridge plus distance dog runs)
As an equation:
(3) {{{(30 + d)*t = 30t + (1/4)*b}}}
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This is 3 equations and 3 unknowns, so it's solvable
(1) {{{d*(t + b/30) = (3/4)*b}}}
(1) {{{dt + db/30 = (3/4)*b}}}
(1) {{{60dt + 2db = 45b}}}
and
(2) {{{dt = (1/4)*b}}}
and
(3) {{{(30 + d)*t = 30t + (1/4)*b}}}
(3) {{{30t + dt = 30t + (1/4)*b}}}
(3) {{{120t + 4*dt = 120t + b}}}
I can substitute (2) in (1)
(1) {{{60*(1/4)*b + 2db = 45b}}}
(1) {{{15b + 2db = 45b}}}
(1) {{{2db = 30b}}}
(1) {{{d = 15}}}
The dog was running 15 mi/hr
check answer:
(2) {{{dt = (1/4)*b}}}
(2) {{{15t = (1/4)*b}}}
(2) {{{b = 60t}}}
Substitute in (3)
(3) {{{120t + 4*dt = 120t + b}}}
(3) {{{120t + 4*15*t = 120t + 60t}}}
(3) {{{120t + 60t = 120t + 60t}}}
OK