Respuesta :
recall your d = rt, distance = rate * time
one car is going at 55kph, at say hmmm "t" time
the second car zooms by at 75kph, an hour later, thus " t + 1 "
notice, overtaking the 1st car, simply means, the second car comes from behind approaches the first car from behind and "gets there", at that very second, both cars have travelled a distance, say "d"
if the first car travelled "d" kilometers
for the second car to "get there" has to had travelled "d" distance as well
[tex]\bf \begin{array}{lccclll} &distance&rate&time\\ &-----&-----&-----\\ \textit{first car}&d&55&t\\ \textit{second car}&d&75&t+1 \end{array} \\\\\\ \begin{cases} d=55t\\ d=75(t+1)\\ -------\\ 55t=75(t+1) \end{cases}[/tex]
solve for "t"
one car is going at 55kph, at say hmmm "t" time
the second car zooms by at 75kph, an hour later, thus " t + 1 "
notice, overtaking the 1st car, simply means, the second car comes from behind approaches the first car from behind and "gets there", at that very second, both cars have travelled a distance, say "d"
if the first car travelled "d" kilometers
for the second car to "get there" has to had travelled "d" distance as well
[tex]\bf \begin{array}{lccclll} &distance&rate&time\\ &-----&-----&-----\\ \textit{first car}&d&55&t\\ \textit{second car}&d&75&t+1 \end{array} \\\\\\ \begin{cases} d=55t\\ d=75(t+1)\\ -------\\ 55t=75(t+1) \end{cases}[/tex]
solve for "t"
Let t equal time. Then:
55t+55=75t
20t=55
t=55/20=2.75 hours before the second car catches the first. ☺☺☺☺
55t+55=75t
20t=55
t=55/20=2.75 hours before the second car catches the first. ☺☺☺☺