Question

A car passes a landmark traveling at a constant rate of 55 kilometers per hour. One hour later, a second car passes the same landmark traveling in the same direction at 75 kilometers per hour. How much time after the second car passes the landmark will it overtake the 2nd car?

Answer 1

Let t equal time. Then:
55t+55=75t
20t=55
t=55/20=2.75 hours before the second car catches the first. ☺☺☺☺

Answer 2

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


`\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}`

solve for "t"