Respuesta :
recall your d = rt, distance = rate * time
the faster car of 65km/h, will "overtake" the first car.... when they both have are next to each other, and when that happens, the distance covered by both, is exactly the same, let's say the first car is A and the second is B
by the time car B catches up to car A, car A has covered "d" kilometers, and car B has to had covered "d" kilometers as well by then
if say... car A took off at "t" time, car B took off 2hrs later, or " t + 2 "
thus [tex]\bf \begin{array}{lccclll} &distance&rate&time\\ &-----&-----&-----\\ \textit{car A}&d&40&t\\ \textit{car B}&d&65&t+2 \end{array} \\\\\\ \begin{cases} \boxed{d}=40t\\ d=65(t+2)\\ ----------\\ \boxed{40t}=65(t+2) \end{cases}[/tex]
solve for "t"
the faster car of 65km/h, will "overtake" the first car.... when they both have are next to each other, and when that happens, the distance covered by both, is exactly the same, let's say the first car is A and the second is B
by the time car B catches up to car A, car A has covered "d" kilometers, and car B has to had covered "d" kilometers as well by then
if say... car A took off at "t" time, car B took off 2hrs later, or " t + 2 "
thus [tex]\bf \begin{array}{lccclll} &distance&rate&time\\ &-----&-----&-----\\ \textit{car A}&d&40&t\\ \textit{car B}&d&65&t+2 \end{array} \\\\\\ \begin{cases} \boxed{d}=40t\\ d=65(t+2)\\ ----------\\ \boxed{40t}=65(t+2) \end{cases}[/tex]
solve for "t"
