The point marked is where the northbound car was located when the eastbound car started. (40∙3 or 120 miles north of where it started). Solve by the quadratic formula and get 5.871020759 hours from the time the second car started, which was 3 hours more than that for the time after the first car started, or 8.871020759 hours. Edwin
Let "t" be the time chronometered after the first car starts. First car traveled t hours; second car traveled (t-3) hours (assuming t >= 3). The distance between the cars is then= 500. It is the equation to solve in order for to find "t". Square both side (40*t)^2 + (60*(t-3))^2 = 500^2 1600t^2 + 3600t^2 - 3600*6t + 9*3600 = 250000 5200t^2 - 21600t + 32400 = 250000 5200t^2 - 21600t - 217600 = 0. 520t^2 - 2160t - 21760 = 0 Use the quadratic formula. The positive root is approximately 8.87 hours. ANSWER. 8.87 hours after the first car started. CHECK. = 500 miles (total distance). ! Correct !