Note that you already asked that question and was answered to you. Yet you refused to understand it and to work it out. You just wanted the answer given to you. That's not how you are going to learn anything. Again, in this problem, YOU need to realize that it starts off from highschool geometry and algebra. Can you do that part on your own? U need to know this so as to know your level of understanding, so that u may take appropriate steps to get better; learn to walk before u learn to run & jump.
Let t = number of hours elapsed Let x = how far east of the truck is the car: starts at 0 when t = 0 and increases at 60 per hour, so we have: x = 60t Let y = how far north of the car is the truck: starts at 100 when t = 0 and decreases and 40 per hour, so we have: y = 100 - 40t Let d is the distance between the car and the truck, so Pythagoras gives us: d^2 = x^2 + y^2 d^2 = (60t)^2 + (100 - 40t)^2 d^2 = 3600t^2 + 10000 - 8000t + 1600t^2 d = sqrt(5200t^2 - 8000t + 10000) Let u = 5200t^2 - 8000t + 10000, so du/dt = 10400t - 8000 So d = sqrt(u) = u^0.5, so dd/du = 0.5u^(-0.5) = 1 / (2 * sqrt(5200t^2 - 8000t + 10000)) By the Chain Rule: dd/dt = dd/du * du/dt dd/dt = (1 / (2 * sqrt(5200t^2 - 8000t + 10000))) * (10400t - 8000) dd/dt = (260t - 200) / sqrt(13t^2 - 20t + 25) At 7:30, half an hour has elapsed, so t = 0.5, so we have: dd/dt = (260*0.5 - 200) / sqrt(13*0.5^2 - 20*0.5 + 25) dd/dt = (130 - 200) / sqrt(13*0.25 - 10 + 25) dd/dt = -70 / sqrt(3.25 - 10 + 25) dd/dt = -70 / sqrt(18.25) dd/dt = -140 / sqrt(73) dd/dt =~ -16.385760607458278952082613228405 Note: A negative number means the distance is decreasing.