Duas Rodovias Retilíneas Cruzam-se Perpendicularmente Na Cidade A

Two Perpendicular Highways: What Actually Happens When You Model It

A lot of people see a problem like duas rodovias retilíneas cruzam-se perpendicularmente na cidade A and immediately go for pure coordinate geometry. That works, but it's not always the fastest route, especially when you're working against a clock and the question starts asking for distances from moving vehicles or times when two cars are closest to each other. I've graded enough of these to know the patterns, and more importantly, I know where students lose points for no reason.

duas rodovias retilíneas cruzam-se perpendicularmente na cidade a

Let me just set it up the way I actually draw it on the board. City A is the intersection point, so it becomes the origin. One highway runs east-west along the x-axis. The other runs north-south along the y-axis. That's it. The coordinate system is chosen specifically to take advantage of the perpendicularity. You don't need anything fancy. The usual follow-up is something like: a car leaves city A heading east at some constant speed, another car leaves city A heading north at another constant speed, and you need to find when they're a certain distance apart, or when that distance is minimized. This shows up constantly in entrance exams and textbooks, and it's almost always tested the same way.

Here's the method that actually works without mistakes. Let the eastbound car have position x(t) = v1·t and the northbound car have position y(t) = v2·t. The distance between them at any time t is the hypotenuse of a right triangle, so d(t) = [(v1·t)² + (v2·t)²]. That simplifies to d(t) = t·(v1² + v2²). If the question asks when the distance reaches a specific value D, you just solve t = D / (v1² + v2²). Straightforward algebra from there. The trickier version is when the cars don't start at the same time or from the same point. Maybe one car is already 30 km down its road when the other begins moving. In that case you write x(t) = d0 + v1·t for the head start car and y(t) = v2·t for the other, then d(t) = [(d0 + v1·t)² + (v2·t)²]. Now you have a quadratic under the square root, and finding minimum distance means taking the derivative of the inside expression and setting it to zero. That gives you t = -(d0·v1)/(v1² + v2²). Yes, the time can come out negative, and that simply means the minimum distance occurred before the second car even started moving. If the question restricts the domain to t 0, you check the boundary point at t = 0 instead of using that negative value.

I ran into a real issue with this last year when a student was working a problem where one highway wasn't perfectly aligned with an axis because the map was rotated. They tried to use slopes and angles unnecessarily, which turned a five-minute problem into twenty minutes of trig. The workaround is always the same: rotate your mental coordinate system so one highway sits on an axis and the other sits on the perpendicular axis. The geometry doesn't change. Only your labels do. I tell students to literally draw a new set of axes on their scratch paper before they write a single equation. That small habit has probably saved a few hundred of them from silly mistakes over the years. There's another subtlety people miss. When both cars move away from the intersection, the distance between them grows monotonically if they both start from A at the same time. The rate of growth is (v1² + v2²), which is just the magnitude of the relative velocity vector. Some questions ask for the rate at which the distance is changing, and the answer is not v1 + v2. It's the vector sum magnitude. I've seen students write v1 + v2 on exams and lose the whole point because they treated velocity as a scalar. That mistake costs easy points and it's completely unnecessary once you see the right triangle relationship.

If the problem involves angles instead of speeds, use tan = opposite/adjacent. Say a car is traveling on the north-south highway and you need the angle between its path and the line connecting it to the eastbound car. At any time t, that angle is arctan(v1·t / v2·t), which simplifies to arctan(v1/v2). The angle is constant. This is another counter-intuitive result that comes up, and students usually try to compute it at specific time values instead of recognizing the simplification immediately. The ratio of the speeds determines the angle, not the elapsed time. One more practical note about graphing. The trajectory of the second car relative to the first, if you plot it on a moving frame, traces a straight line. This is useful when you're checking work quickly. If your relative path curve is bending, you made a setup error. I check this by hand every time now, even on simple problems, because it takes three seconds and catches sign errors that are otherwise easy to miss.

The main downside of the coordinate method is that it gets messy fast if the highways are not perpendicular. Then you're dealing with the law of cosines, extra angle terms, and a much uglier quadratic. For non-perpendicular crossings, I switch to vector notation from the start and keep everything in dot product form until the final step. It's cleaner on paper and harder to mess up algebraically. But for the standard perpendicular case, coordinates are faster and less error-prone. If you're preparing for exams, practice at least ten variations of this problem where one car has a head start and where speeds differ significantly. The algebra is identical, but the numbers will trip you up if you only memorize the formula without understanding the triangle structure underneath it.