Colin, from Newfoundland, Canada wrote in this week to ask:
The Trans Canada Highway runs fairly straight. If I was to start at the border of Ontario and Manitoba and, as soon as the sun came up, began to drive west at 100km/h. How many more hours of sunlight would I be able to gain?
Colin,
Great question—and a fun idea for a roadtrip! Let’s see if it’s practical. There are a few ways to approach this problem; we’ll look at two of them.
Standing still on the longest day of the year at the latitude of the TransCanadian Highway—about 50° from the equator—will net you 16 hours and 19 minutes of sunlight for starters. Let’s be optimistic and take those 19 minutes for refueling, seeing as you’re probably going to run down your tank at least twice, so you’ve got 16 hours of travel time. You’ve specified a speed, so that makes things relatively easy—how far can we get in 16 hours at 100 km/h? Obviously, 1600 km!
Assuming we’re staying on the road and not going “as the crow flies”, that’s enough to get you from Whiteshell—a town along the Trans Canada Highway at the border of Manitoba and Ontario—to Canmore, just past Calgary.

Sunrise and sunset times at various longitudes and cities can be found just by googling, so let’s look at Whiteshell vs. Canmore

We see here that we get an extra 29 minutes! Our original 16 hours and 19 minutes of sunlight is now 16:48. Here’s where it gets interesting, though—if we’re driving until sunset, the 16 hours of roadtime we’ve calculated for isn’t all we get; we can go another 29 minutes—which gets us almost 50 km farther, and buys us another few seconds of sunlight. You might wonder where this line of thinking ends—we could drive for those few extra seconds, and it would get us a few more microseconds—and the answer is that, technically, it doesn’t! The problem is described by an infinite series. That doesn’t mean we can drive forever, though; a series that’s made up of an infinite number of evershrinking terms can still take on a finite value—this is known as the limit.
Hundreds of years ago, a young Isaac Newton was vexed with problems like this one. To address them, and to calculate things like the exact amount of time between sunrise and sunset for a westward traveler without computing an infinite number of terms, he developed calculus. Here, of course, the first few terms in the series—i.e. the 16 hours plus a secondorder term to account for the daylight gained—provides a close enough approximation for our purposes.
Treating this explicitly as a calculus problem would require us to know the function describing the curviness of the road, which we don’t have. However, we can get there by knowing a little algebra and making a simplifying assumption.
Say we’re heading straight west, as the crow flies, rather than following the TransCanadian Highway. Now we can treat this like a “two trains leaving their stations” algebra problem, where your car is one train, and the “terminator“—the line of sunset—is the other.

Although if it helps to picture Arnold Schwarzenegger as the other train, I don’t see why not. 
On the longest day of the year at 50° latitude, there’s 16 hours and 19 minutes of sunlight. This tells us that, at that latitude, it’s daytime across 68% of Earth’s surface at your latitude, with the remaining 32% in night—which tells us where the terminator is when the sun rises at your location. To figure out how fast the terminator is moving, we need to know the circumference of the earth at your latitude. Fortunately, this is pretty simple—just the cosine of your latitude, multiplied by Earth’s circumference at the equator.
Now, we just need to set this up as an algebra problem, to figure out how much daylight we gain by traveling west at 100 km/hr. We’ve got a 17,516.8 km head start, and putting that into an equation looks something like this:
—Stephen Skolnick