Let’s say, theoretically, that there is a perfectly straight road that precisely follows the equator. If I were to get in my car and begin driving west along this road at precisely 60 miles per hour at the exact time that the sun breaches the horizon and I drive until the exact time that the sun drops below the horizon, how much longer would I be able to enjoy the sun than someone who remains stationary?
The Earth at equator spins at 1,040 miles-per-hour.
Crusing along at 60 mph isn’t going to extend the day by much. If you’re flying at Mach 2, then you’ve got something.
Right, the time zones are a little more than 1,000 miles wide at the equator. At 60 miles an hour, it would take you about 17 hours to move the equivalent of one time zone, and the sun is out for 12 hours and a couple of minutes at the equator. So according to my calculations… you just wasted all day in the car when you could’ve been out enjoying the sunshine.
To quantify this, if you’re traveling at 60 mph opposite the Earth’s rotation, then you’re effectively going around the Earth’s center at 980 mph rather than 1,040 mph. So your day would be 1040/980 of a normal solar day, or about 12 hours and 44 minutes.
Guess I should have added that I’d be driving opposite the Earths rotation. What if you were driving with the rotation? 
But it’s a moving car so then you’d have to account for Relativity and add a few picoseconds to that (or maybe not depending on observer.)
Undoubtedly, the easiest way to think about the whole thing is to consider the spinning Earth underneath the tires as a giant treadmill…
Just go to one of the poles at the vernal equinox and stick around for your half year of daylight until the autumnal equinox.