Slow Down, Save Gas: The Simple Physics of Cheaper Summer Road Trips
Summer road trip season is here, but soaring gas prices have turned a fun getaway into a pretty pricy proposition. For most drivers stuck behind the wheel of traditional gas-powered internal combustion vehicles, that means opening your wallet wider than you’d like—and that’s not the only cost, either: our gas-guzzling drives contribute directly to the climate crisis, too. But what if I told you cutting your fuel costs is as simple as taking your foot off the accelerator? The tradeoff, of course, is that you’ll arrive at your destination a little later.
A few years back, I penned a piece for WIRED breaking down the optimal driving speed when extra time on the road translates to lost wages (think: showing up late to an hourly work shift). Think of this as the summer holiday edition, where we throw the “time is money” assumption out the window. Instead, we’ll go back to basic physics to prove exactly why slowing down cuts your gas use. If you’ve ever cruised an interstate only to notice your fuel tank is nearly empty, with the next gas station still 20 miles away, you’ve probably wondered if speeding up to get there faster makes sense. Spoiler: that’s a losing bet, and by the end of this you’ll know exactly why.
Speed and Time
Let’s start with the basics. Suppose you’re heading out on a straight 30-mile highway drive from your home, where the posted speed limit is 70 mph. How does total travel time shift if you go faster or slower than that limit? Let’s frame this in simple physics terms first.
Imagine your route is marked as a number line, with your starting point at home as zero. Your position x at any moment is just how far you’ve traveled from home—physicists call this displacement, but you just call it distance. Velocity v is the rate your position changes over time, written as:
$$v = \frac{\Delta x}{\Delta t}$$
This formula just means velocity equals your change in position ($\Delta x$) divided by your change in time ($\Delta t$). If you drive 100 miles in 2 hours, your average velocity works out to 50 mph, which checks out.
Now plug that into our 30-mile example. We have a total displacement of 30 miles, and a velocity of 70 mph—so how long will the trip take? It’s easy to rearrange the formula to solve for time:
$$\Delta t = \frac{\Delta x}{v}$$
Plugging in our numbers gives 30 divided by 70, which equals 0.428 hours, or 25.7 minutes total. What if you bump your speed up to 75 mph? Using the same formula, the trip takes 24 minutes flat. That means you saved a whole 1.7 minutes.
For context, a graph of time saved versus speed (for speeds from 55 to 85 mph on a 30-mile trip) shows that dropping your speed from 70 to 65 mph only adds 2 minutes to your total travel time. Let’s be real: is 2 extra minutes going to make or break your entire summer day?
Speed and Fuel Consumption
You probably already know you get better gas mileage on the highway than in stop-and-go city driving, which leads many people to assume that fuel economy (measured in miles per gallon) gets better the faster you go. That’s not actually true. The real reason highway driving is more efficient is that you’re moving at a steady pace, with no constant braking and accelerating like you do in the city—getting a heavy car back up to speed after stopping for a red light burns a ton of extra gas.
When you’re sticking to steady highway cruising, the opposite holds: going faster over the same distance always burns more gas. Why? Let’s break down the forces that act on a moving car. There are three horizontal forces at play:
Static friction ($F_f$) between your rear tires and the asphalt—this is the “grip” that pushes your car forward
Rolling friction ($F_{roll}$) between your tires and the road surface, which pulls the car backward
Air resistance (also called drag) ($F_{air}$), which also pushes against the car as it moves forward
If your car is moving at a constant speed, it’s in mechanical equilibrium, which means the forward push from friction has to exactly equal the sum of the two backward forces—this is straight from Newton’s second law of motion. We can write that as an equation:
$$F_f = C_1 + C_2v^2$$
Rolling friction barely changes with speed, so we treat it as a constant $C_1$. Air resistance, on the other hand, doesn’t just depend on speed—it’s proportional to the square of your speed: $F_{air} = C_2v^2$. Double your speed, and air drag quadruples. That’s a huge jump.
The forward friction that moves your car comes from energy generated by burning gasoline to spin the wheels. When you drive at a higher speed v, the right side of the equation gets much bigger, which means you need more forward force to keep that speed up. More force requires more energy from the engine, which means burning gas faster—and that adds up to worse fuel efficiency.
In short, your gas mileage drops at higher speeds because air resistance grows much faster than your speed does. Going back to that earlier example where you’re 20 miles from the next gas station running on fumes: slowing down is the right call, even if it takes longer, because you’ll burn less gas to cover the same distance. The slower you go (at least down to around 50 mph, where air drag stops being the dominant factor), the more distance you get out of each tank of gas.
How much does speed actually change your mileage? Every car is a little different, but the U.S. Department of Energy estimates that for every 5 mph you go over 50 mph, your fuel efficiency drops by 7%. We can write that as:
$$e = e_0 \times (0.93)^\frac{v - v_0}{5}$$
Where $e_0$ is your fuel efficiency at a benchmark speed $v_0$, so this formula gives you your efficiency at any other speed. The 0.93 factor accounts for that 7% drop in efficiency per 5 mph increase.
Let’s test this with a quick example. Suppose your car gets 30 mpg ($e_0$) when driving 70 mph ($v_0$). At 75 mph, that drops to 27.9 mpg, and at 65 mph that jumps up to 32.3 mpg. You can see just how much of a difference a small speed change makes.
Time and Money: The Real Tradeoff
Now let’s put all this together to look at the actual tradeoff between speed, time and cost. Driving faster saves you time, but it burns more fuel, so it costs you more. What does that tradeoff actually look like in real dollars and cents?
Let’s go back to our 30-mile trip example, and assume gas costs $4 per gallon. If you drive 70 mph and get 30 mpg, you’ll use 1 gallon of gas for the trip, costing $4. If you speed up to 75 mph, you use 1.08 gallons, which adds just 32 cents to your total cost. That seems like nothing, right? But remember that 32 cents gets you only 1.7 minutes of saved time, which is 0.028 hours. If we divide the extra cost by the time saved, that works out to paying an extra $11.15 per hour of time you save. That’s a pretty steep premium for getting there a couple minutes earlier.
Let’s scale this up to a typical summer holiday trip: say you’re planning a 500-mile round trip drive for the Fourth of July. If you set your cruise control to 70 mph, you’ll spend 7.14 hours behind the wheel. If you drop that to 60 mph (10 mph slower), you’ll add about 30 minutes of driving time each way. But your fuel efficiency will jump from 30 mpg to 35 mpg. That adds up to using 2.5 fewer gallons of gas for the whole trip, which saves you $10 at current prices.
Put another way: sticking to the slow lane is the equivalent of getting gas for $3.40 a gallon instead of $4.00. Most drivers would go out of their way to find gas that cheap, right? And that’s not the only benefit: you’ll also cut your trip’s carbon dioxide emissions by more than 50 pounds.
So what’s my advice? Take it easy on the gas pedal, lighten your impact on the planet, and take the extra few minutes to enjoy the scenery along the way.
