If you know a car's fuel economy you can estimate its top speed. How? With knowledge of basic physics and 2 simple facts:

SPEED TO POWER

The power needed to move a given object through air increases with the cube of its speed.

Why? It's best explained with one simple fact and an example.

Fact: Air resistance increases with the square of speed - twice the speed has 4 times the air resistance.

Example: consider a car driving at a certain speed, say 60 mph. If that car increases its speed to 120 mph it has 4 times the air resistance. That means the engine must produce 4 times as much forward thrust. For each mile it travels, the engine does 4 times as much work. But since the car is moving twice as fast, it does that work in half the time. So to double the speed, the engine must do 4 times as much work in half the time. That is 8 times the power, since power is work / time, or the rate at which work is performed.

Thus twice the speed requires 8 times the power. 2 cubed is 8. How much extra speed does twice the power give? What number cubed equals 2? The cube root of 2, which is about 1.26. So twice the power increases speed by 26%.

EFFICIENCY OF PISTON ENGINES

Here we are talking not about miles per gallon, but how much fuel the engine requires to do a certain amount of work. Miles per gallon depends on the shape and weight of the car, gearing, etc. But the efficiency we care about here is intrinsic to the engine itself, regardless of what car it is in.

Most piston engines produce 10 to 14 horsepower for each gallon of fuel per hour. Closer to 14 at peak efficiency, partial power cruise operation, and closer to 10 at full power. At full power, the engine is operating less efficiently. Compared to partial power cruise, it is consuming a lot more fuel in order to produce a little more power.

TOP SPEED FROM FUEL ECONOMY

Now let's apply these simple ideas and see how well they work.

EXAMPLE 1: Subaru Forester

Our Subaru Forester has 160 HP and gets about 28 mpg at 60 mph under ideal highway cruise conditions. What is its top speed?

28 mpg at 60 mph means every hour the engine burns 60/28 = 2.14 gallons of fuel.
At 14 HP per gallon per hour, the engine is producing 2.14 gph * 14 HP/gph = 30 horsepower.
So it takes 30 HP to push this car through the air at 60 mph. This is about average for most cars.

Now the car has 160 HP.
That is 160/30 = 5.33 times as much power as it needs to go 60 mph.
Since power is the cube of speed, how much faster the car can go with 5.33 times the power, is the cube root of 5.33.
The cube root of 5.33 is 1.75, so the car can go 1.75 times faster with 5.33 times the power.
1.75 times 60 mph is 105 mph.
So, the car's top speed is (or should be) about 105 mph. If you look this up, it's within 1-2 mph of reality.

EXAMPLE 2: Panoz Roadster

My Panoz Roadster had 320 HP and a top speed of about 135 mph. What is its fuel economy at 60 mph under ideal conditions?

135 / 60 = 2.25, so 135 mph is 2.25 times faster than 60 mph.
2.25 cubed is 11.4, so it takes 11.4 times as much power to go 2.25 times faster.
If it takes 320 HP to go 135 mph, then it takes 320 / 11.4 = 28.1 HP to go 60 mph.
If the engine produces 14 HP per gallon per hour, then it burns 28.1 / 14 = 2 gph at 60 mph.
2 gph at 60 mph is 30 miles per gallon.

Now this matches exactly the factory fuel economy estimate, but it's higher than reality. My Panoz actually got around 25 mpg under ideal cruise conditions at 60 mph. That means its engine is not as efficient as the Subaru - it produced less than 14 HP per gph.

How much less?
25 mpg at 60 mph is 60/25 = 2.4 gph.
We already computed that the car needs 28.1 HP to go 60 mph.
So the engine produces 28.1/2.4 = 11.7 HP per gph.

The comes as no surprise, since the Ford Cobra quad-cam V8 engine in the Panoz Roadster is designed for power, not for efficiency. It's not as efficient as the Subaru engine, but it makes more power per cubic inch. The Subaru makes 160 HP from 152 ci, which is 1.05 HP per cubic inch. The Panoz makes 320 HP from 281 ci, which is 1.14 HP per cubic inch.

EXAMPLE 3: Mazda RX-7 Twin Turbo

My 1995 RX-7 twin turbo had 260 HP and a top speed of 155 mph. What is its fuel economy at 60 mph?

This car had a Wankel rotary engine, which is not a piston engine. It produces a lot of power from a small engine, but is less efficient. Let's see how much less efficient below...

155/60 = 2.583, so 155 mph is 2.583 times faster than 60 mph.
2.583 cubed is 17.24, so it takes 17.24 times as much power go to 2.583 times faster.
If it takes 260 HP go to 155 mph, then it takes 260/17.24 = 15.1 HP to go 60 mph.
[NOTE: the astute reader will observe this is about half the power needed to push the Subaru or the Panoz along at 60 mph, but that's expected as the RX-7 was a much more aerodynamic car]
If the engine produces 14 HP per gallon per hour, then it burns 15.1/14 = 1.1 gph at 60 mph.
1.1 gph at 60 mph is 54 mpg !!!

Now in reality this car got about 23 mpg under ideal cruise conditions at 60 mph. So what is its engine's internal efficiency?

23 mpg at 60 mph is 60/23 = 2.61 gallons per hour
We know from the above it takes 15.1 HP to push this car through the air at 60 mph.
So this engine consumes 2.61 gallons per hour to make 15.1 HP.
That is 15.1/2.61 = 5.8 HP per gph

That means the twin turbo Wankel rotary in the RX-7 is about 2.4 times less efficient than a piston engine! However, it produces 260 HP from 80 ci, which is 3.25 HP per cubic inch. The Panoz engine produced 1.14 HP per cubic inch! So the Wankel is 2.4 times less efficient, but makes 3.25/1.14 = 2.85 times more power per cubic inch.