 So let's fly some rockets. Without going into the derivation, which requires a little bit of physics, we can find the following. As a rocket burns fuel, its acceleration is given by the following formula, where F is the total mass of propellant and K is the rate at which the rocket consumes propellant. I is another constant that has to do with the efficiency of the propellant and M is the mass of the rocket itself without the fuel. So suppose K equals 100 kilograms per second, M is 500 kilograms, and I is 5,000. If our rocket initially carries 5,000 kilograms of propellant and begins with a velocity of zero meters per second, how rapidly will it be moving after the propellant is exhausted? So we can substitute in our values for K, M, and I and get an expression for A of t. Since A of t is measured in meters per second squared and velocity is measured in meters per second, then velocity is going to be the anti-derivative of acceleration. We'll use a u-substitution with u equals 5,500 minus 100t. This gives us, and making our substitution, we can find the anti-derivative. To find the constant of anti-differentiation, we'll use the fact that our initial velocity is zero. And so that tells us that C is 50 log 5,500, and we get our velocity after t seconds. So how can we find the velocity after the propellant is exhausted? Well, we need to know what t is. So here we have to look closely at what our variables and parameters represent. And in this case, K is the rate at which the rocket consumes propellant. And we know that K is 100 kilograms per second, and the mass of propellant is 5,000 kilograms. Since the rocket is consuming propellant at 100 kilograms per second, and it has 5,000 kilograms of propellant, then it will run out after t equals 50 seconds. So we find the velocity at 50, which will work out to around 120 meters per second. Well, maybe that's a little disappointing. So let's throw in a lot more fuel. Let's double the amount of fuel to 10,000 kilograms. So we'll substitute in our new values to find velocity as the anti-derivative of acceleration. We'll do a u substitution. Use the initial velocity of zero to find our constant. And this time, since we have 10,000 kilograms of propellant consuming at 100 kilograms per second, then we'll run out after t equals 100 seconds. And so we want to find v of 100, which works out to be 152 meters per second. And that's a rather disappointing result. Because when we had 5,000 kilograms of propellant, we got to 120 meters per second. Doubling the amount of propellant didn't give us a very large change in our final velocity. And this leads to what might be called the rocket problem. This doubling of our amount of propellant didn't give us a very large increase in our final velocity. And in fact, this is a general problem. It takes a substantial increase in the amount of propellant to produce any significant gains in the final velocity. After all, it is rocket science.