 John drives a 2013 Honda Accord, which has a coefficient of drag of 0.3 and a frontal area of 52 inches by 73 inches. Determine the amount of drag force acting on the car when driving down I-29 at 80 miles an hour, assuming ambient conditions are 20 degrees Celsius in one atmosphere and that there is zero wind on that day. The amount of power the engine would have to expend to fight drag force for par-de. Then if John were to speed, and this is hypothetical of course, John is a responsible adult who would never violate the speed limit by 10 miles an hour, meaning he were traveling down I-29 at 90 miles per hour, how would that affect the amount of power the engine would have to expend to fight drag force? Then lastly, on a completely unrelated note, what is the theoretical maximum speed of the car, assuming 100% of its maximum rated power of 185 horsepower, was expended against air resistance, i.e. neglecting rolling resistance and mechanical resistance. So I have a coefficient of drag, a frontal area, and a velocity of the fluid that allows me to calculate the drag force by taking one-half times the coefficient of drag, times the density of the fluid through which the car is traveling, multiplied by the velocity squared, multiplied by the cross-sectional area of the frontal perspective of the car. So that's going to be one-half times 0.3 times the density of air, which I can get from table A2. If the ambient conditions are 20 degrees celsius and one atmosphere, that means the density of air is going to be 1.2, so air is going to have a density of 1.2 kilograms per cubic meter, and if I need it, a kinematic viscosity of 1.5 times 10 to the negative fifth meter squared per second. So using a density of 1.2 kilograms per cubic meter, a velocity of 80 miles an hour, then I square everything, and a cross-sectional area from the front of 52 inches by 73 inches. So I have to make a decision about what unit to express this answer in. I wasn't told a specified force unit, so let's go with pounds of force, since everything else except for the density so far has been in the imperial unit system. So I will use a pound of force, which is 32.174 pounds feet per second squared. So one pound of force is 32.174 pounds of mass times feet per second squared. I know one pound of mass is 0.4536 kilograms. One meter is 3.2808 feet, and then if I cube everything that will allow me to cancel the cubic meters. One mile is 5,280 feet, and I can square everything. There are 12 inches in most feet, feet squared, and feet squared is going to cancel cubic feet and feet. Square inches is going to cancel inches and inches. There are 3,600 seconds in an hour, square, square, square, then hour squared cancels hour squared, second squared cancels second squared, and that leaves me with pounds of force. So I can pop up my calculator at which point I have 0.5 times 0.3 times 1.2 times 80 squared times 52 times 73 times 1 over 32.174 times 0.4536 times 3.2808 cubed. Then multiply by 5,280 squared divided by 12 squared and I multiply by 1 over 3,600 squared. That gives me 126.75 pounds of force. So the drag force exerted on my car under these conditions is 126.754 pounds of force. For part B, I'm asking how much power does it take to overcome that drag force? Well, we know power is a rate of work with respect to time, which I could express as a work divided by a duration, and I could express work as the integral of force with respect to displacement. The force is constant, so it comes out, so I'm left with force multiplied by dx over dt at which point I have force times velocity. So in the event that my force is constant like the situation, the amount of power required is just force times velocity. So I could take 126.754 pounds of force and multiply by 80 miles an hour, and presumably I want this power in horsepower. So I can recognize that a horsepower is 550 feet pounds of force per second. So 550 feet pounds of force per second, and that's a horsepower. So then I have to use the same mile to feet conversion. One mile is 5,280 feet, pounds of force cancels pounds of force. I need the conversion from hours to seconds. One hour is 3,600 seconds, and that leaves me with horsepower. So this time I'm taking 126.754, let's use the correct parentheses calculator. Come on, you can do it. Multiply it by 80, multiply it by 5,280, it's 5,280, no. One more time calculator, 5,280 divided by 550 times 3,600, and we get 27. So it only takes 27 horsepower to overcome the wind resistance if there is no wind traveling down the interstate, neglecting mechanical and frictional losses. For part C, moving into the entirely hypothetical situation, we are speeding by 10 miles an hour, meaning we're going to take the calculation for part A and change the velocity from 80 to 90. Everything else is going to be the same, so we're multiplying by 9 eighths squared. So drag force is related to the square of the velocity. If we double the velocity and leave everything else the same, we would increase the drag force by a factor of 4. Tripling the velocity would increase the drag force by a factor of 9 and so on. So 80 to 90, that gives us surely an incorrect number. Let's try that again, going from 80 to 90 yields 160 pounds of force. Just for good measure while we're here, again we take 126.754 multiplied by 9 eighths squared, that gives us the same answer. We could have done that as well. And with our new drag force, we can calculate a new power. Note how drag force and velocity are related to power. If I were to plug in symbolically my representation for drag force into the power equation, what I would be left with is a function of velocity cubed. So increasing the velocity by 10 miles an hour is going to increase my power requirement by 9 eighths cubed. But just in the interest of showing that full calculation again. So going from 20 miles an hour to 40 miles an hour is going to multiply my power requirement by a factor of 8. Going from 20 miles an hour to 60 miles an hour is going to multiply my power requirement by 3 cubed, which is 27. Interesting right? So 160.423 multiplied by 90, actually just let me go grab that equation from my calculator. So 160.423 multiplied by 90 gives us 30.5 and again we could just do 27 multiplied by 9 over 8 cubed. Either way, we should get the same thing about 38.5, 38.5 horsepower. So going 90 miles an hour down 929, hypothetically of course, on a still day with no other losses of any kind is going to require a power input of 38.5 horsepower from the engine. And then while we're on a hypothetical train of thought, I wanted us to figure out what the theoretical maximum speed would be if all glorious 185 horsepower from the 2013 Honda Accord were dedicated to overcoming wind resistance. So neglecting mechanical losses all together we are saying power is equal to the drag force multiplied by velocity, which I can express as one half times the coefficient of drag times the density times velocity squared times the cross sectional area from the front perspective multiplied by the velocity, which is one half times the coefficient of drag times the density times the velocity cubed times the cross sectional area. So I can say the velocity is going to be the cube root of two times the power divided by coefficient of drag times density times area. So I'm saying two, no, two times 185 horsepower divided by 0.3 divided by 1.2 kilograms per cubic meter divided by 52 inches times 73 inches. And if I want an answer in miles per hour, then everything inside of my parentheses should be miles cubed per hour cubed. So I will start by recognizing that a horsepower can be represented as 550 feet pounds of force per second. I can write one pound of force is 32.174 pounds of mass feet per second squared. So a pound of force cancels pound of force, horsepower cancels horsepower, I know one kilogram is 2.2046 pounds of mass, pound of mass would cancel pound of mass, kilograms would cancel kilograms. So I have cubic meters times feet times feet in the numerator and I have square inches times seconds cubed in the denominator. So I will start by converting 12 inches squared squared to one squared feet squared. That'll get rid of two of my inches and both feet, but I have to convert meters to feet anyway. I guess I could go meters into kilometers and then kilometers into miles, but potato potato one mile, no one meter is 3.2808 feet cube everything meters cubed cancels meters cubed. And then lastly, I have 5,280 feet in a mile. And then I want to cube everything that'll leave me with miles cubed per second cubed. So one more conversion an hour is 3,600 seconds. And I cube everything there and that'll give me miles cubed per hour cubed. So when I take the cube root of that quantity, I will end up with an answer in miles per hour. The seconds cubed cancels second squared times seconds leaving me with just miles cubed per hour cubed. Okay, no way possible for my calculator to screw this one up. I believe in you calculator. I've never not believed in you. Despite what your own senses say, first up, we have two times 185 times 185 and I will group these in extra parentheses just to be extra cautious times 550 times 32.174 times 12 squared times 3.2808 cubed times 3,600 cubed divided by 0.3 times 1.2 times 52 times 73 times 2.2046 times 5,280 cubed and then one big cube root to unite them. And I get 151.87, a perfectly respectable maximum speed, miles per hour. So if we were to subtract about 10-15% from mechanical losses, we'd probably end up with a number that's really close to the actual maximum speed. But I guess we'll never know what that is because there's no way for me to empirically test that, of course.