A Porsche 944 Turbo has a rated engine power of 217 hp. 30% of the power is lost in the engine and the drive train, and 70% reaches the wheels. The total mass of the car and driver is 1480 kg, and two-thirds of the weight is over the drive wheels. (c) How long does it take the Porsche to reach the maximum power output?

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Hey everyone. So this problem is about power. Let's see what it's asking us. The engine of a passenger vehicle generates a maximum power of hp. At one instance, it is loaded with a total mass of 4500 kg Engine and transmission losses dis dissipate 25% of the power generated by the engine. While 75% is successfully transferred to the drive wheels using a mass distribution where the drive wheels bear 60% of the loaded vehicle mass. We are asked to determine the time the vehicle takes to obtain maximum power output. We are told to assume the vehicle accelerates at An acceleration max of a max on a surface with a static friction coefficient given by us equals 1.12. Our multiple choice answers here are a 0.642 seconds. B 3.77 seconds. C 2.26 seconds, four D 3.2 seconds. OK. So this problem gives us a lot of information and it's going to require us to recall formulas and theories from a few different parts of physics. But we will work through this and figure out how it all fits together. So the problem in the end is asking for a time, we can start with our kinematics equations. So we can recall that V max is equal to V knot plus a max or acceleration maximum multiplied by our delta T or T F minus T I or T final minus T initial knowing that we are starting at rest, our V is zero. And so, and our T initial is going to be zero seconds. So that term also goes to zero. So we can rewrite this as V max equals a max multiplied by T F or T F is what we are looking for. That is gonna be our final answer. So T F equals V max divided by a max. OK. So now the problem is that the problem doesn't give us either velocity or acceleration, but we will get there. It does give us power. So we can recall that power is given by force multiplied by velocity. So we can rewrite that say velocity equals power divided by force, power from the problem. So we're gonna be looking at the power of the engine and the point of the drive wheels that's going to be the point that we'll focus on for this problem. So the power of the maximum power of the engine at the drive wheels Is going to be 75%. So 0.75 multiplied by 200 horsepower. And then we can recall that the Um conversion factor for horsepower to watts is 746 watts per horsepower. And so that gives us r power from the engine at the drive wheels Of 1.1, 2 Times 10 to the 5th watts. Ok. So to find velocity, we needed power and force, we've got power, let's figure out force. So the force we're told That's 60% of the loaded mass is transferred to is distributed to the drive wheels. So we will take 60% or 0. of the weight. And we can recall that weights is given by mass multiplied by gravity. So the force of the drive wheels is going to be 0. multiplied by 4500 kg. That mass was given to us in the problem multiplied by our gravity constant 9.8 m per second squared, plug that into our calculators and we get 2.65 times 10 to the four newtons. And now we can solve for velocity. So velocity equals power divided by force. So 1.12 times 10 to the fifth watts divided by 2.65 times 10 to the four newtons equals 4.2, 3 meters per second. So when we look at this first equation, our T final equals V max over a max, we've now found V max 4.23 seconds. So the next step is finding our acceleration here. We're going to pull in our knowledge of Newton's second law, which tells us the sum of the forces equals the mass multiplied by acceleration. You can draw a free body diagram of the vehicle as it is attempting to move or as it's starting to accelerate. And we have in the positive X direction, our static friction force of our normal force acting in the positive Y and then weight acting in the negative Y. So when we split up our X and Y directional forces, we have the sum of the forces and the X direction is equal to mass times acceleration in the X direction. And that is just our static friction force. That's the only force in the X direction is equal to mass times that X component of acceleration in the Y direction we have are some of our forces and our, the Y direction equals mass multiplied by acceleration, the Y component. So we have our normal force minus our weight is equal to zero because we are not accelerating. There's no vertical acceleration. So that whole term goes to zero. So our normal force is equal to our, we, we can recall then that our fri our static friction is given by the equation us multiplied by N and so we'll divide both sides by the mass to come up with our acceleration, the normal force we will find from the sum of our Y forces. And we can recall here that our weight is given by mass times gravity. But again, we are looking at this from the perspective of the um drive wheels. So the mass is only 60% of the total mass. So that would be 600.6 multiplied by mass multiplied by gravity. And so we can plug that and over here. So our acceleration is going to be Us multiplied by zero or 0.6 multiplied by mass multiplied by gravity all divided by mass. And so then the mass is canceled. The problem gives a us of 1.12. So we can say our acceleration is equal to 1.12, multiplied by 0.6, multiplied by our gravitational constant 9.8 m per second squared, sorry, multiplied by gravity 9.8 m/s squared. And all of that equals when we plug it in 6.5, 9 meters per second squared. So now we have our velocity and we have our acceleration. So we are ready to solve for our final answer for time. So I will rewrite this final or that last equation down here again. So T F equals V max over a max. And so that was V max was 4. m per second, divided by 6.59 m per second squared equals 0.64, Sorry, 6 4, 2 seconds. And that is the final answer to this problem. You can go back up to our answer choices and that aligns with answer choice. A Thanks for sticking with me through this one. We'll see you in the next video.

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Key Concepts

Power and Work

Force and Acceleration

Energy Loss in Systems