How Does Speed Affect Car Fuel Efficiency and Required Forces?

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At higher speeds, such as 75 mph compared to 65 mph, the forces required to maintain speed increase significantly due to factors like aerodynamic drag and rolling resistance. The drag force, which increases with the square of speed, plays a crucial role in reducing fuel efficiency at higher speeds. As speed increases, the engine must work harder to overcome these forces, leading to greater fuel consumption and lower gas mileage. Understanding the balance of forces, including gravity and friction, is essential for estimating fuel efficiency impacts. Overall, maintaining lower speeds can enhance fuel efficiency by reducing the required forces.
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Estimate the ratio of the forces required to keep your car at a speed of 75 mph to that required to keep your car at a speed of 65 mph. What impact does this have on your gas mileage?

Where should I start?
 
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OK so you will have a force of gravity, normal force, drag, and static friction. Gravity will be down, normal will be up, drag will be towards the left and static friction will be towards the right since they are trying to keep the car at a certain speed.

So where to go from here?
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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