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The mass of the car is 1500 kg. The shape of the body is such that its aerodynamic drag coefficient is $C_D=0.330$ and the frontal area is $2.50 m^2$. Assuming that the drag force is proportional to $v^2$ and neglecting other sources of friction, calculate the power required to maintain a speed of $100 km/h$ as the car climbs a long hill sloping is $3.20^\circ$.
Use the formula: $$ F_D=\frac{1}{2}C_D\rho_{air}Av^2 $$ where $\rho_{air}=1.2kg/m^3$

I don't know how to incorporate power into the formula. I get that $F_d$ is proportional to $v^2$ so am I able to cancel them out?

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    $\begingroup$ You just have to use this formula for drag as a given expression for drag. But do you know how to calculate power in general? $\endgroup$
    – fibonatic
    Apr 10, 2014 at 17:52
  • $\begingroup$ P = w/delta t i think $\endgroup$
    – Tim
    Apr 10, 2014 at 17:59

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Power is defined as $P = F \cdot v$ where $F$ is the driving force and $v$ is the velocity of the moving object. In this case, determine the values of both $F$ and $v$, and use this to calculate the power.

If you need additional help, feel free to ask in the comments.

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  • $\begingroup$ How do i solve for the force exactly? Do i substitute v^2 for Fd? $\endgroup$
    – Tim
    Apr 10, 2014 at 18:04
  • $\begingroup$ Not quite. You've provided the equation for force above ($F_D = \frac{1}{2} C_D \rho_{air} A v^2$), and you have the values of each of those variables, so you can use that equation to find the force. $\endgroup$ Apr 10, 2014 at 18:06
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    $\begingroup$ nevermind didn't see that they gave velocity $\endgroup$
    – Tim
    Apr 10, 2014 at 18:06
  • $\begingroup$ ...and don't forget the slope it's climbing... $\endgroup$
    – Joce
    Apr 10, 2014 at 20:49
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$HP = (F*v) = [(1/2)*Cd*ρ*A*v^2]*v = (1/2)*Cd*ρ*A*v^3 $

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    $\begingroup$ Can you expand your answer to explain how you arrived at this equation and the significance of the variables in it. $\endgroup$ Jun 22, 2014 at 5:50

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