(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

A Toyota Prius has a total mass of 1520 kg when carrying passengers. The driver shifts

the car into neutral and coasts down a long, straight section of road with a grade of 9%

(the term "grade" means slope). The car eventually reaches a terminal speed. Ignoring the

effects of "rolling friction" and assuming that the driver does not have her foot on the brake

pedal, compute the terminal speed of the Prius as it coasts down the incline. Assume the

temperature of the air is 30 degrees C.

The drag coefficient of a 2011 Toyota Prius moving forward is 0.25

The frontal cross-sectional area of a 2011 Toyota Prius is 2.1 m^2

note that, when the object is moving at a constant velocity, a = 0, then solve for v.

2. Relevant equations

F=ma

V(terminal)=sqrt(2mg/Cd*density*A)

g=GM/R^2

3. The attempt at a solution

A=2.1m^2

m=1520kg

a=0

Cd=0.25

terminal velocity=?

so far I have A=pir^2

r=sqrt(A/pi)

sqrt((2.1m^2)/pi)=0.6685

V=4pir^3

4pi(0.6685)^3=3.75

g=GM/R^2

(6.67*10^-11)(1520kg)/(0.6685^2)=2.27*10^-7

D=m/v

1520kg/3.75m^3=404.97kg/m^3

Vterm=sqrt((2*1520*(2.27*10^-7))/(0.25*404.97kg/m^3*2.1m^2))=0.0018

I found the terminal velocity but I don't think it is right because I did not know how to use the grade of 9% on the straight section road and the temperature given. My teacher also said to use f=ma but I don't know how to incorporate it into this problem.

thank you for your help

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# Compute the terminal speed of a car as it coasts down the incline

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