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

[vannessa001]

Homework Statement

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.

Homework Equations

F=ma
V(terminal)=sqrt(2mg/Cd*density*A)
g=GM/R^2

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

 

[gneill]

"Percent" means "per hundred" (from the French). A 9% grade implies that the height changes by 9 units of distance when the horizontal distance changes by 100 units. The angle involved will be $tan(\theta) = 9/100$.

 

[vannessa001]

"Percent" means "per hundred" (from the French). A 9% grade implies that the height changes by 9 units of distance when the horizontal distance changes by 100 units. The angle involved will be $tan(\theta) = 9/100$.

So I use tan theta=9/100 to find sin theta. And use the equation V=sqrt(2*mg*sintheta/density*area*drag coefficient)
my answer was 0.0896 when iI plug in all my values but it doesn't seem right

 

[gneill]

Why don't you look at is as a force balance situation? When the car reaches terminal speed acceleration is zero, so it must be because the downslope force due to gravity is balanced by the air resistance force.

The air resistance force, from the formula that you gave, is
$$F_d = \frac{1}{2} C_d \rho v^2 A$$
Find the downslope force due to gravity and equate it with this force. Solve for velocity.

 

[vannessa001]

Why don't you look at is as a force balance situation? When the car reaches terminal speed acceleration is zero, so it must be because the downslope force due to gravity is balanced by the air resistance force.

The air resistance force, from the formula that you gave, is
$$F_d = \frac{1}{2} C_d \rho v^2 A$$
Find the downslope force due to gravity and equate it with this force. Solve for velocity.

how would you find the down slope force?

 

[gneill]

how would you find the down slope force?

Have you done problems with a block sitting on a ramp of a given angle and had to work out the component of the block's weight that is directed down the slope, and the component that is normal to the slope?