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

• vannessa001
In summary: If so, could you please share the problem and solution?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? If so, could you please share the problem and solution?
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

"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 $tan(\theta) = 9/100$.

gneill said:
"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

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.

gneill said:
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?

vannessa001 said:
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?

## 1. What is terminal speed?

Terminal speed, also known as terminal velocity, is the maximum speed that an object can reach as it falls through a fluid, such as air or water. It occurs when the force of gravity on the object is equal and opposite to the drag force of the fluid.

## 2. How is terminal speed calculated?

Terminal speed can be calculated using the formula v = sqrt((2mg)/(ρAC)), where v is the terminal speed, m is the mass of the object, g is the acceleration due to gravity, ρ is the density of the fluid, A is the cross-sectional area of the object, and C is the drag coefficient.

## 3. What factors affect the terminal speed of a car?

The terminal speed of a car can be affected by several factors, including the mass and shape of the car, the density of the fluid it is moving through, and the presence of any external forces, such as wind resistance.

## 4. How does the incline of the surface affect the terminal speed of a car?

The incline of the surface can affect the terminal speed of a car by changing the force of gravity acting on the car. If the surface is inclined, the force of gravity will be greater and therefore the terminal speed will be higher.

## 5. Can the terminal speed of a car be exceeded?

In most cases, the terminal speed is the maximum speed that a car can reach. However, if the car experiences a sudden increase in external forces, such as a strong gust of wind, it is possible for the car to exceed its terminal speed temporarily. However, it will eventually return to its terminal speed once the external forces subside.

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