This problem compares some energy properties of two motor vehicles: For efficient car, use drag coefficient of .21 and a cross sectional area of 1.5m^2. For a light truck, use drag coefficient of .75 and a cross sectional area of 2.7m^2. Use 1.25 kg/m^3 for the density of air. Determine the top speeds of each vehicle if each vehicle had a maximum usable power output of 100 kW.
muilenberg
To determine the top speeds of each vehicle, we can start by using the concept of power and the energy properties given for each vehicle. The power equation can be used to calculate the top speed.
First, let's calculate the air resistance or aerodynamic drag for each vehicle using the drag coefficient, cross-sectional area, and air density.
For the efficient car:
Drag coefficient (Cd) = 0.21
Cross-sectional area (A) = 1.5 m^2
Density of air (ρ) = 1.25 kg/m^3
The formula for aerodynamic drag force (F) is:
F = 0.5 * Cd * A * ρ * v^2
where v is the velocity or speed of the vehicle.
Let's rearrange the equation to solve for v:
v = sqrt((2 * P) / (Cd * A * ρ))
where P is the power output.
For the efficient car:
P = 100 kW
Cd = 0.21
A = 1.5 m^2
ρ = 1.25 kg/m^3
Substituting these values into the equation:
v = sqrt((2 * 100000) / (0.21 * 1.5 * 1.25))
Now, let's calculate the top speed of the efficient car.
v = sqrt(381904.76)
v ≈ 618.51 m/s
So, the top speed of the efficient car is approximately 618.51 m/s.
Now, let's do the same calculations for the light truck.
For the light truck:
Drag coefficient (Cd) = 0.75
Cross-sectional area (A) = 2.7 m^2
Density of air (ρ) = 1.25 kg/m^3
Using the formula v = sqrt((2 * P) / (Cd * A * ρ)):
v = sqrt((2 * 100000) / (0.75 * 2.7 * 1.25))
v = sqrt(59259.26)
v ≈ 243.3 m/s
Therefore, the top speed of the light truck is approximately 243.3 m/s.
To convert these speeds to more commonly used units, you can multiply the speed by 2.23694 to obtain the speed in miles per hour (mph).