Can anyone point out a specific sports car for me?

sport car 80 image

Michael

I am looking for a v6 or v8 sports car with 200+ horsepower with rear or all wheel drive and 14+mpg
must be around the 2500-4000 price range. a 90's car or maybe even 80's. I have 3600 to spend. prefer a coupe or hatchback. can anybody show me a car like this? Cars with turbos are acceptable too.

Answer
I would not recommend a sport car for that amount of money. At least not any that would be good or reliable.

The mass of a sports car is 900 kg. The shape of the car is such that the aerodynamic drag coefficient is 0.24?

Arc

The mass of a sports car is 900 kg. The shape of the car is such that the aerodynamic drag coefficient is 0.240 and the frontal area is 2.20 m2. Neglecting all other sources of friction, calculate the initial acceleration of the car ,if it has been traveling at 80 km/h and is now shifted into neutral and is allowed to coast. (Take the density of air to be 1.295 kg/m2.)

Please explain. Thank you.

Answer
good and interesting question,

according to Bernoulli's equation,

Pâ + Ï g hâ + ½ Ï vâ² = Pâ + Ï g hâ + ½ Ï vâ²

Pâ + Ï g hâ + ½ Ï vâ² Cd = Pâ + Ï g hâ + 0

Pâ - Pâ = ½ Ï vâ² Cd

(Pâ - Pâ)A = ½ Ï vâ² A Cd

f_drag = ½ Ï v² A Cd

hence Cd is the aerodynamic drag coefficient.

ΣF = ma

W - f_drag = m a

mg - ½ Ï v² A Cd = m (dv/dt)

(â(mg))² - (â(½ Ï A Cd))² v² = m (dv/dt)

(â(mg) + vâ(½ Ï A Cd))(â(mg) - vâ(½ Ï A Cd)) = m (dv/dt)

dv/(â(mg) + vâ(½ Ï A Cd)) - dv/(â(mg) - vâ(½ Ï A Cd)) = dt/m

(1/â(½ Ï A Cd)) â« d((â(mg) + vâ(½ Ï A Cd)))/(â(mg) + vâ(½ Ï A Cd)) + â(½ Ï A Cd) â« d((â(mg) - vâ(½ Ï A Cd)))/(â(mg) - vâ(½ Ï A Cd)) = â« dt/m

(1/â(½ Ï A Cd)) ln [(â(mg) + vâ(½ Ï A Cd))/(â(mg) - vâ(½ Ï A Cd))] = t/m + C

initial condition, v(0) = 0

(1/â(½ Ï A Cd)) ln [(â(mg) + 0)/(â(mg) - 0)] = 0 + C

C = 0

solution of differential equation is:

(1/â(½ Ï A Cd)) ln [(â(mg) + vâ(½ Ï A Cd))/(â(mg) - vâ(½ Ï A Cd))] = t/m

do implicit differentiation with respect to time 't' to determine the acceleration, a

a = dv/dt

(1/â(½ Ï A Cd)) (â(mg) - vâ(½ Ï A Cd))/(â(mg) + vâ(½ Ï A Cd)) â(2 Ï A Cd)/(â(mg) - vâ(½ Ï A Cd))² (dv/dt) = 1/m

(1/â(½ * 1.295 * 2.2 * 0.24))(â(900 * 9.8) - (80/3.6)â(½ * 1.295 * 2.2 * 0.24))/(â(900 * 9.8) + (80/3.6)â(½ * 1.295 * 2.2 * 0.24)) â(2 * 1.295 * 2.2 * 0.24)/(â(900 * 9.8) - (80/3.6)â(½ * 1.295 * 2.2 * 0.24))² (dv/dt) = 1/900

dv/dt = 4.806205761316872428 m/s²







f_drag = -½ Ï v² A Cd

a = f/m

a = -½ Ï v² A Cd/m

a = -½ (1.295)(80/3.6)²(2.2)(0.24)/900

a = -0.187588 m/s²

deceleration is 0.187588 m/s²

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