Question

The mass of a sports car is 1200 kg. The shape of the car is such that the aerodynamic drag coefficient is 0.260 and the frontal area is 2.00 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.)

Answer 1

Answer:

The acceleration is -0.138 m/$s^{2}$

Explanation:

The drag force of a body can be expressed using the relationship below;

$Drag force = -0.5CADV^{2}$...........................1

where C is the aerodynamic drag coefficient = 0.260

A is the  frontal area = 2.00 $m^{2}$

D is the density of air =1.295 kg/$m^{3}$

V is the speed of the sports car = 80 km/hr =  

                    80 km/hr 1 hr/ 3600 sec x 1000 m / km = 22.22 m/sec

using newtons second law;

f = ma

a = f/m..................................................2

Substituting f in equation 1 into equation 2 we have;

$a = \frac{-0.5CADV^{2}}{m}$

$a = \frac{-0.5(0.260)(2.00)(22.22)^{2}}{1200}$

a = -0.138 m/$s^{2}$

Therefore the acceleration is -0.138 m/$s^{2}$, which is a deceleration of

0.138 m/$s^{2}$