Answer:
The acceleration is -0.138 m/

Step-by-step explanation:
The drag force of a body can be expressed using the relationship below;
...........................1
where C is the aerodynamic drag coefficient = 0.260
A is the frontal area = 2.00

D is the density of air =1.295 kg/

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 = -0.138 m/

Therefore the acceleration is -0.138 m/
, which is a deceleration of
0.138 m/
