Question

Hint: Use 1D relative velocity to find the velocity of the car with respect to the air, which is used to find the drag force.

A 2520 lb car has a frontal cross-sectional area A of 20.0 ft2, a drag coefficient Cd of 0.25, and a drive (or tractive) force of 250 lbs. Assume an air density of 2.3x10-3 slug/ft3, Here is a list of frontal areas and drag coefficients for cars. This web site has interesting information on how the physics we are learning is incorporated into video games with cars. NOTES IMAGES DISCUSS UNITS STATS HELP Part Description Answer Chk History Determine the acceleration of the car when it is driving ft/s^2 at 43 mph on a level road (include units with answer) ft/s^2 А: 2.9 А. tries: 1 Show Details C: 2.902 Hints: 0,0 Points (D/T) Time Feedback Ans 2019-09-12.22 Wrong 2.8 18 ft/s 2 09:15: 14 (1.1*1) WARNING: 2 SF given, 3 SF expected. 2019-09-11.98 2.884 18 Determine the acceleration ft/s^2 11:07:13 (1.1*0.98) Wrong of the car when it is driving at 43 mph on a level road into a 5 mph head wind. (include units with answer) 2.898 ft/s 2 11.24 pts. 101% В. Format Wrong Check 2019-09-11.73 2% try penalty 2.9 18 WARNING: 2 SF given, 3 SF expected. ft/sA2 11:08:26(1.1*0.96) Hints: 1,0 2.898 2019-09-11.49 18 ft/s 2 18:05:11 (1.1*0. 94) Wrong Current problem value: 101.2% of 11.11 # tries: 4 Hide Detaile

Answer 1

Drag force is given by
D = 1/2*P*A*Cd*v^2

a)
43 mph = 63.07 fps
Fnet = F-D
= 250-1/2(PACdv^2)
= 250-1/2(2.3e-3*lbs^2/ft^2*ft*20*0.25)*63.07^2
= 227.1 lb
a = Fnet / mass
= 227.1lb /(2520 lb/32.2 ft/s^2)
= 2.9 ft/s^2

b)
Net force is F-D-fw where fw is force due to wind on the Car.
fw=0.5*(P)(v^2)A

So fw= 1.235lb
So a2= (F-D-fw)/(W/g)
=((F-D)g/W)-g(fw)/W
a2=a1-0.0158
=2.9-0.0158
=2.884
= 2.89 in 3 significant digits.