Answer:
[tex]\Delta V = 209.151\,L[/tex], [tex]\Delta C = 217.517\,USD[/tex]
Explanation:
The drag force is equal to:
[tex]F_{D} = C_{D}\cdot \frac{1}{2}\cdot \rho_{air}\cdot v^{2}\cdot A[/tex]
Where [tex]C_{D}[/tex] is the drag coefficient and [tex]A[/tex] is the frontal area, respectively. The work loss due to drag forces is:
[tex]W = F_{D}\cdot \Delta s[/tex]
The reduction on amount of fuel is associated with the reduction in work loss:
[tex]\Delta W = (F_{D,1} - F_{D,2})\cdot \Delta s[/tex]
Where [tex]F_{D,1}[/tex] and [tex]F_{D,2}[/tex] are the original and the reduced frontal areas, respectively.
[tex]\Delta W = C_{D}\cdot \frac{1}{2}\cdot \rho_{air}\cdot v^{2}\cdot (A_{1}-A_{2})\cdot \Delta s[/tex]
The change is work loss in a year is:
[tex]\Delta W = (0.3)\cdot \left(\frac{1}{2}\right)\cdot (1.20\,\frac{kg}{m^{3}})\cdot (27.778\,\frac{m}{s})^{2}\cdot [(1.85\,m)\cdot (1.75\,m) - (1.50\,m)\cdot (1.75\,m)]\cdot (25\times 10^{6}\,m)[/tex]
[tex]\Delta W = 2.043\times 10^{9}\,J[/tex]
[tex]\Delta W = 2.043\times 10^{6}\,kJ[/tex]
The change in chemical energy from gasoline is:
[tex]\Delta E = \frac{\Delta W}{\eta}[/tex]
[tex]\Delta E = \frac{2.043\times 10^{6}\,kJ}{0.3}[/tex]
[tex]\Delta E = 6.81\times 10^{6}\,kJ[/tex]
The changes in gasoline consumption is:
[tex]\Delta m = \frac{\Delta E}{L_{c}}[/tex]
[tex]\Delta m = \frac{6.81\times 10^{6}\,kJ}{44000\,\frac{kJ}{kg} }[/tex]
[tex]\Delta m = 154.772\,kg[/tex]
[tex]\Delta V = \frac{154.772\,kg}{0.74\,\frac{kg}{L} }[/tex]
[tex]\Delta V = 209.151\,L[/tex]
Lastly, the money saved is:
[tex]\Delta C = \left(\frac{154.772\,kg}{0.74\,\frac{kg}{L} }\right)\cdot (1.04\,\frac{USD}{L} )[/tex]
[tex]\Delta C = 217.517\,USD[/tex]
