Answer:
15.065ft
Explanation:
To solve this problem it is necessary to consider the aerodynamic concepts related to the Drag Force.
By definition the drag force is expressed as:
[tex]F_D = -\frac{1}{2}\rho V^2 C_d A[/tex]
Where
[tex]\rho[/tex] is the density of the flow
V = Velocity
[tex]C_d[/tex]= Drag coefficient
A = Area
For a Car is defined the drag coefficient as 0.3, while the density of air in normal conditions is 1.21kg/m^3
For second Newton's Law the Force is also defined as,
[tex]F=ma=m\frac{dV}{dt}[/tex]
Equating both equations we have:
[tex]m\frac{dV}{dt}=-\frac{1}{2}\rho V^2 C_d A[/tex]
[tex]m(dV)=-\frac{1}{2}\rho C_d A (dt)[/tex]
[tex]\frac{1}{V^2 }(dV)=-\frac{1}{2m}\rho C_d A (dt)[/tex]
Integrating
[tex]\int \frac{1}{V^2 }(dV)= - \int\frac{1}{2m}\rho C_d A (dt)[/tex]
[tex]-\frac{1}{V}\big|^{V_f}_{V_i}=\frac{1}{2m}(\rho)C_d (\pi r^2) \Delta t[/tex]
Here,
[tex]V_f = 60mph = 26.82m/s[/tex]
[tex]V_i = 120.7m/s[/tex]
[tex]m= 1600lbf = 725.747Kg[/tex]
[tex]\rho = 1.21 kg/m^3[/tex]
[tex]C_d = 0.3[/tex]
[tex]\Delta t=7s[/tex]
Replacing:
[tex]\frac{-1}{26.82}+\frac{1}{120.7} = \frac{1}{2(725.747)}(1.21)(0.3)(\pi r^2) (7)[/tex]
[tex]-0.029 = -5.4997r^2[/tex]
[tex]r = 2.2963m[/tex]
[tex]d= r*2 = 4.592m \approx 15.065ft[/tex]
