Respuesta :
Solution :
The road gradient is given as = tan θ = 0.08
So, θ = 457 degrees, sin θ = 0.08
a). While moving downhill, the terminal velocity the forward force which acts due to the gravitation = backward force due to drag
[tex]$Mg\sin \theta = C_d \times \frac{1}{2} \rho AV^2$[/tex]
Taking the air density, [tex]$\rho = 1.2 kg/m^3$[/tex] and putting the values we get
[tex]$100 \times 9.8 \times 0.08 = C_d \times \frac{1}{2} \times 1.2 \times 0.46 \times 15^2$[/tex]
Therefore, [tex]$C_d = 1.26$[/tex]
This is approximately the same as the value of coefficient of drag given in the question.
Hence verified.
b). Drag force on level road, F = [tex]$C_d \times \frac{1}{2}\rho A V^2$[/tex]
Hence, deceleration :
[tex]$F/m = C_d \times \frac{1}{2m} \rho AV^2$[/tex]
[tex]$dV/dt = C_d \times \frac{1}{2m} \rho AV^2$[/tex]
[tex]$(dV/ds)(ds/dt) = C_d \times \frac{1}{2m} \times \rho AV^2$[/tex]
[tex]$V(dV/ds) = C_d \times \frac{1}{2m} \times \rho AV^2$[/tex]
[tex]$ dV/V= C_d \times \frac{1}{2m} \times \rho A \times ds$[/tex]
Integrating both the sides, we get
[tex]$\ln (V_2/V_1) = C_d \times \frac{1}{2m} \times \rho A \times (s_2-s_1)$[/tex]
Hence, [tex]$s_2-s_1 = \frac{2m}{C_d} \times \ln (V_2/V_1) / \rho A$[/tex]
Putting the values,
Distance = [tex]$2 \times \frac{100}{1.2} \times \frac{ \ln(10/15)}{(1.2 \times 0.46)}$[/tex]
= -122.5 m
Therefore, ignoring the negative sign, we get distance = 122.5 m
