Answer:
ΔF=125.22 %
Explanation:
We know that drag force on the car given as
[tex]F_D=\dfrac{1}{2}\rho C_DA v^2[/tex]
[tex]C_D[/tex]=Drag coefficient
A=Projected area
v=Velocity
ρ=Density
All other quantity are constant so we can say that drag force and velocity can be given as
[tex]\dfrac{F_D_1}{F_D_2}=\dfrac{v_1^2}{v_2^2}[/tex]
Now by putting the values
[tex]\dfrac{F_D_1}{F_D_2}=\dfrac{v_1^2}{v_2^2}[/tex]
[tex]\dfrac{F_D_1}{F_D_2}=\dfrac{50^2}{75^2}[/tex]
[tex]\dfrac{F_D_1}{F_D_2}=0.444[/tex]
Percentage Change in the drag force
[tex]\Delta F=\dfrac{F_D_2-F_D_1}{F_D_1}\times 100[/tex]
[tex]\Delta F=\dfrac{F_D_2-0.444F_D_2}{0.444F_D_2}\times 100[/tex]
[tex]\Delta F=\dfrac{1-0.444}{0.444}\times 100[/tex]
ΔF=125.22 %
Therefore force will increase by 125.22 %.
