91Ó°ÊÓ

$a=\frac{2mg}{Cp{v^2}_t}$

which illustrates the inverse proportionality between the area and the speed-squared

thus

$$\begin{gathered} A_{slow}=310\mathrm{km}/\mathrm{h} \\ {A}_{fast=}160\mathrm{km}/\mathrm{h} \end{gathered}$$

This problem is based on the drag force which is a type of friction. This is the force acting opposite to the relative motion of an object moving with respect to the surrounding medium. Using the concept of the drag force and terminal speed the ratio of the effective cross-sectional area A in the slower position to that in the faster position.

Formula:

The terminal speed is given by

$v_t=\sqrt{\frac{2F_g}{CpA}}$

Where C is the drag coefficient,${}^p$is the fluid density, A is the effective cross-sectional area, and ${}^{F_g}$is the gravitational force.

solve for the area

$A=\frac{2mg}{Cp{v^2}_t}$

which illustrates the inverse proportionality between the area and the speed-squared

When a ratio of areas has been setup of the slower case to the faster case,

$$\begin{gathered} \frac{A_{slow}}{A_{fast}}={\left(\frac{110\mathrm{km}/\mathrm{h}}{160\mathrm{km}/\mathrm{h}}\right)}^2 \\ =3.75 \end{gathered}$$

Hence, the ratio of the effective cross-sectional area A in the slower position to that in the faster position is ${}^{3.75}$