Chapter 7, Problem 50P

To determine

Find the frontal area of the car in SI unit and U.S. Customary unit.

Answer to Problem 50P

The frontal area of the car in inches is $3018{\text{in}}^2$, expressed in SI units and U.S. Customary units as $1.9466{\text{m}}^2$ and $20.945{\text{ft}}^2$.

Explanation of Solution

Given data:

The height of the car $\left(H\right)$ is $50.9\text{in}$,

The width of the car $\left(W\right)$ is $68.5\text{in}$.

Formula used:

Formula to calculate the frontal area of the car is,

$A=A_1-A_2$ (1)

Here,

$A_1$ is the total frontal area of the car,

$A_2$ is the area of the squares in which the parts of the car is not touched.

Calculation:

Total frontal area:

Formula to calculate the total frontal area of the car is,

$A_1=H\times W$ (2)

Here,

$H$ is height of the car,

$W$ is width of the car.

Substitute $50.9\text{in}$ for $H$, and $68.5\text{in}$ for $W$ in equation (2) to find $A_1$.

$\begin{array}{c}{A}_1=\left(50.9\text{in}\right)\times \left(68.5\text{in}\right)\\ =3486.65{\text{in}}^2\end{array}$

Area of the squares in which the parts of the car is not touched:

Refer to the figure 7.50, the height and width of the car is divided as slots. Therefore,

$\text{The height of the car = 50}\text{.9 in = 15}\text{.5 slots}$

For 1 slot, the height (h) is:

$\begin{array}{c}1\text{slot}=\frac{50.9}{15.5}\text{in}\\ =3.28\text{in}\end{array}$

$\text{The width of the car}=\text{68}\text{.5 in}=\text{23 slots}$

For 1 slot, the width (w) is:

$\begin{array}{c}1\text{slot}=\frac{68.5}{23}\text{in}\\ =2.98\text{in}\end{array}$

From the figure 7.50, approximately the total number of slots which is not touching the parts of car is 48.

Formula to calculate the area of the squares which is not touched by the parts of car is,

$A_2=48\left(h\times w\right)$ (3)

Here,

$h$ is height of the square,

$w$ is width of the square.

Substitute $3.28\text{in}$ for $h$, and $2.98\text{in}$ for $w$ in equation (3) to find $A_2$.

$\begin{array}{c}{A}_2=\left(48\right)\left(3.28\text{in}\times 2.98\text{in}\right)\\ =469.17{\text{in}}^2\end{array}$

Substitute $3486.65{\text{in}}^2$ for $A_1$ and, $469.17{\text{in}}^2$ for $A_2$ in equation (1) to find $A$.

$\begin{array}{c}A=\left(3486.65{\text{in}}^2\right)-\left(469.17{\text{in}}^2\right)\\ =3018{\text{in}}^2\end{array}$

The SI unit of the frontal area of the car is,

$\begin{array}{c}A=3018{\text{in}}^2\left[\therefore 1{\text{in}}^2=\text{0}\text{.000645}{\text{m}}^2\right]\\ =\left(3018\right)\left(\text{0}\text{.000645}{\text{m}}^2\right)\\ =1.9466{\text{m}}^2\end{array}$

The U.S. Customary unit of the frontal area of the car is,

$\begin{array}{c}A=3018{\text{in}}^2\left[\therefore 1{\text{in}}^2=\text{0}\text{.00694}{\text{ft}}^2\right]\\ =\left(3018\right)\left(\text{0}\text{.00694}{\text{ft}}^2\right)\\ =20.945{\text{ft}}^2\end{array}$

Therefore, the frontal area of the car in SI unit and U.S. Customary unit is $1.9466{\text{m}}^2$ and $20.945{\text{ft}}^2$.

Conclusion:

Thus, the frontal area of the car in SI unit, U.S. Customary unit and in inches is $1.9466{\text{m}}^2$, $20.945{\text{ft}}^2$ and $3018{\text{in}}^2$ respectively.