A car travels at $55\mathrm{mi}/\mathrm{h}$and gets $11\mathrm{km}/\mathrm{L}$of gasoline. How many gallons of gasoline are needed for a $3.0$-h trip? (2,6)

The given is the distance travelled by a car with gasoline.

The objective is to determine the number of gallons of gasoline.

The top speed of the car is 55 mi/hr.

Amount of gasoline it gets as : 11 km/L

The car's top speed is $55mi/h$. The rate of use of gasoline is $11km/L$.

We need to figure out how many gallons of gasoline are needed for a $3.0$-hour drive.

We'll utilize the provided data to calculate the conversion factors. We must also convert between miles and kilometers, as well as liters and gallons.

The speed of the car is $55\mathrm{mi}/\mathrm{h}$. So, the conversion factors are

$\frac{55mi}{1h}and\frac{1h}{55mi}$

The consumption of gasoline is increasing. $11\mathrm{km}/\mathrm{L}$. So, the conversion factors are $\frac{11km}{1L}and\frac{1L}{11km}$

The relationship between miles and kilometers is equal $1\mathrm{km}=0.621\mathrm{mi}$

Hence, the conversion factors are $\frac{1\mathrm{km}}{0.621\mathrm{mi}}$and $\frac{0.621\mathrm{mi}}{1\mathrm{km}}$

The relationship between liters and gallons is equal.

$1\mathrm{L}=0.265\mathrm{gal}$

Hence, the conversion factors are,

$\frac{1\mathrm{L}}{0.265\mathrm{gal}}\text{and}\frac{0.265\mathrm{gal}}{1\mathrm{L}}$

The relevant conversion factors are employed, and each factor is set up to cancel the unit before it until the desired unit is produced.

Consequently, we have

$3.0\mathrm{h}\times \frac{55\mathrm{mi}}{1\mathrm{h}}\times \frac{1\mathrm{km}}{0.621\mathrm{mi}}\times \frac{1\mathrm{t}}{11\mathrm{km}}\times \frac{0.265\mathrm{gal}}{1\mathrm{t}}=6.4\mathrm{gal}$

Hence, a$3.0h$trip requires $6.4\mathrm{gal}$of gasoline.