College Algebra, Chapter 1, 1.2, Section 1.2, Problem 60

Suppose that a woman driving a car $14 ft$ long is passing a truck $30 ft$ long. The truck is traveling at $50 mi/hr$. How fast must a woman drive her car so that she can pass the truck completely in $6s$?

To be consistent with the units, let us use feet and seconds instead of miles and hours.

Since both cyclists are traveling towards each other, the effective speed will be $2x + x = 3x$.

So the speed of the truck, $\displaystyle 50 \frac{mi}{hr} \left( \frac{5280 ft}{mi} \right) \left( \frac{1 hr}{3600 s} \right) = \frac{220}{3} ft/s$

Thus, in $6$ seconds, the truck travels $\displaystyle \left( \frac{220}{3} \right) \left( \frac{ft}{s} \right) (6s) = 440 ft$

Therefore, in order for the woman's car to pass the truck completely, (up to the back bumper), the distance it should traveled is $440 ft + 30 ft$ (length of the truck) $+ 14 ft$ (length of the car) $= 484 ft$.

Hence, the speed of the car must be..


$
\begin{equation}
\begin{aligned}

V = \frac{d}{dt} =& \frac{484 ft}{6s} \left( \frac{1 mi}{5280 mi} \right) \left( \frac{3600 s}{1 h} \right)
\\
\\
=& 55 \frac{mi}{hr}


\end{aligned}
\end{equation}
$