College Algebra, Chapter 1, 1.3, Section 1.3, Problem 92

Christian drove from Tortula to Cactus, a distance of $250 mi$. He increased his speed by $10 mi/hr$ for the $360 mi$ trip from Cactus to Dry Junction. If the total trip took $11 h$, what was his speed from Tortula to Cactus?

Recall that the formula for speed is $\displaystyle V = \frac{d}{t}$, so $\displaystyle t = \frac{d}{v}$. Let $t_1$ be the time consumed from Tortula to Cactus and $t_2$ be the time from Cactus to Dry Junction. Thus,


$\begin{equation} \begin{aligned} t_T =& t_1 + t_2 && \text{Model} \\ \\ t_T =& \frac{d_1}{V_1} + \frac{d_2}{V_2} && \text{Substitute the formula for } t \\ \\ 11 =& \frac{250}{V_1} + \frac{360}{V_1 + 10} && \text{Substitute the given} \\ \\ 11 =& \frac{250(V_1 + 10) + 360 (V_1)}{V_1 (V_1 + 10)} && \text{Take the LCD} \\ \\ 11 =& \frac{250 V_1 + 2500 + 360V_1}{V_1 (V_1 + 10)} && \text{Simplify} \\ \\ 11V_1^2 + 110 V_1 =& 610V_1 + 2500 && \text{Simplify the numerator and apply cross multiplication} \\ \\ 11V_1^2 - 500V_1 - 2500 =& 0 && \text{Combine like terms and subtract } 2500 \\ \\ (V_1 - 50) \left(V_1 + \frac{50}{11} \right) =& 0 && \text{Factor} \\ \\ V_1 - 50 =& 0 \text{ and } V_1 + \frac{50}{11} = 0 && \text{ZPP} \\ \\ V_1 =& 50 \text{ and } V_1 = \frac{-50}{11} && \text{Solve for } V_1 \\ \\ V_1 =& 50 \frac{mi}{hr} && \text{Choose } V_1 > 0 \end{aligned} \end{equation}$


Thus, Christian's speed from Tortula to Cactus is $V_1 = 50 mi/hr$.