Problem 47 - Introductory Algebra for College Students

Question

A snowstorm causes a bus driver to decrease the usual average rate along a 60 -mile route by 15 miles per hour. As a result, the bus takes two hours longer than usual to complete the route. At what average rate does the bus usually cover the 60 -mile route?

Step-by-Step Solution

Answer

The average speed the bus usually covers the 60-mile route is given by the positive solution of the equation \(X^{2} - 15X + 450 = 0\).

Step 1: Identify the variables

Let's denote the normal speed of the bus as \(X\) miles per hour. After the snowstorm, the speed of the bus is \(X-15\) miles per hour.

Step 2: Setup the equation

We know that distance equals speed times time. Therefore, we can write down the following equations based on the information given in the problem: Normal condition: \[\frac{60}{X}\ =\ t\] Condition after the decrease in speed due to the snowstorm: \[\frac{60}{X-15}\ =\ t\ +\ 2\] The '2' in the second equation represents the additional 2 hours the bus takes to complete the route after the speed decrease.

Step 3: Solve for X

Now, we can solve for \(X\) by setting the two equations equal to each other and solving for \(X\). This gives: \[\frac{60}{X} = \frac{60}{X-15} + 2\] Multiplying through by \(X(X-15)\) to clear the fractions gives: \[60(X-15) = 60X + 2X(X-15)\] Expanding and simplifying gives: \[60X - 900 = 60X + 2X^{2} - 30X\] Simplifying further gives the quadratic: \[2X^{2} - 30X + 900 = 0\] Dividing this equation by 2 gives: \[X^{2} - 15X + 450 = 0\] Solving this quadratic equation provides the possible values for \(X\).

Step 4: Calculate the average speed

The quadratic equation can be solved using the quadratic formula: \[X = \frac{-b ± \sqrt{b^{2} - 4ac}}{2a}\] Where \(a = 1\), \(b = -15\) and \(c = 450\). Calculating gives: \[X_{1,2} = \frac{15 ± \sqrt{15^{2} - 4*1*450}}{2*1}\] This gives two solutions, but since a speed cannot be negative, we discard the negative solution. Therefore, \(X\), the normal bus speed, is the positive solution to the equation.