Problem 34 - Algebra for College Students

Question

Solve each problem. David and Keith are route drivers for a fast-photo company. David's route is 80 miles, and Keith's is 100 miles. Keith averages 10 mph more than David and finishes his route 10 minutes before David. What is David's speed?

Step-by-Step Solution

Answer

David's speed is 50 mph.

Step 1: Define the problem

Let David's speed be denoted by the variable \(d\) (in mph). Keith, who drives faster, has a speed of \(d + 10\) mph.

Step 2: Setup equations based on travel times

The travel time formula is \( \text{Time} = \frac{\text{Distance}}{\text{Speed}} \). Therefore, the time it takes David to complete his route is \( \frac{80}{d} \) hours, and the time it takes Keith is \( \frac{100}{d+10} \) hours.

Step 3: Calculate the time difference

Given that Keith finishes his route 10 minutes (or \( \frac{1}{6} \) hour) before David, the equation can be setup as: \( \frac{100}{d+10} + \frac{1}{6} = \frac{80}{d} \)

Step 4: Simplify the equation

Convert both times to a common form: \( \frac{100}{d+10} + \frac{1}{6} = \frac{80}{d} \). Multiply through by \(d(d+10)6\) to clear the denominators: \( 600d + d(d+10) = 480(d+10) \)

Step 5: Solve for \(d\)

Expand and rearrange terms: \( 600d + d^2 + 10d = 480d + 4800 \). Combine like terms and simplify: \( d^2 + 130d - 4800 = 0 \). Use the quadratic formula to solve for \( d \): \( d = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) with \( a=1, b=130, c=-4800 \). This gives \( d = \frac{-130 \pm \sqrt{16900 + 19200}}{2} \)

Step 6: Calculate the roots

Find the solutions for \( d \). Solve: \( d = \frac{-130 \pm 230}{2} \). This yields two values, \( d = 50 \) or \( d = -180 \). Since speed cannot be negative, \( d = 50 \).

Key Concepts

distance-speed-time relationship

In solving word problems involving distance, speed, and time, it's essential to understand their relationship. The fundamental formula connecting these three variables is:
\[ \text{Time} = \frac{\text{Distance}}{\text{Speed}} \]
This equation states that the time required to travel a distance depends on how fast you're going.

• Distance is how far you travel, usually measured in miles or kilometers.
• Speed is how fast you travel, measured in miles per hour (mph) or kilometers per hour (kph).
• Time is how long the travel takes, measured in hours or minutes.

In our exercise, David and Keith are traveling different distances at different speeds. To find their travel times, we apply the formula:

• David's travel time: \(\frac{80}{d}\), where \(d\) is David's speed.
• Keith's travel time: \(\frac{100}{d+10}\), where his speed is \(d+10\) mph because he is 10 mph faster than David.

They have different times because their speeds and distances are different, and this relationship helps us set up our equations correctly.

quadratic equations

A quadratic equation is an equation of the form:
\[ ax^2 + bx + c = 0 \]
Here, \(a\), \(b\), and \(c\) are constants, and \(x\) represents the variable we want to solve for.
In solving the word problem, we reach a point where we have a quadratic equation. To form this equation, we first clear the denominators by multiplying throughout, which simplifies our times difference equation:
\[ \frac{100}{d+10} + \frac{1}{6} = \frac{80}{d} \]
Multiply both sides by \(d(d + 10)6)\), and we get:
\[ 600d + d^2 + 10d = 480d + 4800 \]
This simplifies to:
\[ d^2 + 130d - 4800 = 0 \]
We use the quadratic formula to solve for \(d\):
\[ d = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Where \(a=1\), \(b=130\), and \(c=-4800\).
Substituting these values, we find:
\[ d = \frac{-130 \pm 230}{2} \]
The solutions are \(d = 50\) or a negative value, which is not physically meaningful in this context. Therefore, \(d = 50\) mph is David's speed.

problem-solving steps

Approaching word problems can be systematic if you follow these steps:
Step 1: Define the problem. Clearly state what you need to find. E.g., David's speed.
Step 2: Set up equations. Use the relationships you know (like distance-speed-time) to write equations. For our problem:

• David's travel time: \(\frac{80}{d}\)
• Keith's travel time: \(\frac{100}{d+10}\)

Step 3: Calculate time differences. State the known differences or relationships explicitly.
\( \frac{100}{d+10} + \frac{1}{6} = \frac{80}{d} \)
Step 4: Simplify the equation. Combine like terms and clear denominators for easier manipulation.
Step 5: Solve. Use algebra to isolate the variable. Here, we expanded to a quadratic form and used the quadratic formula.
Step 6: Validate your solution. Ensure the solution is practical (e.g., speeds must be positive). Always cross-check with the initial problem conditions.
These steps make complex problems manageable and ensure you don't overlook key elements.