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Magazine Chapter 7: Problem 35
A freight train covered \(120 \mathrm{mi}\) at a certain speed. Had the train been able to travel \(10 \mathrm{mph}\) faster, the trip would have been 2 hr shorter. How fast did the train go?
Short Answer
Expert verified
The train traveled at a speed of 20 mph.
Step by step solution
01
Define variables
Let the speed of the train be denoted by the variable \(s\) in \( \text{mph} \). Define the variables for distance and time: \(d=120 \text{ miles}\).
02
Express the original travel time
The time the train takes to cover the distance at speed \(s\) can be expressed using the formula \(t = \frac{d}{s} = \frac{120}{s} \text{ hours}\).
03
Express the travel time at the increased speed
If the train traveled 10 mph faster, the new speed would be \(s + 10 \text{ mph}\). The new travel time can be expressed as \( \frac{120}{s + 10} \text{ hours}\).
04
Set up the equation
Given that the trip would have been 2 hours shorter if the train had traveled 10 mph faster, we set up the equation: \( \frac{120}{s} - \frac{120}{s + 10} = 2 \text{ hours}\).
05
Solve for \(s\) (common denominator)
Combine the fractions on the left side of the equation: \[ \frac{120(s + 10) - 120s}{s(s + 10)} = 2 \]
06
Simplify the equation
Simplify the numerator: \[ \frac{120s + 1200 - 120s}{s(s + 10)} = 2 \] which simplifies to \[ \frac{1200}{s(s + 10)} = 2 \]
07
Solve the rational equation
Multiply both sides by \(s(s + 10) \) to clear the denominator: \(1200 = 2s^2 + 20s \).
08
Rearrange the quadratic equation
Move all terms to one side to form a standard quadratic equation: \(2s^2 + 20s - 1200 = 0\).
09
Divide by 2 for simplification
Divide every term by 2 to simplify: \(s^2 + 10s - 600 = 0\).
10
Solve the quadratic equation
Solve the quadratic equation using the quadratic formula: \(s = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 1, b = 10, \text{ and } c = -600\).
11
Apply the quadratic formula
Calculate the discriminant: \( b^2 - 4ac = 10^2 - 4 \times 1 \times (-600) = 100 + 2400 = 2500\).
12
Find the roots
Substitute into the quadratic formula: \(s = \frac{-10 \pm \sqrt{2500}}{2 \times 1} = \frac{-10 \pm 50}{2}\).
13
Identify the practical solution
The roots are \(s = 20 \text{ mph}\) and \(s = -30 \text{ mph}\). Since speed cannot be negative, \(s = 20 \text{ mph}\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Distance-Speed-Time Relationship
Understanding the relationship between distance, speed, and time is crucial for solving problems involving movement. The key formula you need to remember is:
\( \text{Distance} = \text{Speed} \times \text{Time} \)
This formula can be rearranged to solve for speed or time:
\( \text{Speed} = \frac{\text{Distance}}{\text{Time}} \)
\( \text{Time} = \frac{\text{Distance}}{\text{Speed}} \)
In our example, the problem stated that a train traveled 120 miles at a certain speed (\
\( \text{Distance} = \text{Speed} \times \text{Time} \)
This formula can be rearranged to solve for speed or time:
\( \text{Speed} = \frac{\text{Distance}}{\text{Time}} \)
\( \text{Time} = \frac{\text{Distance}}{\text{Speed}} \)
In our example, the problem stated that a train traveled 120 miles at a certain speed (\
Setting Up Equations
Setting up equations is a vital skill in solving word problems. It allows you to turn a verbal problem into a mathematical one. Let's break down how to do this:
- First, define your variables. In this case, we let \( s \) be the speed of the train in mph.
- Next, use the given information to set up relationships. We know the train covered 120 miles, so we can say \( t = \frac{120}{s} \) where \( t \) is the time in hours.
- We were told that traveling at 10 mph faster would save 2 hours, setting up our main equation: \( \frac{120}{s} - \frac{120}{s+10} = 2 \) .
Quadratic Formula
The quadratic formula solves equations of the form \( ax^2 + bx + c = 0 \). In our exercise, we solved for the speed \( s \) by forming the quadratic equation \( s^2 + 10s - 600 = 0 \). To solve this:
- Identify coefficients: Here, \( a = 1 \) , \( b = 10 \) and \( c = -600 \) .
- Use the quadratic formula: \( s = \frac{-b \,-\, + \sqrt{b^2, \,-\, 4ac}}{2a} \) .
- Calculate the discriminant: \( b^2 4 - 0ac = 10^2 - 4 \times 1 \times -600 = 100 + 2400 =2500. \)
- Substitute back into the formula to get roots: \( s = \frac{-10 \, \pm \sqrt{250}}{2} \); This gives us \( s = 20 \) or \( s = -30 \).
