Solving Train Rate Problems with Mathematical Equations

In this article, we will delve into a classic train rate problem, exploring the steps required to solve it using mathematical equations. This problem not only illustrates the application of basic algebra but also highlights the practical significance of understanding distance, speed, and time relationships.

Problem Statement

Two trains leave towns 468 miles apart at the same time and travel toward each other. One train travels 16 mi/h faster than the other. If they meet in 3 hours, what is the rate of each train?

Setting Up the Equations

To solve this problem, we need to set up equations based on the given information.

Let the speed of the slower train be x miles per hour. Then the speed of the faster train will be x 16 miles per hour. Since the two trains are traveling toward each other, their combined speed is x (x 16) 2x 16 miles per hour.

They meet after 3 hours, covering a distance of 468 miles in that time. Therefore, we can set up the equation:

[ (2x 16) times 3 468 ]

Solving the Equation

Simplify the equation:

2x 16 frac{468}{3}

2x 16 156

2x 156 - 16

2x 140

x 70

Thus, the rates of the trains are:

Slower train: 70 mi/h Faster train: 70 16 86 mi/h

Another Example: Different Units and Numbers

Consider a similar problem where two trains are 532 miles apart and travel toward each other. Train A travels 12 mph faster than train B. They meet in 2.5 hours. Let's solve this:

Let the speed of train B be x miles per hour. Then the speed of train A will be x 12 miles per hour. Their combined speed is x (x 12) 2x 12 miles per hour. They meet in 2.5 hours, so the equation is:

(2x 12) times 2.5 532

Simplify the equation:

2x 12 frac{532}{2.5}

2x 12 212.8

2x 212.8 - 12

2x 200.8

x 100.4

Thus, the rates of the trains are:

Train B: 100.4 mi/h Train A: 100.4 12 112.4 mi/h

A Third Example: Cyclists Moving Toward Each Other

Two cyclists start from two different towns 120 miles apart and travel towards each other. Cyclist A travels at 4 mph less than Cyclist B. They meet after 4 hours. Let's solve this:

Let the speed of Cyclist B be x mph. Then the speed of Cyclist A is x - 4 mph. Since they meet after 4 hours, the total distance covered by both cyclists is 120 miles. So, we have:

(x - 4) times 4 x times 4 120

Simplify the equation:

4x - 16 4x 120

8x - 16 120

8x 120 16

8x 136

x 17

Thus, the rates of the cyclists are:

Cyclist B: 17 mph Cyclist A: 17 - 4 13 mph

Metric Example: Converting Units

Consider a problem where a train travels 600 miles in 8 hours. We need to find the speed in feet per second and then convert it back to miles per hour.

Convert miles to feet: 600 miles 3168000 ft. Convert hours to seconds: 8 hours 28400 seconds. Calculate speed in feet per second: 3168000 ÷ 28400 111.549 ft/s. Convert ft/s to mph: 111.549 ÷ 1.466 76 mph. Calculate the difference in speeds: 76 - 9.5 66.5 mph.

Thus, the speeds are:

First train: 85.5 mph Second train: 66.5 mph The difference in speeds is: 85.5 - 66.5 19 mph.