Algebra: Speed and Distance Problems
Speed and Distance Problems
Have you ever heard of a word problem like this one? "Train A heads north at an average speed of 95 miles per hour, leaving its station at the precise moment as another train, Train B, departs a different station, heading south at an average speed of 110 miles per hour. If these trains are inadvertently placed on the same track and start exactly 1,300 miles apart, how long until they collide?"
If that problem sounds familiar, it's probably because you watch a lot of television (like me). Whenever TV shows talk about math, it's usually in the context of a main character trying but failing miserably to solve the classic "impossible train problem." I have no idea why that is, but time and time again, this problem is singled out as the reason people hate math so much.
Make sure that the units match in a travel problem. For instance, if the problem says you traveled at 70 miles per hour for 15 minutes, then r = 70 and t = 0.25. Since the speed is given in miles per hour, the time should be in hours also, and 15 minutes is equal to .25 hours. I got that decimal by dividing 15 minutes by the number of minutes in an hour: 1560 = 14 = 0.25.
In fact, it's not so hard. This, like any distance and rate of travel problem, only requires one simple formula:
- D = r t
Distance traveled (D) is equal to your rate of speed (r) multiplied by the time (t) you traveled that speed. What makes most distance and rate problems tricky is that you usually have two things traveling at once, so you need to use the formula twice at the same time. In this problem, you'll use it once for Train A and once for Train B.
To keep things straight in your mind, you should use little descriptive subscripts. For example, use the formula DA = rA tA for Train A's distance, speed, and time values and use the formula DB = rB tB for Train B.
The little A's in the formula DA = rA tA don't affect the values D, r, and t. They're just little labels to ensure that you only plug values corresponding to Train A into that formula.
Example 4: Train A heads north at an average speed of 95 miles per hour, leaving its station at the precise moment as another train, Train B, departs a different station, heading south at an average speed of 110 miles per hour. If these trains are inadvertently placed on the same track and start exactly 1,300 miles apart, how long until they collide?
Solution: Two trains means two distance formulas: DA = rA tA and DB = rB tB. Your first goal is to plug in any values you can determine from the problem. Since Train A travels 95 mph, rA = 95; similarly, rB = 110.
Notice that the problem also says that the trains leave at the same time. This means that their travel times match exactly. Therefore, instead of denoting their travel times as tA and tB (which suggests they are different), I will write them both as t (which suggests they are equal). At this point, your formulas look like this:
|DA = 95t||DB = 110t|
Even though you added the distances in this problem, you won't always do soit depends on how the problem is worded. In Problem 3, for example, you will not calculate a sum.
Here's the tricky step. The trains are heading toward one another on a track that's 1,300 miles long. Therefore, they must collide when, together, both trains have traveled a total of 1,300 miles. Of course, Train B is going to travel more of those 1,300 miles than Train A, since it's traveling faster, but that doesn't matter. You don't even have to figure out how far each train will go. All that matters is that when DA + DB = 1300, it's curtains. Luckily, you happen to know what DA and DB are (95t and 110t, respectively) so plug those into the equation and solve.
You've Got Problems
Problem 3: Dave rode his bike from home to a 7-11 at an average speed of 17 mph, and the trip took 1.25 hours. However, as he pulled up to the store, he rode over some glass, causing both tires to go flat. Because of this rotten luck, he had to push his bike back home at an average speed of 3 mph. How long did the trip home take?
- DA + DB = 1300
- 95t + 110t = 1300
- 205t = 1300
- t 6.341 hours
So, the trains will collide in approximately 6.341 hours.
Excerpted from The Complete Idiot's Guide to Algebra 2004 by W. Michael Kelley. All rights reserved including the right of reproduction in whole or in part in any form. Used by arrangement with Alpha Books, a member of Penguin Group (USA) Inc.