"Distance" Word Problems (page 1 of 2)
"Distance" word problems, often also called "uniform rate" problems, involve something travelling at some fixed and steady ("uniform") pace ("rate" or "speed"), or else moving at some average speed. Whenever you read a problem that involves "how fast", "how far", or "for how long", you should think of the distance equation, d = rt, where d stands for distance, r stands for the (constant or average) rate of speed, and t stands for time.
Warning: Make sure that the units for time and distance agree with the units for the rate. For instance, if they give you a rate of feet per second, then your time must be in seconds and your distance must be in feet. Sometimes they try to trick you by using the wrong units, and you have to catch this and convert to the correct units.
First I'll set up a grid: Copyright © Elizabeth Stapel 2000-2011 All Rights Reserved
Using "d = rt", the first row gives me d = 105t and the second row gives me:
555 – d = 115(5 – t)
Since the two distances add up to 555, I'll add the two distance expressions, and set their sum equal to the given total:
555 = 105t + 115(5 – t)
Then I'll solve:
555 = 105t
+ 575 – 115t
According to my grid,
stands for the time spent on the first part of the trip, so my answer
is "The plane
flew for two hours at 105
mph and three hours at 115
You can add distances and you can add times, but you cannot add rates. Think about it: If you drive 20 mph on one street, and 40 mph on another street, does that mean you averaged 60 mph?
As you can see, the actual math involved is often quite simple. It's the set-up that's the hard part. So what follows are some more examples, but with just the set-up displayed.
The first row gives me the equation d = 30t. Since the first part of his trip accounted for d miles of the total 150-mile distance and t hours of the total 3-hour time, I am left with 150 – d miles and 3 – t hours for the second part. The second row gives me the equation:
150 – d = 60(3 – t)
Adding the two "distance" expressions and setting their sum equal to the given total distance, I get:
150 = 30t + 60(3 – t)
(As it turns out, I won't need the "total" row this time.) The first row gives me:
d + 20 = 2(2r – 30)
This is not terribly helpful. The second row gives me:
d = 2r
(As it turns out, I won't need the "total" row this time.) Why is the distance just "d" for both trains? Partly, that's because the problem doesn't say how far the trains actually went. But mostly it's because they went the same distance as far as I'm concerned, because I'm only counting from the depot to wherever they met. After that meet, I don't care what happens. And how did I get those times? I know that the passenger train drove for three hours to catch up to the freight train; that's how I got the "3". But note that the freight train had a two-hour head start. That means that the freight train was going for five hours.
Now that I have this information, I can try to find my equation. Using the fact that d = rt, the first row gives me d = 3r (note the revised table above). The second row gives me:
d = 5(r – 20)
Since the distances are equal, I will set the equations equal:
3r = 5(r – 20)