"Distance" word problems, often also called "uniform rate" problems, involve something travelling at some fixed and steady ("uniform") pace ("rate", "velocity", or "speed"), or else you are told to regard to object as moving at some average speed.
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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.
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 mismatched units, and you have to catch this and convert to the correct units.
In case you're wondering, this type of exercise requires that the rate be fixed and steady (that is, unchanging) for the d = rt formula to work. The only way you can deal with a speed that might be changing over time is to take the average speed over the time or distance in question. Working directly with changing speeds will be something you'll encounter in calculus, as it requires calculusbased (or more advanced) methods.
A fixedspeed exercise is one in which the car, say, is always going exactly sixty miles an hour; in three hours, the car, on cruisecontrol, will have gone 180 miles. An averagespeed exercise is one in which the car, say, averaged forty miles an hour, but this average includes the different speeds related to stop lights, highways, and back roads; in three hours the car went 120 miles, though the car's speed was not constant. Most of the exercises you'll see will be fixedspeed exercises, but obviously they're not very "real world". It's a simplification they do in order to make the situation feasible using only algebraic methods.
There is a method for setting up and solving these exercises that I first encountered well after I'd actually been doing them while taking a class as an undergraduate. But, as soon as I was introduced to the method, I switched over, because it is *so* way easier.
First I set up a grid, with the columns being labelled with the variables from the "distance" formula, and the rows being labelled with the "parts" involved:

d 
r 
t 

first part 



second part 



total 



In the first part, the plane covered some distance. I don't know how much, so I'll need a variable to stand for this unknown value. I'll use the variable they give me in the distance equation:

d 
r 
t 

first part 
d 


second part 



total 



They gave me the speed, or rate, for this part, so I'll add this to my table:

d 
r 
t 

first part 
d 
105 

second part 



total 



The plane flew for some amount of time during this first part, but I don't know how long that was. So I need a variable to stand for this unknown value; I'll use the variable from the distance equation:

d 
r 
t 

first part 
d 
105 
t 
second part 



total 



For the second part, the plane travelled the rest of the total distance. I don't know the exact distance that was flown during this second part, but I do know that it was "however much was left of the 555 miles, after the first d miles were flown in the first part. "How much was left after [some amount] was taken out" is expressed with subtraction: I take the amount that has been taken care of already, and subtract this from whatever was the total. Adding this to my table, I get:

d 
r 
t 

first part 
d 
105 
t 
second part 
555 − d 


total 



They've given me the speed, or rate, for the second part, and I can use the same "How much is left?" construction for whatever was the time for this second part. So now my table looks like this:

d 
r 
t 

first part 
d 
105 
t 
second part 
555 − d 
115 
5 − t 
total 



For the "total" row, I add down (or take info from the exercise statement):

d 
r 
t 

first part 
d 
105 
t 
second part 
555 − d 
115 
5 − t 
total 
555 
 
5 
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Why did I not add down in the "rate" column? Because I cannot add rates! In this exercise, adding the rates would have said that the average rate for the entire trip was 105 + 115 = 220 miles per hour. But obviously this makes no sense.
The genius of this tablebased method of setup is that I can now create equations from the rows and columns. In this exercise, there is more than one way to proceed. I'll work with the "distance" equation to create expressions for the distances covered in each part.
Multiplying across, the first row tells me that the distance covered in the first part of the flight was:
1st part distance: 105t
Again multiplying across, the second row tells me that the distance covered in the second part of the flight was:
2nd part distance: 115(5 − t)
I can add these two partialdistance expressions, and set them equal to the known total distance:
105t + 115(5 − t) = 555
This is an equation in one variable, which I can solve:
105t + 115(5 − t) = 555
105t + 575 − 115t = 555
575 − 10t = 555
20 = 10t
2 = t
Looking back at my table, I see the I had defined t to be the time that the plane spent in the air on the first part of its journey. Looking back at the original exercise, I see that they want to know the times that the plane spent at each of the two speeds.
I now have the time for the first part of the flight; the time was two hours. The exercise said that the entire trip was five hours, so the second part must have taken three hours (found by subtracting the firstpart time from the total time). They haven't asked for the partial distances, so I now have all the information I need; no further computations are necessary. My answer is:
first part: 2 hours
second part: 3 hours
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When I was setting up my equation, I mentioned that there was more than one way to proceed. What was the other way? I could have used the table to create an expression for each of the two partial times, added, set the result equal to the given total, and solved for the variable d. Since the distance equation is d = rt, then the expressions for the partial times would be created by solving the equation for t =. My work would have looked like this:
Looking back at my table, I would have seen that this gives me the distance covered in the first half of the flight. Looking back at the exercise, I would have seeing that they are wanting times, not distances. So I would have backsolved for the time for the first part, and then done the subtraction to find the time for the second part. My work would have had more steps, but my answer would have been the same.
There are three things that I hope you take from the above example:
(My value for the distance, found above, is correct, but was not what they'd asked for.) But even more important to understand is this:
NEVER TRY TO 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? Of course not.
Can I even average the rates? If I drove at 20 mph for one hour, and then drove 60 mph for two hours, then I would have travelled 140 miles in three hours, or a little under 47 mph. But 47 is not the average of 20 and 60.
As you can see, the actual math involved in solving this type of exercise is often quite simple. It's the setup that's the hard part. So what follows are some more examples, but with just the setup displayed. Try your hand at solving, and click on the links to get popups from which to check your equations and solutions.
I will start in the usual way, by setting up my table:

d 
r 
t 

driving 



flying 



total 



I have labelled my rows so it's clear how they relate to the exercise. Now I need to fill in the rows. As before, I don't know the distance or the time for the part where the executive was driving, so I'll use variables for these unknowns, along with the given rate.

d 
r 
t 

driving 
d 
30 
t 
flying 



total 



For the flying portion of the trip, I'll use the "how much is left" construction, along with the given rate, to fill in my second row. I'll also fill in the totals.

d 
r 
t 

driving 
d 
30 
t 
flying 
150 − d 
60 
3 − t 
total 
150 
 
3 
The first row gives me the equation d = 30t. The second row is messier, giving me the equation:
150 − d = 60(3 − t)
There are various ways I can go from here; I think I'll solve this second equation for the variable d, and then set the results equal to each other.
150 − d = 60(3 − t)
150 − 60(3 − t) = d
Setting equal these two expressions for d, I get:
30t = 150 − 60(3 − t)
Both vehicles travelled for the same amount of time.

d 
r 
t 

car 


2 
bus 


2 
total 

 

The car's values are expressed in terms of the bus' values, so I'll use variables for the bus' unknowns, and then define the car in terms of the bus' variables. This gives me:

d 
r 
t 

car 
d + 20 
2r − 30 
2 
bus 
d 
r 
2 
total 
 
 
 
(As it turns out, I won't need the "total" row this time.) The car's row gives me:
d + 20 = 2(2r − 30)
This is not terribly helpful. The second row gives me:
d = 2r
I'll use the second equation to simplify the first equation by substituting "2r" from the second equation in for the "d" in the first equation. Then I'll solve the equation for the value of "r". Finally, I'll need to interpret this value within the context of the exercise, and then I'll state the final answer.
(Remember that the expression for the car's speed, from the table, was 2r − 30, so all you need to do is find the numerical value of this expression. Just evaluate; don't try to solve — again — for the value of r.)
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