"Distance" Word Problems: More Examples
Purplemath
When working with the distance equation, d = rt, you will likely also come across exercises where you're working with two moving objects (like two cars approaching each other from opposite directions) or one object whose speed is being influenced in different ways (such as a boat moving upstream against the current and then downstream with the current). But the equation and set-up method will remain the same.
As long as you label things clearly (so you don't lose track of which variable or expression stands for what), you should be okay.
- A passenger train leaves the train depot 2 hours after a freight train left the same depot. The freight train is traveling 20 mph slower than the passenger train. Find the rate of each train, if the passenger train overtakes the freight train in three hours.
The freight train left two hours before the passenger train, and the passenger (or "pax") train took three hours to catch up. This means that the freight train travelled for 2 + 3 = 5 hours.
For the purposes of this exercise, I care only about the distance between the depot (where each train started) and the point at which the passenger train caught up. So, for the purposes of this exercise, the distance travelled by the trains is the same. (I neither know nor care how far the trains might have gone once the second train passed the first.)
Whatever was the speed r of the passenger train, the freight train was 20 mph slower, or r − 20.
Using variables for the unknowns of rate and distance, I create this table:
|
d |
r |
t |
|---|---|---|---|
pax |
d |
r |
3 |
freight |
d |
r − 20 |
5 |
total |
|
--- |
|
Multiplying across the two rows, I get these equations:
pax: d = 3r
freight: d = 5(r − 20)
Setting the right-hand sides equal to each other (since each is equal to the same distance d) gives me a one-variable equation:
3r = 5(r − 20)
Solve for r; interpret the value within the context of the exercise, and state the final answer.
In the grid I made for the above exercise, it turned out that I didn't need the "total" row. Sometimes you will and sometimes you won't.
- Two cyclists start at the same time from opposite ends of a course that is 45 miles long. One cyclist is riding at 14 mph and the second cyclist is riding at 16 mph. How long after they begin will they meet?
However long it takes the two cyclists to meet, they will have taken the same amount of time to get to their meeting point. I don't know (yet) how much time this is, so I'll use a variable.
However much distance d the slower guy covers, the faster guy covered how much was left of the course, or 45 − d.
They have given me the two speeds. So my grid looks like this:
|
d |
r |
t |
|---|---|---|---|
slow |
d |
14 |
t |
fast |
45 − d |
16 |
t |
total |
45 |
--- |
|
Using "d = rt", I get d = 14t from the first row, and 45 − d = 16t from the second row. Since these distances add up to 45, I will add the distance expressions and set equal to the given total:
45 = 14t + 16t
Solve for t.
There is a class of "distance" exercises that refers to things like "a boat's speed in still water" or "a plane's speed in calm air". These require the adding and subtracting of speeds.
What do they mean by "the speed in calm/still water/air"?
Considering a boat, the "in calm water" phrase refers to the power that the engine is putting out; it's how fast the boat would be going if there were no current affecting it. In simple terms, the "in still water" speed is what the boat's speedometer is showing. When the boat is going with the current (that is, when it's going downstream), then the river is adding its own speed to that of the boat; when the boat is going against the current (that is, when it's going upstream), then the river is pushing against the boat, subtracting its own speed from that of the boat. If the boat is moving across a lake with no current, then the speedometer reading is the actual speed of the boat.
This means that, for the downstream part, the boat's speedometer reading and the speed of the river's current will be added to find how fast the boat was moving (with respect to the shore, which is not moving). This makes sense, if you think about it: if you cut the engine, the boat would still continue to move downstream because that's the way that the current is moving. When the boat is going up against the current, the water's speed is subtracted from the boat's speedometer reading. This makes sense, too: in the extreme case, if the water were to be flowing fast enough, the boat would still be going downstream (that is, it will have a "negative" speed, because the boat would be going backwards at this point), because the water is more powerful than the boat. (Think of a cartoon boat in a cartoon river that's heading toward a waterfall. The guy paddles like crazy, but he still goes backwards and over the edge.)
The situation is the same (in math classes) for a plane, helicopter, etc. If there's no wind (that is, if it's flying in still or calm air), then the plane's speedometer reading is how fast the plane is going. If there is a wind, then either it will be pushing the plane foreward to speed it up, or else pushing against the plane to slow it down. You've probably seen an extreme case of this in real life. Think of a very windy day and a bird is trying to fly, against the wind, from your yard to the yard across the street. The bird can flap like crazy, but the wind pushes it right back into your yard. It wasn't that the bird wasn't flapping hard; it's just that the speed of the wind was greater than whatever speed the poor little bird's wings were trying to generate.
(In real life, a tailwind will reduce a plane's wings' lift, meaning that the plane's engines will actually have to work harder — that is, spin faster — to achieve the same desired airspeed. We ignore this fact in algebra exercises.)
- A boat travels for three hours with a current of 3 mph and then returns the same distance against the current in four hours. What is the boat's speed in calm water? How far did the boat travel one way?
Whatever the distance was, it was the same in each direction. I'll use a variable to represent this. I don't know what the boat's speedometer was saying; I'll use the variable b (for "boat") to stand for this rate. They've given me the times in each direction, and they've given me the speed of the current, so I can do addition and subtraction to get expressions for the net speed upstream and downstream. My grid looks like this:
|
d |
r |
t |
|---|---|---|---|
downstream |
d |
b + 3 |
3 |
upstream |
d |
b − 3 |
4 |
total |
2d |
--- |
7 |
Using "d = rt", the downstream row gives me:
d = 3(b + 3)
The upstream row gives me:
d = 4(b − 3)
Since these distances are the same, I will set them equal:
3(b + 3) = 4(b − 3)
Solve for b. Then back-solve for d.
In this case, I didn't need the "total" row. But it was helpful that I'd picked a variable that reminded me of what it stood for. Don't feel like you have to stick to the "regular" variables; pick whatever you find most useful.
- With the wind, an airplane travels 1120 miles in seven hours. Against the wind, it takes eight hours. Find the rate of the plane in still air and the velocity of the wind.
Just as with the last problem, I am really dealing with two rates together: the plane's speedometer reading, and the wind's speed. When the plane turns around, the wind is no longer pushing the plane to go faster, but is instead pushing against the plane to slow it down. I don't know either of the speeds; to keep them straight, I'll use the variables p and w for the speed of the plane and the wind, respectively.
They have given me the distances and the times. Adding in this info, my table looks like this:
|
d |
r |
t |
|---|---|---|---|
tailwind |
1120 |
p + w |
7 |
headwind |
1120 |
p − w |
8 |
total |
2240 |
--- |
15 |
The tailwind row gives me:
1120 = 7(p + w)
The headwind row gives me:
1120 = 8(p − w)
The temptation is to just set these equal, like I did with the last problem, but that just gives me:
7(p + w) = 8(p − w)
...which doesn't help much. I need to get rid of one of the variables.
I'll take that tailwind equation:
1120 = 7(p + w)
...and divide through by 7 and solve for [*flipping a coin*] the variable p:
160 = p + w
160 − w = p
I'll do the same thing with headwind equation, this time dividing through by 8:
140 = p − w
140 + w = p
Then I'll set the two left-hand sides equal to each other:
160 − w = 140 + w
Then I'll solve for w. Then I'll back-solve to find p.
- A spike is hammered into a train rail. You are standing at the other end of the rail. You hear the sound of the hammer strike both through the air and through the rail itself. These sounds arrive at your point six seconds apart. You know that sound travels through air at 1100 feet per second and through steel at 16,500 feet per second. How far away is that spike? (Round to one decimal place.)
However far away that spike is, the distance d travelled by the sound, either through the air or through the rail, is the same. They've given me the rates. However long the sound took to reach me through the air, the sound reached me through the steel six seconds earlier. (The travel-speed is greater, so the travel-time must be lesser.) Putting this together, my grid looks like this:
|
d |
r |
t |
|---|---|---|---|
air |
d |
1,100 |
t |
steel |
d |
16,500 |
t − 6 |
total |
--- |
--- |
--- |
Multiplying across, the air and steel rows give me these two equations:
d = 1100t
d = 16500(t − 6)
Since the distances are the same, I set the distance expressions equal to get:
1100t = 16,500(t − 6)
Solve for the time t, and then back-solve for the distance d by plugging t into either expression for the distance d.
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