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"Distance" Word Problems: More Examples (page 2 of 2)

• 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?
     

	d	r	t
slow guy	d	14	t
fast guy	45 – d	16	t
total	45	---	---

Why is t the same for both cyclists? Because we are measuring from the time they both started to the time they met somewhere in the middle. And how did we get "d" and "45 – d" for the distances? Because the slow guy went one "half" of the way, and the fast guy went the other "half" of the way to meet him. (I say "half" in quotes, because it wasn't exactly halfway, since they were going different speeds.) The two guys together covered the whole 45 miles.

Using "d = rt", we get d = 14t from the first row, and 45 – d = 16t from the second row. Since these distances add up to 45, add the distances:

45 = 14t + 16t

Solve for t.

• A boat travels for three hours with a current of 3mph 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?
     

	d	r	t
downstream	d	b + 3	3
upstream	d	b – 3	4
total	2d	---	7

(It may turn out that we don't need the "total" row.)

I have used "b" to indicate the boat's speed. Why are the rates "b + 3" and "b – 3"? Because we actually have two speeds combined into one on each trip. The boat has a certain speed (the "speed in calm water" that we're looking for; this is the speed that registers on the speedometer), and the water has a certain speed (this is the "current"). When the boat is going with the current, the water's speed is added to the boat's speed. This makes sense, if you think about it: even if you cut the engine, the boat would still be moving, because the water would be carrying it downstream. When the boat is going against the current, the water's speed is subtracted from the boat's speed. This makes sense, too: if the water is going fast enough, the boat will still be going downstream (a "negative" speed, because the boat would be going backwards at this point), because the water is more powerful than the boat!

	d	r	t
downstream	d = 3(b + 3)	b + 3	3
upstream	d = 4(b – 3)	b – 3	4
total	2d	---	7

Using "d = rt", the first row (of the revised table above) gives us:

d = 3(b + 3)

The second row gives us:

d = 4(b – 3)

Since these distances are the same, set them equal:

3(b + 3) = 4(b – 3)

Solve for b. Then solve for d.

In this case, we didn't need the "total" row.

• With the wind, an airplane travels 1120 miles in seven hours. Against the wind, however, it takes eight hours. Find the rate of the plane in still air and the velocity of the wind.
     

	d	r	t
tailwind	1120	p + w	7
headwind	1120	p – w	8
total	2240	---	15

(We probably won't need the "total" row.) Just as with the last problem, we are really dealing with two rates together. The plane, which changes direction, and the wind, which doesn't. So, when the plane turns around, the wind is no longer pushing the plane faster, but is instead pushing against the plane to slow it down.   Copyright © Elizabeth Stapel 2006-2008 All Rights Reserved

The first row gives us:

1120 = 7(p + w)

The second row gives us:

1120 = 8(p – w)

The temptation is to just set these equal, like we did with the last problem, but that just gives us:

7(p + w) = 8(p – w)

...which doesn't help much. We need to get rid of one of the variables.

Let's take that first equation:

1120 = 7(p + w)

Dividing by seven, we get:

160 = p + w

Then, subtracting w from either side, we get that p = 160 – w. Substitute "160 – w" for "p" in the second equation:

1120 = 8([160 – w] – w)

1120 = 8(160 – 2w)

...and solve for w. Then 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?
    

	d	r	t
air	d = 1100t	1100	t
steel	d = 16,500(t – 6)	16,500	t – 6
total	---	---	6

However long the sound took to travel through the air, it took six seconds less to propagate through the steel. (Since the speed through the steel is faster, then the time must be shorter.) Multiply the rate by the time to get the values for the distance column. (Once again, we didn't need the "total" row.)

Since the distances are the same, we 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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