For distance problems in which the bodies are moving away from each other or toward each other at right angles (for example, one heading east, the other north), the Pythagorean Theorem is used. This topic will be covered in the last chapter.
Some distance problems involve the
complication of the two bodies starting at different times. For these, you need
to compute the head start of the first one and let t represent the time they
are both moving (which is the same as the amount of time the second is moving).
Subtract the head start from the distance in question then proceed as if they
started at the same time.
Examples
A car
driving eastbound passes through an intersection at 6:00 at the rate of 30 mph.
Another car driving westbound passes through the same intersection ten minutes
later at the rate of 35 mph. When will the cars be 18 miles apart?
The eastbound driver has a 10-minute head
start. In 10 minutes (10/60 hours), that driver has traveled 30(10/60) = 5
miles. So when the westbound driver passes the intersection, there is already 5
miles between them, so the question is now ‘‘How long will it take for there to
be 18 – 5 = 13 miles between two bodies moving away from each other at the rate
of 30 + 35 = 65 mph?’’
Let t represent the number of hours after
the second car has passed the intersection.
In ⅕ of an
hour or 60(⅕) = 12 minutes, an additional 13 miles is between them. So 12
minutes after the second car passes the intersection, there will be a total of
18 miles between the cars. That is, at 6:22 the cars will be 18 miles apart.
Two
employees ride their bikes to work. At 10:00 one leaves work and rides
southward home at 9 mph. At 10:05 the other leaves work and rides home
northward at 8 mph. When will they be 5 miles apart?
The first employee has ridden 9(5/60) = ¾ miles
by the time the second employee has left. So we now need to see how long, after
10:05, it takes for an additional 5 – ¾ = 17/4 miles to be between them. Let t
represent the number of hours after 10:05. When both employees are riding, the
distance between them is increasing at the rate of 9 + 8 = 17 mph.
After ¼ hour,
or 15 minutes, they will be an additional 4 ¼ miles apart. That is, at 10:20,
the employees will be 5 miles apart.
Two boys are
1250 meters apart when one begins walking toward the other. If one walks at a
rate of 2 meters per second and the other, who starts walking toward the first
boy four minutes later, walks at the rate of 1.5 meters per second, how long
will it take for them to meet?
The boy with the head start has walked for 4(60)
= 240 seconds. (Because the rate is given in meters per second, all times will
be converted to seconds.) So, he has traveled 240(2) = 480 meters. At the time the
other boy begins walking, there remains 1250 – 480 = 770 meters to cover. When
the second boy begins to walk, they are moving toward one another at the rate
of 2 + 1.5 = 3.5 meters per second. The question becomes ‘‘How long will it
take a body moving 3.5 meters per second to travel 770 meters?’’
Let t represent the number of seconds the
second boy walks.
The boys
will meet 220 seconds, or 3 minutes 40 seconds, after the second boy starts
walking.
A plane
leaves City A towards City B at 9:10, flying at 200 mph. Another plane leaves
City B towards City A at 9:19, flying at 180 mph. If the cities are 790 miles
apart, when will the planes pass each other?
In 9 minutes the first plane has flown 200(9/60)
= 30 miles, so when the second plane takes off, there are 790 – 30 = 760 miles
between them. The planes are traveling towards each other at 200 + 180 = 380
mph.
Let t
represent the number of hours the second plane flies.
Two hours
after the second plane has left the planes will pass each other; that is, at 11:19
the planes will pass each other.
Practice
1. Two
joggers start jogging on a trail. One jogger heads north at the rate of 7 mph.
Eight minutes later, the other jogger begins at the same point and heads south
at the rate of 9 mph. When will they be two miles apart?
2. Two
boats head toward each other from opposite ends of a lake, which is six miles
wide. One boat left at 2:05 going 12 mph. The other boat left at 2:09 at a rate
of 14 mph. What time will they meet?
3. The
Smiths leave the Tulsa city limits, heading toward Dallas, at 6:05 driving 55
mph. The Hewitts leave Dallas and drive to Tulsa at 6:17, driving 65 mph. If
Dallas and Tulsa are 257 miles apart, when will they pass each other?
Solutions
1. The
first jogger had jogged 
miles when the other jogger began.
So there is 
miles left to cover. The distance between them
is growing at a rate of 7 + 9 = 16 mph. Let t represent the number of hours the
second jogger jogs.
In 1/15 of an hour, or 60(1/15) = 4 minutes after the second jogger began,
the joggers will be two miles apart.
2. The
first boat got a 12(4/60) = ⅘ mile head start. When the second boat leaves, there
remains 6 – ⅘ = 5 ⅕ = 26/5 miles between them. When the second boat leaves the
distance between them is decreasing at a rate of 12 + 14 = 26 mph. Let t
represent the number of hours th second boat travels.
In ⅕ hour, or ⅕ (60) = 20 minutes, after the second boat leaves, the boats
will meet. That is, at 2:29 both boats will meet.
3. The
Smiths have driven 55(12/60) = 11 miles outside of Tulsa by the time the
Hewitts have left Dallas. So, when the Hewitts leave Dallas, there are 257 – 11
= 246 miles between the Smiths and Hewitts. When the Hewitts leave Dallas, the
distance between them is decreasing at the rate of 55 + 65 = 120 mph. Let t
represent the number of hours after the Hewitts have left Dallas.
2 1/20 hours, or 2 hours 3 minutes (60 – 1/20 = 3), after the Hewitts leave Dallas, the Smiths and Hewitts will pass each other. In other words, at 8:20, the Smiths and Hewitts will pass each other.
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