Distance Problems
Another common word problem type is the distance problem, sometimes called the uniform rate problem. The underlying formula is d = rt (distancemequals rate times time). From d = rt, we get two other relationships: r = d/t and t = d/r. These problems come in many forms: two bodies traveling in opposite directions, two bodies traveling in the same direction, two bodies traveling away from each other or toward each other at right angles. Sometimes the bodies leave at the same time, sometimes one gets a head start. Usually they are traveling at different rates, or speeds. As in all applied problems, the units of measure must be consistent throughout the problem. For instance, if your rates are given to you in miles per hour and your time is given in minutes, you should convert minutes to hours. You could convert miles per hour into miles per minute, but this would be awkward.
Examples
When the
bodies move in the same direction, the rate at which the distance between them
is changing is the difference between their rates.
A cyclist starts at a certain point and
rides at a rate of 10 mph. Twelve minutes later, another cyclist starts from
the same point in the same direction and rides at 16 mph. How long will it take
for the second cyclist to catch up with the first?
When the second cyclist begins, the first
has traveled 10(12/60) = 2 miles. The ‘‘10’’ is the rate
and the ‘’12/60’’ is the twelve minutes converted to hours. Because the
cyclists are moving in the same direction, the rate at which the distance
between them is shrinking is 16 – 10 = 6 mph. Then, the question boils down to
‘‘How long will it take for something traveling 6 mph to cover 2 miles?’’
Let t
represent the number of hours the second cyclist is traveling.
It will take
the second cyclist ⅓ of an hour or 20 minutes to catch up the first cyclist.
A car passes an intersection heading north
at 40 mph. Another car passes the same intersection 15 minutes later heading
north traveling at 45 mph. How long will it take for the second car to overtake
the first?
In 15 minutes, the first car has traveled 40(15/60)
= 10 miles. The second car is gaining on the first at a rate of 45 – 40 = 5
mph. So the question becomes ‘‘How long will it take a body traveling 5 mph to
cover 10 miles?’’
Let t
represent the number of hours the second car has traveled after passing the
intersection.
It will take
the second car two hours to overtake the first.
Practice
1. Lori
starts jogging from a certain point and runs 5 mph. Jeffrey jogs from the same
point 15 minutes later at a rate of 8 mph. How long will it take Jeffrey to
catch up to Lori?
2. A
truck driving east at 50 mph passes a certain mile marker. A motorcyclist also
driving east passes that same mile marker 45 minutes later. If the motorcyclist
is driving 65 mph, how long will it take for the motorcyclist to pass the
truck?
Solutions
1. Lori
has jogged
miles before Jeffrey began. Jeffrey is catching up to Lori at the rate of
8 – 5 = 3 mph. How long will it take a body traveling 3 mph to cover 5/4 miles?
Let
t represent the number of hours Jeffrey jogs.
Jeffrey will catch up to Lori in 5/12 hours or (5/12)(60) = 25 minutes.
2. The
truck traveled 50(45/60) = 752 miles. The motorcyclist is catching up to the
truck at a rate of 65 – 50 = 15 mph. How long will it take a body moving at a
rate of 15 mph to cover 752 miles?
Let
t represent the number of hours the motorcyclist has been driving since passing
the mile marker.
The motorcyclist will overtake the truck in 2 ½ hours or 2 hours 30 minutes.
“Sumber Informasi”
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