4. Linda drove to Eureka. She drove 240 miles. On the way back,
she was having car troubles and she had to drive 20 mi/hr slower. If
it took her 2 hours longer to get back, how fast was she driving on
the trip there, and on the trip back?
The first step in solving word problems is to define the
unknown. In this problem, there are two unknowns, Linda's speed
driving to Eureka and her speed driving back. When it says that she
was driving 20 mi/hr slower on the way back, that gives us the
speed on the way back as a function of her speed on the way there.
Your author suggests that we make use of d = rt tables. We
know that the distance both there and back is 240 miles. We also
have expressions for the rate. That gives us this much of the table.
Once we have two columns for the table filled in we can use the
appropriate form of the d = rt equation to fill ion the third column.
Since the unfilled column is the one for time, we need to use the
to fill in the time. That enables us to fill in the table as follows.
Now the sentence "It took her 2 hours longer to get back." can
be translated as
Since we now have an equation, we can take the steps to solve
it. First, clear denominators. The smallest common denominator is
x(x-20), so multiply both sides by that.
The x(x-20) needs to be distributed through the two terms on
This will clear all of our denominators. We assemble the
and first notice that we can subtract 240x from both sides of the
equation. That leaves us with the quadratic
We already have a 0 on one side, so we are ready to see if it
factors. First notice that there is a common factor of 2 in all the terms
which can be factored out.
That makes it a little easier to factor the expression inside. It will
factor because there are factors of 2400 which have a difference of
20: 40 and 60.
This product to be equal to 0, if and only if one of these factors
is equal to 0. The factor of 2 is never equal to 0, so one of the other
factors must be equal to zero. Set them equal 0
If she was driving at 60 mi/hr going to Eureka, then it would
have taken her 4 hours to get there. If she drove 20 mi/hr slower on
the way back, then she was driving at only 40 mi/hr, and it would
have taken her 6 hours, which is 2 hours longer than it took her to
But if she was driving at -40 mi/hr, then the computation for
the time it would take would be
This could be interpreted as meaning that if she was driving at
-40mi/hr, then she was driving 40 mi/hr backward and at that rate
she would have been in Eureka 6 hours ago. 20 mi/hr slower than
-40 mi/hr would be - 60 mi/hr and at that rate she would have been
in Eureka only 4 hours ago which is 2 more hours than -6 hours.
However, many authors will simply reject negative velocities as
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