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Marabyn
Posted May 20,
2001 as a Math Forum EPoW
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Introduction: In this problem,
students use the simulation to change
the
distance Marabyn rides the bus and
the distance
she walks to fit constraints about
those distances.
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Where’s the Math: This
problem deals primarily with the
distance formula, d = rt. For one
question, students also formulate
their own equation to represent
additional constraints. Using the
slope of lines on the graph, students
can also calculate Marabyn’s
walking and riding speeds. With
this data, an inequality can be
formulated which represents the
minimum and maximum walking distances.
Standards: Algebra,
measurement
Role of Components: SimCalc
Graph is used to display functions
representing the motion of
the actors: the bus and Marabyn.
An invisible FunctionEntry component
specifies the function. SimCalc
World shows the motion of the bus
and
Marabyn. ESCOTClock is used to animate
the actors. Standard Swing
components are used, for example,
JButtons
are used for the "Try" and "Predict"
buttons; a JSlider presents a list
of
time intervals; JComboBoxes
allow
the user to set display options
for the data table. A subclass of
the
Swing JTable
displays
the results of each prediction.
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Try the applet!
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Sample submitted solutions:
From: Esther, age 14
School: Taipei American School, Taipei,
Taiwan
1. How far (distance) can Marabyn travel
on the bus if she wants to walk at least
10 minutes but not more than 2 kilometers
to get home? How long (time) will she walk?
Marabyn can travel 7 km on the bus, and
walk for 2 km. It would take her 20 minutes.
I solved this problem by finding the maximum
and minimum "bus travel distance".
Because Marabyn's walking speed is 0.1 km/min,
it would also equal 10 min = 1 km. In that
case, the maximum bus ride she gets is 8
km, because she would like to walk for at
least 10 minutes which equals to 1 km. 9
km total - 8 km bus ride = 1 km walking
Because she doesn't want to walk more than
2 km, her bus ride would be:
9 km total - 2 km walk = 7 km bus ride.
As long as she rides the bus for no longer
than 8 km no less than 7 km, walks no farther
than 2 km no less than 1 km, walks no longer
than 20 minutes no less than 10 minutes,
she would be fine.
2. List two more bus-travel distances and
walking times that also get Marabyn home.
Marabyn can also travel:
- 7.5 km on the bus, walk 1.5 km for 15
minutes
- 8 km on the bus, walk 1 km for 10 minutes
3. Write an equation that uses Marabyn's
walking speed of 0.1 km/min and the variables
"t" for walking time and "d"
for bus-travel distance to express the distance
of Marabyn's journey home.
0.1t + d = 9 (the "9" is the "distance
of Marabyn's journey home)
From: Andy, age 14
School: Taipei American School, Taipei,
Taiwan
1. How far (distance) can Marabyn travel
on the bus if she wants to walk at least
10 minutes but not more than 2 kilometers
to get home? How long (time) will she walk?
She can travel 8 kilometers on the bus if
she wants to walk at least 10 minutes but
no more than 2 kilometers to get home. She
will walk for 10 minutes. I got this answer
by using the java applet and calculating.
Since we want to know how far she can travel
on the bus if she wants to walk for at least
10 minutes, we want her to walk the least
possible, so I set the applet to 10 minutes
for her walking time. Then we need to know
how far she travels in 10 minutes. Using
the applet, it shows that the rate of her
walking is 1 kilometer per 10 minutes. So
then we can set the traveled distance of
the bus to 8 kilometers, and the walking
time of Marabyn to 10 minutes, and once
we predict, it shows that it is correct!
2. List two more bus-travel distances and
walking times that also get Marabyn home.
Marabyn will also get home if the bus travels
7.5 kilometers, and she walks for 15 minutes.
When the bus travels for 7 kilometers and
she walks for 20 minutes, she will also
get home. I got these answers because Marabyn
walks 1 kilometer every 10 minutes. This
means that she walks 0.5 kilometers every
5 minutes. Using this rate, we know that
every 0.5 kilometer added or subtracted
from the bus-travel distances, we can just
add or subtract a 5 minute walking time,
and she would still get home.
3. Write an equation that uses Marabyn's
walking speed of 0.1 km/min and the variables
"t" for walking time and "d"
for bus-travel distance to express the distance
of Marabyn's journey home.
I will name the distance of Marabyn's journey
home x. First, I started out with this,
because she will only walk the distance
of the difference of the total distance
of her school to her house and the bus-traveled
distance. Then we multiply it by 10, since
the rate of her walk speed is 0.1 km/min.
t = (x - d) * 10
Then we use the distributive property and
multiply both x and d by 10.
t = 10x - 10d
Then we take the variable and switch it
to the left side. We also take t and put
it to the right side, since the variable
must be alone.
-10x = -10d - t
Since x, the variable, cannot be negative,
we multiply both sides of the equation by
-1 to turn it positive.
10x = 10d + t
And then we divide both sides by 10, because
we want the variable to be standing by itself.
x = d + (t/10)
And that is the equation we end up with.
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Reflections:
I was impressed by how the students continued
to think about the problem and send in their
revised answers. In the end, 10 out of 27
students were given credit for their answers.
Everyone who revised, but did not receive
credit for all three questions, showed improvement
with each try.
Some of the problems students had included
giving the same answer for question 1 and
2. Students showed that they understood
that if Marabyn rode the bus to the 8 kilometer
mark it would mean that she would walk 1
kilometer. They also showed that they knew
that she would walk for 10 minutes. I'm
not sure why, but students gave this answer
for question 1 and then gave it again for
question 2.
To have correct responses for question
1 and 2 students needed to provide three
specific answers that met the conditions
given in the problem. They needed to understand
that the total distance was 9 km, and combine
that with the thought that to walk "no
more than 2 km." meant that Marabyn
would get off the bus "not before"
the 7 km. mark.
Next, if students took the time into account
they understood that Marabyn would get off
the bus "not after" the 8 km.
mark. After establishing the range of distances,
it should have been clear that the answers
had to be 7.0 km, 7.5 km. and 8.0 km. with
the accompanying times of 20 minutes, 15
minutes, and 10 minutes. Students who originally
gave answers including 6.5 km. or 8.5 km.
quickly revised their thinking once the
"at least 10 minutes" and "not
more than 2 km." parts of the problem
were pointed out.
To figure out the answer to question 3
it helped to first think of this familiar
equation:
d = r x t
where d = distance, r = rate, and t = time.
Using the information in the problem would
give: d = 0.1 x t. Then, since Marabyn was
traveling a total distance of 9 km. and
she was riding the bus for a part of that
distance, the following equation is derived:
9 - d = 0.1 x t
where t = time she walked and d = distance
she rode on the bus.
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