Mathematics: Modeling Our World
Unit 7: MOTION
ARE WE THERE YET?
SUPPLEMENTAL ACTIVITY
S7.1
Imagine that you are behind the wheel of a car and that you
are stopped at a red light (Location 0). There are traffic lights
at every intersection on this straight city street. Imagine your
drive for the next three blocks. Decide how far apart the
traffic lights are, how fast you’ll drive, and how many traffic
lights you’ll have to wait through.
1.Draw a distance-versus-time graph for your trip. Assume that distance is
measured from the first traffic light (Location 0). Describe in words how your
axes are labeled and what your graph indicates about your velocity during
your drive.
2.What was your average velocity as you traveled from the first red light
(Location 0) to the intersection of the third block? Approximately what was
your fastest velocity?
3. You’ve had some time to imagine how the velocity of a real car might change
during a three-block drive. During an actual drive, you could look down at the
speedometer to check how fast you were going. But what happens when you
take a battery-operated toy car and turn the switch on? How fast does the toy
car move? Does the car move at a constant velocity? How could you find out?
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LOCO-MOTION
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Railroad companies have the worrisome task of balancing the
demands of safety and ensuring efficient and predictable
train scheduling. Computerized network control systems
connected to dispatch centers help railway companies handle
this task. This system allows dispatchers to track the progress
of each train and communicate with train engineers.
1. Figure 1 displays a portion of the network
representing railway track. The forks at either
end of the straight track represent switches
from one track to another. The dots indicate
the location of a train traveling a section of
Figure 1. Portion of track
track between the two switches. The
network showing train’s location.
dispatcher observes the train at t = 0 as it
crosses the first switch in the track. The dot corresponding to t = 5 indicates the
train’s location 5 minutes later. (The computer display will represent the train
with a single dot and will update the train’s location as it travels down the
track by repositioning the dot.)
a) Indicate on Figure 1 the location of the train at t = 15 and t = 25 minutes.
Although you cannot be sure of the behavior of the train, you may make a
reasonable assumption. What is the assumption?
b) Estimate how long it will take the train to travel between the two switches
in the track.
c) Suppose that the distance between the two switches is 48 miles. What is the
approximate velocity of the train in miles per hour?
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d)Draw a distance-versus-time graph for the train’s motion as it travels
between the two switches. What equation describes the relationship between
the train’s distance from the first switch and the time elapsed since the train
crossed the first switch?
e) Use your equation to estimate how far the train has traveled 27.5 minutes
after it crosses the first switch.
2. Figure 2 represents another train traveling the same section of track at a later
time.
Figure 2. Portion of track network showing another train’s location.
a) Predict the location of the train at t = 20 minutes and mark this location on
Figure 2 with a dot. Defend your prediction.
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b) Draw a graph representing the relationship between distance and time for
this train. Why does this graph have a different shape than the one that you
drew for Item 1(d)?
3. Figure 3 presents a graphical record of two trains traveling the section of track
between the switches. Tell the story of the two trains. Include the following
details in your story: How fast are the trains traveling? How far apart are they?
From the time of the initial observation, how long will it take for each train to
reach the second switch? Will the trains ever collide?
Figure 3. Distance-versus-time graphs for two trains.
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Unit 7: MOTION
SIMULATING A NEAR-COLLISION STUNT
SUPPLEMENTAL ACTIVITY
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Now it is time to apply what you have learned about motion
along a line to the problem of stunt design. Here’s the stunt:
Two cars are driving down roadways that intersect at right
angles. The key to this stunt is to cause some anxiety for the
spectators—the vehicles should pass through the intersection
as closely as possible without colliding.
To enter distance-time data for a toy car and toy truck, you will need to run the
program STUNT. This program will load three lists: the times go to L1, car’s
distances to L2, and truck’s distances to L3.
DESCRIPTIONS OF THE VEHICLES
1.The first vehicle, a red fire chief’s car, is 5 inches wide and 12 inches long. The
second vehicle, a monster truck, is 7 inches wide and 8 inches long. Both
vehicles are battery-operated. To start the car, you press a button. The message
“We’re on our way” plays and then the fire chief’s car starts moving. The truck
has an on/off switch and begins moving as soon as the switch is turned on.
a) For each vehicle, write an equation modeling the distance-versus-time data
collected by the motion detector. Justify your equations. Include sketches of
the graphs produced by the motion detector readings as part of your
justification. It may be helpful to edit portions of your data when
determining the models.
b) What is your best estimate for how fast each vehicle moves?
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SIMULATING A NEAR-COLLISION STUNT
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THE STUNT DESIGN
Assume that the truck will be traveling east and the car north along these
intersecting roadways (see Figure 1). Instead of staging this stunt with toy vehicles,
you will simulate the stunt using your calculator. All you have to do is determine
the positions F (the distance from the front of the truck to the intersection) and E
(the distance from the front of the car to the intersection).
Figure 1. The intersection.
2.Determine where to position the car and truck by specifying the distances E
and F in feet. Remember, the key to this stunt is to cause some anxiety for the
spectators—the vehicles should pass at the intersection as closely as possible
without colliding. Explain how you determined the values for E and F.
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SIMULATING A NEAR-COLLISION STUNT
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THE SIMULATION
You will run the program DRIVE and test your
design. Before running the program, check that mode
settings are the default settings (see Figure 2).
Note: Each time you run DRIVE, the program turns
your axes off. To restore the axes on a TI-83, press
Figure 2. Mode screen showing default settings.
2nd FORMAT, highlight AxesOn and press ENTER.
(FORMAT is an option after pressing WINDOW on the TI-82.)
Then set a window appropriate for this stunt. Remember, the intersection is the
point (0, 0). Select a window that will allow you to see all the action. For example,
you might set Xmin = –F and Ymin = –E. Then set Xmax and Ymax so that the
intersection is visible.
Now run the program. Here’s the information that you’ll be asked to provide:
1.The dimensions (ft.) of the vehicles. CAR 1 refers to the north-bound vehicle
(car) and CAR 2 to the east-bound vehicle (truck).
CAR 1 WIDTH
CAR 1 LENGTH
CAR 2 WIDTH
CAR 2 LENGTH
2.Positions of the vehicles (distance (ft.) from the intersection to the front of the
vehicle).
POS. CAR 1: value of E
POS. CAR 2: value of F
3. Velocity of the vehicles (ft./sec.).
VEL. CAR 1
VEL. CAR 2
4. Time
STOP: Number of seconds you want the action to run.
TIME INCREMENT: Time increment between views of vehicles’ positions.
START VIEW: Enter 0 to view the action from the start. However, you may
specify a larger number (less than the number that you entered for STOP) if
you want to pick up the action closer to the intersection.
After entering the information above, you should see two rectangles on your
screen. These represent the two vehicles. The motion will advance by one TIME
INCREMENT each time you press ENTER.
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3.Evaluate the success of your stunt. Watch as the vehicles near the intersection.
Do your vehicles avoid collision? Do they pass very close as they cross the
intersection? (If you had trouble telling whether or not the vehicles collided,
change to a smaller viewing window—perhaps [-2, 2] x [-2,2]—select a smaller
value for TIME INCREMENT, and perhaps increase the value you enter for
START VIEW. Then run the stunt again.) If your vehicles collided, or there
wasn’t a near collision at the intersection, rework your design and try again.
4.Suppose the north-bound toy vehicle travels at 2ft./sec. and the east-bound
vehicle at 2.6 ft./sec. The north-bound vehicle is 6 inches wide and 18 inches
long. The east-bound vehicle is 4 inches wide and 6 inches long.
a) Determine possible starting positions, E and F, for your vehicles. Explain
how you determined these values.
b) Now, test your stunt by running the simulation program DRIVE. How did
you do? If your stunt didn’t work, rework the mathematics in part (a) and
try again.
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Unit 7: MOTION
UPSIDE DOWN
SUPPLEMENTAL ACTIVITY
S7.4
The suggested method of collecting data in Lesson 2 was to
use a protective frame around an upward-pointing motion
detector. An alternative, mounting the detector pointing
down, was also mentioned. Try this alternative method of
gathering data. Then discuss how these data differ from those
obtained from a detector lying on the floor. Create a
transformation that converts the data of the new method to
actual heights, as were obtained in Lesson 2. Then try your
transformation with real data from BALLDROP.
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Unit 7: MOTION
BLOWING IN THE WIND
SUPPLEMENTAL ACTIVITY
S7.5
EXPERIMENTAL SET-UP
1.Set a hair dryer to its highest speed. Turn the heating element off if possible.
2.Point the hair dryer straight up.
3.Place a ping-pong ball about one foot above the dryer in the center of the
column of air created by the dryer. Release the ball.
4.If the ball floats on the air, measure how far above the nozzle it floats. If it
blows out of the cone of moving air, try placing it in the cone again until it does
float.
5.Replace the ping-pong ball with a ball of the same size but with a different
weight (try a decorative styrofoam ball), and float it in the cone of air. Measure
how far above the nozzle it floats.
6.Remove the ball from the cone of air. Place your open hand in the air cone and
move it slowly up and down in front of the nozzle.
Answer the following questions based on your hair-dryer experiment.
1.Which ball floats the highest?
2.Did the air exert the same force when your hand was close to the nozzle as
when it was far away? If not, how did it vary?
3.Comment on the connection between what you felt when you put your hand
over the hair dryer and what you observed with the ping-pong or styrofoam
ball. Explain why you were able to float the ping-pong ball (and thus overcome
the force of gravity).
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Unit 7: MOTION
AIR RESISTANCE
SUPPLEMENTAL ACTIVITY
S7.6
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PART 1: PREDICTION
Consider the following thought experiment. Suppose you have several objects to
drop:
• the book you dropped in class
• a book having the same size pages but twice as thick
• a sheet of cardboard the same size as the covers of the books
• another piece of cardboard (half the length and width) with pennies taped to its
back so that it weighs exactly one-fourth as much as the book that you dropped
in class.
If all these objects are dropped from the same height at the same time, what is the
order in which they will land? Why?
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PART 2: DATA COLLECTION
For this experiment, you will need three objects that have different weights but
present the same target area for the motion detector. You will also need two objects
that have different target areas but whose weights per square inch of target area are
the same. Use your imagination in finding materials that meet these criteria.
(For example, if your teacher has some old paperback workbooks, you might put
one, two, and then four together with a loose rubberband. Or you might try coffee
filters and put two, four, and then eight inside one another. Use your imagination
for finding materials to use.)
The target area is the area of the object that the motion detector “sees.” On a book
it would be the area of the cover, if the book were dropped flat. On a ball, it would
be the area of a cross-section of the ball taken at its widest part.
The motion detector set-up is the same as in Activity 5 (Lesson 2). Remember to
protect the motion detector by encasing it in a frame.
Drop each object, one at a time, from directly over the motion detector. Each object
should be dropped from approximately the same height. (A height of about five
feet works well.) Save the data from each object’s drop; separate calculators or a
computer will be needed for this.
PART 3: ANALYSIS
1. Write an equation for the piece of each object’s height-versus-time graph that
corresponds to the actual fall. To make comparisons easier, put each equation in
translated form, with t = 0 as the release time.
2.Are the equations for the heights of the falling objects roughly the same?
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3.Use your equations to estimate the velocity of each object after 0.4 seconds. (If
one of the objects hit the floor before t = 0.4, compare the speeds at the last time
for which the fastest object was still completely above the floor.) Which object
was moving the fastest?
4.Find the ratio of each object’s weight to its cross-sectional (target) area. Use
these ratios as x-coordinates for data points; for y-coordinates use the velocities
you found in Item 3. Plot these points. Discuss what the graph shows. What is
the unit of measurement for the weight to cross-sectional area ratio?
5.Look back at the points that correspond to your two objects that had the same
weight-to-area ratio. Discuss what you see.
6.How can we slow a falling object? How does this relate to parachuting out of
planes?
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7. Take the heaviest object you dropped and make a parachute large enough to
slow its fall to be equal to the slowest falling object. Test your parachute with
an actual drop.
8.If the force pulling the object down is gravity, what force is holding the object
up and thus slowing it down?
9.Just for fun: Watch a movie on sky diving and then read about terminal
velocity.
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COMPLETELY SQUARE
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S7.7
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In Course 1, Unit 8, Testing 1, 2, 3, you learned that a parabola
is the graph of a quadratic equation and that such equations
can be written either in standard form, y = ax2 + bx+c, or in
vertex form, y = a(x-h)2 + k. What you may have forgotten
was how to transform the first form into the second.
1.Recall that an expression in the form x2 + 2bx + b2 can be written in factored
form as (x + b)2. (Note that the value for b can be positive, negative, or zero.)
Quadratic expressions that can be written in this form are called perfect
squares. Identify which of the expressions below are perfect squares. Write each
expression that is a perfect square in factored form.
a) x2 – 6x + 9
b) x2 – 12x + 24
c) x2 + 8x + 16
You can use a method called completing the square to convert quadratic models
such as y = –2x2 + 12x – 13 to the vertex form. Here’s how:
Step 1: Group the terms containing the variable x:
(-2x2 + 12x) – 13
Step 2: Factor the coefficient of the x2 term:
-2(x2 – 6x) – 13
Step 3: Complete the square: Take half of the –6 and then square it to get +9. Add
this amount inside the parentheses. Notice that, since 9 is inside the
parentheses, you have effectively added –2(9) to the expression. So, to
balance what you have added, you need to subtract –2(9). (Subtracting –18
is the same as adding +18.)
-2(x2 – 6x + 9) –13 + 18
Step 4: Rewrite the contents of the parentheses (the perfect square) in factored
form. Combine the constants.
–2(x – 3)2 + 5
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Now you have it: y = –2x2 + 12x – 13 can be re-expressed in the form
y = –2(x – 3)2 + 5.
2.Here’s an example for you to try.
a) Convert y = –5x2 + 10x + 5 to vertex form by completing the square.
b) How can you use your equation from part (a) to determine the maximum
value for y? What x-value will produce the maximum y-value? Explain.
c) You can also find the maximum y-value using CALC/maximum on a TI-82
or TI-83. Graph y = –5x2 + 10x + 5 in a viewing window that clearly shows
the parabola’s peak. Press 2nd CALC 4 for maximum. Your calculator will
then ask you to specify a left bound. If necessary, press the left arrow key to
position the cursor to the left of the parabola’s peak and press ENTER. Then
press the right arrow key and position the cursor to the right of the peak
and press ENTER. Last, specify a guess by positioning the cursor at the top
of the peak (in other words, as close to the vertex as possible) and press
ENTER. Do your calculator results confirm your answer to part (b)? Do you
get the exact answer to part (b) using your CALC/maximum?
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3.Recall that the x-coordinate of the vertex of the parabola y = ax2 + bx + c can be
determined by x = –b/2a.
a) Use this formula above to determine the exact coordinates of the vertex of
y = 2x2 + 16x + 27. How could you use the knowledge of the exact
coordinates of the vertex to convert y = 2x2 + 16x + 27 into vertex form
without completing the square?
b) Quadratic equations are easy to solve symbolically when the quadratic is
written in vertex form. Show how you could solve 2x2 + 16x + 27 = 13 by
converting the quadratic on the left to vertex form.
c) You can also approximate the solution to the equation in part (b) using the
CALC/intersect feature on your calculator. To do that enter y = 2x2 + 16x + 27
and y = 13 in your “Y =” list. Then press 2nd CALC 5 for intersect. Your
calculator will ask you to specify which is the first curve and which is the
second. If you only have two functions entered in the “Y =” list, press
ENTER twice to answer these questions. Next, use the right or left arrow
keys to move the cursor close to one of the points of intersection for your
guess, and then press ENTER. If you don’t get the exact answer, the
approximation that your calculator gives should be very close. Repeat the
process to get the other point of intersection. Do your calculator results
confirm your results in part (b)?
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4.Answer the Items below using the equation y = –2x2 + 12x – 10. Choose
whatever method that you want.
a) Write the quadratic in vertex form.
b) What is the maximum y-value on this parabola? What value for x produces
this value?
c) Find the approximate solution to the equation –2x2 + 12x – 10 = –2.
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Unit 7: MOTION
SOLVING SYSTEMS OF EQUATIONS
SUPPLEMENTAL ACTIVITY
S7.8
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You have seen systems of equations in Course 1, Unit 5,
Animation, and Course 2, Unit 2, Strategies. In solving a
system of equations such as
y = 2x + 3
y = –3x + 8
you are trying to find a single set of values for (x, y) that satisfies both equations.
Satisfying both equations means that if the point (x, y) is substituted into the two
equations, both equations will be true.
The point (1, 5) is a solution to the above system because
5 = 2(1) + 3, and
5 = –3(1) + 8.
But the point (2, 7) is not a solution to that system:
7 = 2(2) + 3, but
7 ≠ –3(2) + 8.
The point (2, 7) satisfies one of the equations but does not satisfy the other. Thus it
is not a solution to the system.
The above discussion shows how to check that (1, 5) is a solution. You have seen
two methods for finding such solutions.
The first is the method of substitution, which you first saw in Course 1, Unit 5,
Animation, in the context of looking for collisions. The basic idea in that unit was
that for a collision to occur, the y-coordinates must be the same. In the system
shown above, that would mean:
2x + 3 = –3x + 8.
This equation can then be solved to get x = 1. Then, using one of the original
equations above with this value of x gives, for example, y = 2(1) + 3, so y = 5.
The second method, weighted averages (or weighted sums), came up in Unit 2,
Strategies. There, the basic idea was to find “weights” (multipliers) for the
equations so that when they were averaged (or added) one of the variables would
drop out—that is, have a 0 coefficient. In the system here, using weights of 1 and –1
will cause the ys to drop out. Then,
(y = 2x + 3)(1)
(y = –3x + 8)(–1)
becomes
y = 2x + 3
–y = 3x – 8
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SOLVING SYSTEMS OF EQUATIONS
S7.8
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and adding the equations in the new system gives
0 = 5x – 5.
This also has the solution x = 1, and again, substitution back into the original
equations gives y = 5.
These methods for solving systems of two equations with two variables will also
allow you to solve larger systems. As an example, consider the following system.
2x + 3y – 4z = 25
x – 4y + z = –2
–5x + y – 3z = –15
If you prefer the method of substitution, rewrite each of your equations with the
same terms isolated, then substitute one equation into another. Doing this twice
will give you two equations with only two variables—a two-by-two system, which
you know how to solve. The disadvantage to this method is that it usually involves
fractional expressions. For example, isolating the ys in this system gives:
y = (25 – 2x + 4z)/3
y = (x + z + 2)/4
y = 5x + 3z – 15
Then,
(25 – 2x + 4z)/3 = 5x + 3z – 15 and
(x + z + 2)/4 = 5x + 3z – 15.
Finish the solution from here. You should get x = 5, y = 1, and z = –3.
If you prefer using weighted averages, note that using weights of 1 and 4 with the
first pair of equations eliminates the z terms, and using 3 and 1 with the second
pair also eliminates z. So you end up with two equations having only xs and ys.
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PRACTICE
1.Solve the system given in the first example of this review reading by finding
weights to eliminate the x terms instead of the y terms.
2.Solve the system 2x – 3y = 11 and –1x + 5y = 14.
3.Solve the system y = 3x + 5 and 4x – 2y = –16.
4.Solve the system
x+y+z=5
5x – 2z = –18
y = 2x + 3z – 5
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JEFF LATTIMORE’S “THE LEAP FOR LIFE”
SUPPLEMENTAL ACTIVITY
S7.9
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Recall the description of Jeff’s specialty stunt, “The Leap for
Life.” When we last saw Jeff, he was stranded atop his eightfoot stool—or so he thought, for a prankster had cut it down
to seven feet. Next, Jeff notices a car speeding toward him. A
moment before the car crashes into his stool, Jeff jumps. His stool is snapped out
from under him as the car whizzes beneath. The question is whether the stunt is
successful and Jeff lands safely on the ground, or the stunt goes tragically wrong
and Jeff smashes into the top of the car.
Assume that the car approaching Jeff is a Chevy wagon, 5 feet high and 17 feet
long. Over the bumpy fair grounds, the car reaches a speed of 45 m.p.h. as it
crashes into Jeff’s stool. Jeff times his jump so that he leaves the stool when the car
is approximately three feet away. Jeff jumps so that he reaches a maximum height
of approximately two feet above the stool.
1.Suppose that Jeff was on the ground and he jumped so that his maximum
height was two feet above the ground. Specify the form of a distance-versustime model for this jump. Use your model to determine the initial velocity
needed to jump two feet off the ground.
2.Based on the assumptions stated above, write a model that describes Jeff’s
height with respect to time when he jumps from his eight-foot stool into the air.
Let t = 0 represent the instant that he jumps. Explain how you determined your
model. State any additional assumptions that you make.
3.Determine how long it takes for Jeff to land on the ground. How long does it
take for him to reach the height of the top of the car?
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4.If Jeff starts his jump when the car is three feet in front of him, how many feet
away will the back of the car be when Jeff drops back to a height of five feet?
(You’ll need to convert the car’s velocity into feet per second.) Explain how you
know that Jeff will land safely.
5.Now, change the height of Jeff’s stool to seven feet and leave everything else
the same. Does Jeff land safely? Justify your answer mathematically.
6.Return Jeff’s stool to its original eight-foot height. What is the slowest that the
car can be driven for a successful outcome to the stunt? Report your answer in
miles per hour.
7.Does Jeff really have to jump in order to land safely? Repeat Item 6 for the “no
jump” possibility.
8.Go back to 45 m.p.h. Three feet away seems pretty close for the jump. Based on
the calculations you did with this set-up, give Jeff some advice on how he
could time his jump better.
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STRONGEST PERSON CONTEST
SUPPLEMENTAL ACTIVITY
S7.10
PART I: DISCUSSION
In this activity, you will determine the strongest person by
staging a basketball throwing contest.
1.First, plan how you will use the data you generate to decide which toss
represents the greatest strength. Remember, the motion detector only has a
range of about 25 feet. Discuss possible plans with your group, reach an
agreement on what you believe is best, and write your procedure.
PART II: DATA COLLECTION
Set up the motion detector apparatus. Then place the motion detector on a box or
flat surface about two feet off the ground so it will make measurements of objects
tossed up directly above it. After you execute BALLDROP, don’t press the
TRIGGER button until the ball is actually being tossed.
• The contestant should hold the ball over the motion detector.
• Whoever is in control of the TI-CBL will say “Go” as they push the TRIGGER
key on the CBL.
• On the command “Go,” the contestant will toss the ball as high as they can
without letting their feet leave the ground.
Let the contest begin!
2.Collect the data for each contestant.
PART III: ANALYSIS
3.Use the method you outlined in Part I to decide on the strongest class member.
Compare the results of your group’s method to methods used by other groups.
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This activity is a replacement for Parts 3 and 4 of Activity 13.
PART 3: DATA COLLECTION
1. a)Determine the acceleration due to gravity. To do this, run the ball-drop
experiment as you did in Lesson 2, this time with your car (or a racquetball,
a golf ball, etc.). Drop the car (ball) away from the motion detector so the
detector is not damaged. Remember, though, to change the sign of the
computed acceleration to negative for use with the ramp.
b) Measure and record the height from the floor to the jump point on the ramp.
Be sure to record units, too.
Take-off height: ___________
2.Find the velocity at take-off:
a) Measure the length of the car (or ball) with a pair of calipers. Record this
number.
b) Get time data for take-off velocity:
Method 1: For use with a light probe.
• Attach the light probe to a CBL that is attached to a TI-82 or TI-83. Run
the program GATE.
• Release the car from the top of the ramp. As the car rolls off the ramp, it
will break the flashlight beam, and the CBL will record the amount of
time the light was blocked.
Method 2: For use with a Vernier photogate.
• Attach the photogate to Channel 1 of a CBL and then attach the CBL to a
TI-82 or TI-83.
• Run the program TIMER.
• Follow the instructions of the program. Release the car from the top ramp.
As the car rolls off the ramp, it will break the photogate beam, and the
CBL will record the amount of time the beam was blocked.
For each run, mark the floor where the car hits. (The carbon paper may be
placed carbon side down in the landing area to mark the floor for you if
you have a hard floor.)
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Run three such trials. If the car does not land in approximately the same
place each run, you need to refine your release methods. If it is not hitting
in the same area, continue practicing your release until you are consistent.
Then run your three trials for the photogate data.
Record here the photogate data for each run.
Run 1: ____________
Run 2: ____________ Run 3: ____________
c) Find the direction of the flight: You marked the floor every time the car hit
the ground. Find the middle of the three points. Place a mark there. Mark a
line from the point on the floor directly below the jump point to the landing
point determined by these trial jumps. (A chalk line or line of masking tape
works well for this.) This line will be the line on which you will place your
landing ramp or can.
PART 4: ANALYSIS—FORMULATING THE MODEL
3.What is the acceleration due to gravity for your car?
4. a)Use your photogate data and the measured length of the car to determine
the velocity of the car as it takes off.
b) For which component of the car’s motion is this velocity important?
c) What is the initial velocity for the other component direction?
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5.Use your work in Items 3 and 4 to write the component equations for the
motion of the car after it leaves the ramp.
6.Graph each component equation as a distance-versus-time graph. Then use the
parametric mode of your calculator to sketch a graph of the actual path of the
car, height versus distance. Discuss what the shapes of the three graphs tell you
about the motion of the car. Select some particular instant during the motion,
locate that instant on each graph, and interpret the graphs.
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A LITTLE MATH PRACTICE BEFORE THE BIG EVENT
SUPPLEMENTAL ACTIVITY
S7.12
Consider this sample car-jump situation. The car will travel
down its ramp toward the jump point. On one side of the end
of the ramp is a flashlight, aimed across the end of the ramp
so that the car must pass through its light. On the other side
of the ramp from the light is a light meter that measures how bright the light is.
Just before it leaves the ramp, the car passes between the light and the meter,
breaking the light’s beam and greatly decreasing the reading of the meter. All the
while, the light-sensor set-up (or what ever set-up you are using) is recording the
time-light information. This configuration is called a “photogate.”
1.Suppose the car is 4 cm long at the height at which the light beam is located.
The photogate shows that the light is blocked for 0.019987 seconds. How fast
was the car going?
2.Suppose after you drop a car vertically and collect your data using a motion
detector, you find the (translated) equation h = –14.76t2 + 5.53, where h is the
height (ft.) above the floor and t is the elapsed time (sec.) since the car was
released. What is the acceleration due to gravity, in m/sec.2? (The motion
detector has recorded the measurements in feet. 2.54 cm = 1 in.)
3.The “jump point” on the ramp in Item 1 is 1.32 m above the floor. Use your
information from Items 1 and 2 to answer parts (a)–(d).
a) Write an equation for the horizontal distance (in meters) that the car has
gone from the end of the ramp t seconds after it takes off.
b) Write an equation for the height (in meters) of the car above the floor t
seconds after it takes off.
c) When the car has sailed 0.5 m in front of the ramp, how high above the floor
is it?
d)When the car hits the ground, how far from the ramp is it?
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PROJECT 2: WHAT’S YOUE ANGLE?
SUPPLEMENTAL ACTIVITY
S7.13
When you did the ramp jump in class, you made sure that
the car left the ramp exactly horizontally. And when you
placed your landing ramp, you probably did not worry as
much about its angle as you did about its location. Of course,
that’s not what Evel Knievel or the monster truck jumpers really do. Instead, they
take off at an upward angle, and the landing ramp is set to catch them somewhat
smoothly.
Devise a method of modifying your work in Lesson 5 to be able to deal with more
realistic jumps. Then try it out—on toys, of course!
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