Mini-Project
for
Calculus and Mathematica
Workshop

Name:   Shirley Barrette

Title:    The Distance Function

Description:  This lesson introduces the Distance Function (d = rt) to 8ÊÓ graders.  It also introduces the function for free falling objects (d=16t€) and an acceleration function v = rt + at.  Solving systems of equations is shown by graphing.

IT KEEPS ON TICKING-Introduction

D = RT

Contrary to popular belief, D = RT is not an abbreviation for

Disaster equals a Raving Tiger.

Au contraire, mon ami.
 Distance = (Rate) x (Time)
 IS THE

WONDERFUL
	 DISTANCE FUNCTION.

WHAT does this WONDERFUL function do?
HOW does it WORK?

Let's use Mathematica and find out.

The function D = RT tells us that the DISTANCE an object travels VARIES DIRECTLY as the product of the constant RATE at which it travels and the TIME it travels.   Therefore, DISTANCE is a function of RATE and TIME.

WHOOPEE, YOU SAY?

Well, try this on for size.

SOME FUN WITH D = RT

A car travels at a steady rate of 55 miles per hour.  How far will the car travel in 4 hours?
To solve, evaluate the function  for r = 55 and t = 4.

f[t] = 55 * 4

The car will travel 220 miles in 4 hours.

How far will the car travel in 90 minutes?
Change 90 minutes to hours, since the rate is given in miles per hour, and evaluate the the function.

f[h] = N[90/60]

Clear f[t]
f[t] = 55 f[h]

The car will travel 82.5 miles in 90 minutes.

EXTENSIONS

The function for distance D in feet traveled by a free-falling object in T seconds is:

D = 16T€

How far will an object fall in 2 seconds? in 4 seconds? in 8 seconds? 
A GOOD TIME FOR A GRAPH!

Clear[d,r,t]
f[t_] = 16 t^2
Plot[f[t],{t,0,10},
AxesLabel->{"TIME","DISTANCE"},
PlotStyle->{RGBColor[0,0,1]},
PlotRange->All,
Ticks->{Range[0,10,1],Range[64,1500,96]}];

Clear[d,r,t]

f[t] = 16 * t^2

Now, substitute for T in the function.

f[T]

2.  Does the object fall the same distance each second?  Explain your answer.

No, the 1ÍÊ second the object falls 16 feet; the 2ˆÎ second  it falls 64 feet.

Well, enough of this.  Let's do some GOOD STUFF!

You know how to graph a linear equation, so now let's graph TWO of them in order to solve problems.

Here is a problem that can be solved by using a graph.

At noon, Kelly begins a brisk walk at 4 miles per hour.  At 1 P.M., Melissa leaves and runs after her at 6 miles per hour.  At what time will Melissa catch up with Kelly?

Make a graph showing the distances

Clear[f,r,t]
plot1 = Plot[4 t,{t,0,10},PlotRange -> {0,60},
        PlotStyle->{RGBColor[0,0,1]},DisplayFunction->Identity];
plot2 = Plot[6(t-1),{t,0,10},PlotRange->{0,60},
        PlotStyle->{RGBColor[0,1,0]},DisplayFunction->Identity];      
Show[{plot1,plot2},
     AxesLabel->{"TIME","DISTANCE"},
     Ticks->{Range[0,10,1],Range[5,60,5]},
     DisplayFunction->$DisplayFunction];

THINK AND TRY

1.  A train leaves a station at 3 p.m. traveling at 40 mph.  Another train leaves at 5:00 p.m. from the same station on a parallel track.  The second train travels at 60 mph.  When will the second train overtake the first train?

Clear[f,r,t]
plot1 = Plot[40(t),{t,0,10},PlotRange -> {0,600},
        PlotStyle->{RGBColor[0,0,1]},DisplayFunction->Identity];
plot2 = Plot[60(t-2),{t,0,10},PlotRange->{0,600},
        PlotStyle->{RGBColor[0,1,0]},DisplayFunction->Identity];      
Show[{plot1,plot2},
     AxesLabel->{"TIME","DISTANCE"},
     Ticks->{Range[0,10,1],Range[5,600,50]},
     DisplayFunction->$DisplayFunction];

WA-HA-TIZZY!  The second train overtakes the first train at 9:00 pm.
The function D = RT, can be solved for R or T.  Divide both sides of the function by R AND THE RESULT IS:

D/R = T {Distance/Rate = Time}
or
T = D/R {Time = Distance/Rate}

1.  A satellite travels at 12,500 miles per hour.  How far will it travel in a week?

f[r] = 12500 * 7 * 24

2.  A missle travels at 1,725  miles per hour.  Find the distance traveled in 90 seconds.

f[t] = N[1725 * 90/60]

3.  Light travels at 186,000 miles per second.  Find the distance traveled in 1 minute.

f[t] = 186000 * 60

4.  The formula for the distance traveled in meters by a freely falling object on the moon in t seconds is:
d = 0.8t€

The planet Earth's gravity is 6 times stronger than that of the moon.  In the same period of time, objects will fall 6 times farther on the Earth than theyu  will on the moon.  How far in meters will an object on the Earth fall in 5 5  5 seconds?, 12 seconds?

Clear[t,f]
f[t] = 6 .8 5^2

Clear[t,f]
f[t] = 6 .8 12^2

Plot[6 .8(t)^2,{t,0,12},PlotRange -> {0,600},
     PlotStyle->{RGBColor[0,0,1]},  
     AxesLabel->{"TIME","DISTANCE"}, 
     Ticks->{Range[0,12,1],Range[5,600,50]},
     DisplayFunction->$DisplayFunction];

I LOVE IT!

5.  A car traveling 30 feet per second accelerates by increasing it speed 8 feet per second each second.  Find its new speed in 6 seconds.  Use the formula v = rt + at.

Clear[v,r,t,a]
f[v] = t(r + a)
f[v] = 6(30 + 8)

This car's new speed is 228 ft/s.

CHECK THIS OUT, DUDES!

I can't help it.  I have to BOGGLE your BRAIN.   I heard the following conversation the other day.

Alicia:  How old are your 3 brothers?
 Kelly:  The product of their ages is 36.
              The sum of their ages equals my age. 
              Alicia:  I know your age, but I need more facts to 
your brothers' ages.
Kelly:  The oldest boy is not a twin.
 Alicia:  Now I can tell you their ages!
 FIND THE BOYS' AGES.

CHECK MARQUEE FOR
COMING
ATTRACTIONS!