RGBT Task 1

RGBT Task 1 5

Running head: RGBT TASK 1

RGBT Task 1

Lori Jacobson

Western Governors University

RGBT Task 1

In this series of problems, I will write as though I am teaching a class of middle schoolers in a basic algebra class.

A Fastest Mile Problem

1 To find out the linear decrease in the fastest timed mile records over time t, where t=0 corresponds to 1850, first we look at the two dates for which we know the speed.

In 1850, the mile was run in 5 minutes, thus R(0)=5

In 1950, the mile was run in 4 minutes, thus R(100)=4

Since we know that the time to run the mile is DECREASING over time, we set up the problem like this: R(t)=5-xt, where x is the constant by which we have to multiply the time variable. What is x? Well, we need to figure that out using the information we have. Lets use the 1950 information to figure out the constant x:

R(t)=5-xt ( when t=100, R(t)=4=5-x(100) ( -1=-100x ( 1/100=x

Plug that constant into our original equation:

R(t)=5-1/100(t) or R(t)=5-t/100 and t = year - 1850

2 Now we will use the above function to predict the following:

a The record for the fastest mile run in the year 2000. In this problem, we are asked to find the value of the function when the year is 2000.

Remember that that t = year-1850, so t = 2000-1850 = 150

R(150)=5-150/100 = 3.5 minutes.

So, using our model, we predict that the mile will be run in 3.5 minutes in the year 2000.

b The year in which a mile will be run in 3 minutes.

In this problem, we first need to find value t when the function is equal to 3, then add t to 1850 to find out the year we expect the mile to be run in 3 minutes.

R(t) = 3 ( 5-t/100 = 3 ( -t/100=-2 ( -t=-200 ( t=200

Thus, the year we expect the mile to be run in 3 minutes is 1850+200, or 2050.

c The year in which a mile will be run in 2 minutes.

In this problem, we first need to find value t when the function is equal to 2, then add t to 1850 to find out the year we expect the mile to be run in 2 minutes.

R(t) = 2 ( 5-t/100 = 2 ( -t/100=-3 ( -t=-300 ( t=300

Thus, the year we expect the mile to be run in 3 minutes is 1850+300, or 2150.

d The year in which a mile will be run in 1 minute.

In this problem, we first need to find value t when the function is equal to 1, then add t to 1850 to find out the year we expect the mile to be run in 1 minute.

R(t) = 1 ( 5-t/100 = 1 ( -t/100=-4 ( -t=-400 ( t=400

Thus, the year we expect the mile to be run in 3 minutes is 1850+400, or 2250.

e The year in which a mile will be run in 0 minutes.

In this problem, we first need to find value t when the function is equal to 0, then add t to 1850 to find out the year we expect the mile to be run in 0 minutes.

R(t) = 0 ( 5 - t/100 = 0 ( -t/100 = -5 ( -t = -500 ( t = 500

Thus, the year we expect the mile to be run in 3 minutes is 1850+500, or 2350. HOWEVER, this is an impossible goal how can the mile be run in zero minutes? Lets think about the limitations of our model, below.

3 The limitations of the model.

Models are only approximations of natural phenomenon. Thus, we have to view them as possibilities. Is it really POSSIBLE for a human being to run a mile in zero minutes? I think not. So in this case, part e above is not possible. In fact, as time goes on, we might find that the graph of this function will become more of a curve than a line. In that case, we would have to change our function definition and formula.

B Launching Object Problem, h = -16t2 + 48t + 10

1 How much time has elapsed from the time of launch to the time the object hits the ground? For this question, we need to figure out the point at which h = 0 (ground level).

Thus, we use the quadratic formula to find t where h=0.

a = -16 b = 48 c = 10

t = [-b +/- sqrt(b2 4ac)]/2a

t = [-48 +/- sqrt(2304 4(-16)(10))]/-32

t -.1956 or 3.1956

Now we evaluate each of these two answers. Is it possible for t = -.1956, for time to be negative? No. So the answer is: The object hits the ground in approximately 3.1956 seconds.

2 How much time has elapsed when the object is 30 feet above the ground? This is similar to the last question, except this time we have to set up the problem to take into account the 30 feet. So, the original equation becomes:

30 = -16t2 + 48t + 10 ( 0 = -16t2 + 48t 20

then use the quadratic formula as before:

t = [-b +/- sqrt(b2 4ac)]/2a

t = [-48 +/- sqrt(2304 4(-16)(-20))]/-32

t 2.5 or .5

Why are there two answers? Think about shooting a projectile up into the air. It goes up and passes the 30 foot mark on its ascent, then when it comes down it passes the 30 foot mark again on its descent. Cool! So it passes 30 feet in .5 seconds and again in 2.5 seconds.

3 The maximum height reached by the projectile.

We have the original equation h = -16t2 + 48t + 10. To find the maximum height, we can use the formula b/2a to find the time at which the projectile is at its maximum height. Plugging into this equation our values of a and b from above, we get:

Time when at maximum height = -48/-32 or 1.5 seconds. So lets find out the height when t = 1.5:

h = -16t2 + 48t + 10 ( h = -16 (1.52) + (48)(1.5) +10 ( h = 46 feet.

The maximum height reached by the projectile is 46 feet.

4 The projectiles distance from the ground at the instant that the projectile is fired.

This is a fun one! Did you, like I, assume that the projectile was fired from ground level? Well, lets see if that assumption was correct. When the projectile is fired, t=0. So lets plug 0 into the equation for t:

h = -16t2 + 48t + 10 ( h = -16(02) + 48(0) + 10 ( h = 10 !!!

So, at the time the projectile is fired, it sits at 10 feet above the ground!