mth122 mod4 2 peer responses 200 words each

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Please respond to both POST1 and POST2 in at least 200 words each. I have included the original post as a reference when answering the post.

Original Post:

For this discussion, complete the following tasks:

  1. Use an outside source to search for a quadratic equation that models something from your daily life.
  2. Solve the equation in two ways.
  3. Discuss which method you liked better and why.
  4. In your responses to peers, compare and contrast your preferences for how to solve quadratic equations.

POST1:

The quadratic equation is used to describe the relationship between objects that are thrown and how long it takes to hit the ground.

In spirit of playoffs, I decided to determine how long a punt takes to hit the ground. As I discovered in combing through ways of finding relevant information, the science of a punt is pretty fascinating. For instance, using a right triangle you can determine the necessary angle for a kick to account for distance. That is a function by itself. For this discussion I will just focus on the quadratic equation to determine how long the average punt takes to hit the ground. This would be relevant to know because members of a special team should be able to cover this distance before the ball hits the return man.

The formula:

y=-4.9t^2+V0(t)+H0 where -4.9t^2 is the constant of gravity, V0 is the initial velocity in Meters/Second and H0 is the starting height.

This formula is created due to the effect of the football traveling in a parabola. When the ball is first kicked it travels upward. Starting from the initial launch, the effect of gravity begins to counteract the initial velocity, eventually causing the football to fall. The actual function output is height.

Y=-4.9t^2+39.2t+2.45

Solving by factoring: I don’t like this method. It usually doesn’t work to solve complex equations.

Solving by completing the square:

y=-4.9t^2+39.2t+2.45

-4.9t^2+39.2t+2.45=0

-2t^2+16t+1=0

-2t^2+16t+1=0 put into the form t^2+2at+a^2+b

-2(t^2-8t+16-.5-16)=0

-2(t-4)^2-.5-16=0

-2(t-4)^2-16.5=0

(t-4)^2=-16.52/2

t= +/-(√16.52+4)/2

t=8.062

t=-0.062

In this case, total time is equal to 8.062 seconds when the ball lands (ignoring -0.062 because it isn’t relevant)

Solving using the quadratic equation:

y=-4.9t2+39.2t+2.45 we set the equation equal to 0 and divide by negative like terms and to give us whole and positive term value 0=2t^2-16t-1

Using the quadratic formula we plug in the values of a, b and c

t=(-(-16)^2 +/-√(-16)2-42(-1)22)/2*2

Solving this equation we get the same answers: t=-0.062 and t=8.062

I like completing the square and quadratic formula because they are similar and allow to manipulate numbers that don’t necessarily factor easily

Source:

Quadratic Applications: Projectile Motion (n.d). Retrieved from https://www.brainfuse.com/jsp/alc/resource.jsp?s=gre&c=36714&cc=108826

POST2:

Over the weekend, I went on a canoe trip. I spent the morning traveling upstream and the afternoon traveling downstream. Overall, I spent 4 hours traveling 10 miles. I know that the stream I was canoeing flows at 3 MPH. Now I need to find out at what speed I traveled.

(distance upstream/speed – current) + (distance downstream/speed + current) = 4 hours

(10 / x -3) + (10 / x+3) = 4 hours

10 (x+3) + 10 (x-3) = 4 (x+3)(x-3)

10x + 30 + 10x – 30 = 4(X^2 – 9)

20X = 4x^2 – 9

F(x) = 4x^2 – 20x – 9

Quadratic Formula:

X = (-b +/- √(b^2-4ac)) / 2a

X = (-(-20)+/- √(-(-20)-4(4)(-20))) / ((2)(4))

X = (20 +/- √(400 + 144))) / 8

X = (20 +/- √(544))/8

X = 5.41

X = -0.41

Therefore, since I cannot travel at a negative speed, I traveled at 5.41 MPH.

Completing the square:

4x^2 – 20x – 9 = 0

(4/4)x^2 – (20/4)x – (9/4) = (0/4)

X^2 -5x – (9/4) = 0

X^2 – 5x = (9/4)

X^2 – 5x + (25/4) = (9/4) + (25/4)

(X – (5/2))^2 = (34/4)

X – (5/2) = +/-√(17/2)

X = (5/2) +/- √(17/2)

X = 5.41

X = -0.41

Overall, I preferred to use the quadratic formula. I found that it was easier for me to plug numbers into a formula versus what I saw as a massive amount of work by completing the square.

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