Find a real-life application of a quadratic function and state the application.?

Question

ALSO: Give the equation of the quadratic function, and state what the x and y in the application represent. Choose at least two values of x to input into your function and find the corresponding y for each. State, in words, what each x and y means in terms of your real-life application.

Answer 1

yes, the path of a projectile (in other words, a trajectory) is the most important application of quadratic functions.  equations of motion involving quadratic functions include:

s = Vit + 1/2 (at^2)

where s = displacement

         Vi = initial velocity

         t = time taken

         a = acceleration

your x value is time taken.

y value is displacement.

Answer 2

Quadratic functions are used alot in error analysis.

Suppose you have a set of data and you are to map a rough function that estimates how the data changes.  For that you can choose from a linear, logarithmic, exponential, or quadratic function.

Most common though is the application of physics.

Specifically in Kinematics ( equations of motion).

This is also know as projectile motion.

Suppose you have a target x meters from your current position.

x represents the change in distance required to hit the target.

Now suppose you want to know how to actually hit the target at a specific point in time.  For that you need to have a couple of pieces of data...

Fistly I will briefly cover kinematic equations...

v = velocity in m/s

a = acceleration due gravity ( -9.8 m/s²)

t = time in sec. or s

x = displacement on the horizontal axis

Here are the three basic laws of kinematics

1)

Velocity is a function of acceleration over time...

V = V_0 + a * t

this means that your final velocity depends on two things.  Your inital velocity and your rate of acceleration each second thereafter.

2)

Position can be determined as a function of time given the inital velocity.

X - X_0 = V_0 t + 1/2 a t²

This states that given an inital velocity we know the change in position (X-X_0)

3)

Final Velocity of impact can be determined by knowing the change in position.

V² = V_0 ² + 2 a ( X - X_0)

This states that if we drop a ball V_0 = 0m/s, we can calculate the velocity just before impact by knowing how far it has traveled (X-X_0).

Now that you briefly understand kinematics lets go back to the original question, How does this relate to quadratic equations...

Notice the 2nd equation has a t².

We can create time as a quadratic equation.

Why would we do this?  Simple....

Lets pretend we have a projectile.

To take the simplest example lets use a ball that is dropped from a height then bounces...

The ball is dropped from a position.

We use EQ3 to calculate the velocity just before impact.

This is the velocity the ball will be launched at as it bounces.

In most models we apply a frictional force here and say that the ball will lose a percentage of its velocity each bounce.

For our example lets assume that is 20%.

So when the ball bouces it now has an initial velocity that is 20% of its impact velocity.

Now comes the fun part...

Ask yourself... "how high is the ball going to bouce???"

Well what do we know from basic physics???

What goes up must come down.

In other words the ball will go up until it has a velocity of 0.

Then begin to fall back to the Earth until it bounces again.

So how do we figure out that height???

Simple...

We use EQ3, accept this time we are referring to the vertical height.

V² = V_0² + 2 a ( Y - Y_0)

We have an impact velocity V_0,

We know the final velocity is 0.

We know the acceleration of gravity is -9.81 m/s².

So we can solve for the change in position or height.

Given this we have the height the ball will travel for a single bounce.

Suppose we then want to find out how long it will take to reach this height.  Why do we want that data, simple, we need to keep track of how much time has elapsed between bounces...

Anyhow...

Lets setup a qudratic function to calculate the time involved in reaching the maximum height...

Y - Y_0 = V_0 * t + 1/2 a * t²

Or written as a quadractic form..

0 = 1/2 a * t² + V_0 * t  + ( Y - Y_0 )

for ax² + bx + c = 0

a = 1/2 a

b = V_0

c = ( Y - Y_0 )

This will give us the time it takes for the ball to reach maximum height.

Thus we know is takes 2t to reach the ground.

We can use this to keep track of how long the ball takes to bounce before coming to rest...

I hope you enjoyed this physics lesson...

Answer 3

Projectile motion.

You can do the ALSO part.