+
3 Tutorials that teach Using Quadratic Equations to Represent Motion
Take your pick:
Using Quadratic Equations to Represent Motion

Using Quadratic Equations to Represent Motion

Author: Colleen Atakpu
Description:

This lesson covers using quadratic equations to represent motion.

(more)
See More

Try Our College Algebra Course. For FREE.

Sophia’s self-paced online courses are a great way to save time and money as you earn credits eligible for transfer to over 2,000 colleges and universities.*

Begin Free Trial
No credit card required

25 Sophia partners guarantee credit transfer.

221 Institutions have accepted or given pre-approval for credit transfer.

* The American Council on Education's College Credit Recommendation Service (ACE Credit®) has evaluated and recommended college credit for 20 of Sophia’s online courses. More than 2,000 colleges and universities consider ACE CREDIT recommendations in determining the applicability to their course and degree programs.

Tutorial

Video Transcription

Download PDF

Today, we're going to talk about using quadratic equations to represent motion. So we're going to look at a real-world situation of an object flying through the air, and we'll talk about how you can use methods for solving quadratic equations to answer questions about the object. So let's start by looking at an example of a rocket being launched through the air.

I've got a graph that represents the path of the rocket. On my horizontal axis, I have time in seconds. And on my vertical axis, I have height in feet. So as we launch the rocket at time 0, our height is 0. We go up into the air and reach a maximum height of 400 after 5 seconds. And then we come back down because of gravity. And after 10 seconds, we're back down at the ground.

So from our graph, we can see that it took 10 seconds for our rocket to go up and come back down. And we have an equation that matches the graph y equals negative 16x squared plus 160x. So we could use our equation to figure out how long it will take for the rocket to reach the ground. Either if we didn't have a graph or if we wanted to verify it algebraically, as well as looking at it graphically.

So if I want to know how long it will take for the rocket to reach the ground, I want to see the value of x when y is equal to 0. When the height is equal to 0 when it hits the ground. So I'm going to solve this quadratic equation by replacing 0 for y.

So my equation will become 0 equals negative 16x squared plus 160x. Now, there's a few different methods that I can use to solve this quadratic equation. But I noticed that both terms here have a common factor, so I'm going to use factoring to solve.

The common factor between these two terms is going to be negative 16x. So factoring this out, I'm going to factor out negative 16x. I'll have to multiply by another x to get negative 16x squared. And to get positive 160x, I'll have to multiply this by negative 10.

So now I can use my zero product property of multiplication to find the solutions for x. I know that if it's equal to 0, then I can set each of these two factors equal to 0. So I'll have negative 16x equals 0 and x minus 10 equals 0.

Solving here for x, I find that x equals 0. And solving here for x, I find that x equals positive 10. So I've found that the amount of time, which is my x-variable, that it takes for the rocket to get to the ground is 10 seconds. And we also had our other solution of 0, which we already know that a height of 0, the time was at 0 also. That's in the beginning before we had launched the rocket.

So let's see how we can answer a different question using the same real-world situation. Let's say now I want to find the maximum height of my rocket.

So again, I can look at my graph and see that the maximum height will be the maximum point of the graph. And I know that because this is a quadratic, the maximum point of the quadratic or the parabola is called the vertex. So in looking for my maximum height, I'm looking for the vertex of this parabola.

And the vertex, the x-coordinate of the vertex, is going to be how long in seconds it takes to get to my maximum height. And the y-coordinate of the vertex is going to be the maximum height. So let's see how we can solve for the maximum height algebraically as well.

So again, I'm going to start with my equation. And I know that I can find the vertex by using the formula for the x-coordinate of the vertex, which is x equals negative b over 2a. I know that my value for b is 160 and my value for a is negative 16. So substituting those values into my formula, I have negative 160 over 2 times negative 16.

Simplifying that, I have negative 160 over negative 32. Divided will give me 5. So I found that the x-coordinate of my vertex is 5. In other words, it's going to take 5 seconds to reach the maximum height.

And I'll also find the y-coordinate of my vertex by substituting 5 in for x in my equation. So y will equal negative 16 times 5 squared plus 160 times 5.

Simplifying this, negative 16 times 25. Negative 16 times 25 is going to give me negative 400. And 160 times 5 will give me 800. Negative 400 plus 800 will give me 400. So I've verified that my maximum height is 400, or 400 feet, for my rocket going through the air.

So let's go over our key points from today. A quadratic equation can be used to model the path of an object rising and falling due to gravity. If x represents time and y represents height, the quadratic equation and/or its graph can be used to answer questions about the object's path such as the time to travel a certain horizontal distance, or the height of the object at a certain time.

So I hope that these key points and examples helped you understand a little bit more about using quadratic equations to represent motion. Keep using your notes and keep on practicing and soon you'll be a pro. Thanks for watching.