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Using Quadratic Equations to Represent Motion

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Hi and welcome. My name is Anthony Varela. And today, we're going to be using quadratic equations to represent motion. So we're going to solve quadratic equations. And we're going to solve these within the context of an object in motion. And typically, this involves solving for a height or a time or some other distance.

So our first problem is a baseball player throws a ball into the air. The relationship between the ball's height and the amount of time it is in the air can be represented by this equation, y equals negative 16x squared plus 32x plus 6. And our question is, after how many seconds does the ball hit the ground?

So let me draw a picture of our situation. Here is a sketch of y equals negative 16x squared plus 32x plus 6. Now, our x-axis is going to be time, measured in seconds. And our y-axis is going to be height, measured in feet.

And now the interesting thing about our constant term, 6, is that this is the y-intercept of the parabola. So at a time 0, it's at a height of 6. So we can think of this then maybe being the height of the baseball player. And so he's throwing the ball.

And as time goes on, our ball is approaching a maximum height. But then at some point, due to gravity, it starts to fall. And it eventually hits the ground. So after a certain number of seconds, x, our height is 0. And that's-- we'd like to solve.

So with our motion problem, this involves an object hitting the ground. Typically, we see the y-axis as being height, the x-axis as being time. And we're solving for x when y equals 0. And we can do this, one, by factoring or by using the quadratic formula.

So using the quadratic formula, we have to set our equation equal to 0. We have to do this for factoring as well. But I'm going to be using the quadratic formula. So now I need to then identify a-, b-, and c-values to plug into our quadratic formula.

Now, one thing that I always encourage you to do is factor out a common factor, if possible. So I noticed that all of these are even numbers, or they share a factor of 2. So I can write this as 2 times negative 8x squared plus 16x plus 3. And I'm going to be using these as my values for a, b, and c. You're still going to get the same value of x. I just like to do this because you don't have to deal with such large numbers in your quadratic formula.

So plugging these into our quadratic formula, we get x equals negative 16 plus or minus the square root of 16 squared minus 4 times negative 8 times 3, all over 2 times negative 8. This is just using our values for a, b, and c.

Well, we can simplify our denominator right away. That's negative 16. Simplifying our numerator, I have negative 16 plus or minus 16 squared. I'm going to write that as 256.

Now, 4 times negative 8 times 3 is negative 96. But we're subtracting this. So I'm going to write plus 96. Well, 256 plus 96 is 352.

Now, here's a tricky part. What I'm going to do is take 352. And I'm going to divide it by the largest perfect square that I can that would result in an integer. So I'm going to divide 352 by 4. I'll divide it by 9, divide it by 16, 15, so on and so forth, and see what's that largest perfect square that divides evenly. And that would be 16.

So I'm going to write 352 as 16 times 22. This will allow me to express my square roots of 16 times 22 as 4 times root 22 because 16-- the square root of 16 is 4. 4 squared is 16.

Well, now I'm going to divide 16 by 16 to get 1. And I can simplify 4 over negative 16 as negative 1 over 4. And then I still have my square root of 22.

Now, using my calculator then to evaluate 1 plus negative square root 22 over 4 and 1 minus negative square root 22 over 4, I get two x-values. One would be negative 0.173 when we round. And another x-value would be 2.173 when we round.

But take a look at which x-value makes sense. Remember that x is a unit of time. So we're only going to accept our positive value. So after 2.173 seconds, our ball hits the ground. That's how we can interpret our solution.

Our second problem involves the same equation, the same scenario of this ball being thrown into the air. But we're interested in calculating that maximum height. So what is the maximum height of the ball before it starts to hit-- return towards the ground?

And so looking at a picture, this represents our maximum heights. And we can see that this is going to occur then at the vertex of this parabola because the vertex is at that maximum or minimum point if it was a U-shaped parabola.

So our maximum height occurs at the vertex of the parabola. And the vertex lies on our axis of symmetry, which is a vertical line here. And we can find the equation to this line using x equals negative b over 2a.

So once again, I need to find a- b-values. Once again, I could factor out a 2 and use negative 8 as a and 16 as b. So negative b would be negative 16. And then 2a would be negative 16 as well.

So we know that x equals 1. So this means that after 1 seconds, the ball has reached its maximum height. But what is that maximum height? We'll just plug 1 back into our original equation and solve for y.

So I'm going to be using our original equation here, negative 16x squared plus 32x plus 6. So when x equals 1, we have negative 16 times 1 squared plus 32 times 1 plus 6. So this evaluates to negative 16 plus 32 plus 6. So our y-value is 22. So this means then after 1 second in the air, the ball reaches its maximum height of 22 feet.

So let's review using quadratic equations to represent motion. With our motion problem involving an object hitting the ground, our height was in the y-axis. Our time was on the x-axis. And we solved for x when y equals 0 using either factoring or the quadratic formula.

To find the maximum height, you need to identify then the vertex of the parabola. And you can find the x-coordinate of the vertex using negative b over 2a and just substitute that back into your original equation and solve for y.

Thanks for watching this tutorial and using quadratic equations to represent motion-- hope to see you next time.