Position, velocity, and acceleration Suppose the position of a...

Question

Position, velocity, and acceleration Suppose the position of an object moving horizontally along a line after t seconds is given by the following functions s = f(t), where s is measured in feet, with s > 0 corresponding to positions right of the origin.
a. Graph the position function.
$f (t) = 6 t^{3} + 36 t^{2} - 54 t; 0 \leq t \leq 4$

a. Graph the position function.

$f (t) = 6 t^{3} + 36 t^{2} - 54 t; 0 \leq t \leq 4$

Accepted Answer

Hi everyone, let's take a look at this practice problem. This problem says the function below describes the position of a ball moving horizontally along the line, where S is in meters and T is in seconds. We're given the functions of T is equal to T cubed minus 3T2 minus T + 8. Provide the graph of the position function given that S is greater than 0 and 0 is less than or equal to T is less than or equal to 3. Now, in order to create the graph, we need to first plot some points based upon our function. And so, since T has to be between 0 and 3, we'll pick 4 different points to plot 012, and 3. So we're gonna calculate S of 0 1st. And that's gonna be equal to, and we were given that function, as of T in our problem. So it's gonna be 0 cubed minus 3 multiplied by 0 squared. 0 + 8, which is equal to 8. Or as of one. It's going to be equal to 1 cubed minus 3 multiplied by 1 squared. -1 + 8. And this is going to be equal to 5. For as of 2. It's going to be equal to 2 cubed. -3 multiplied by 2 squad, minus 2 + 8. I want to calculate that, that's going to be equal to 2. And then finally for S of 3. This is gonna be equal to 3 cubed minus 3 multiplied by 3 squared. -3 + 8, and that is equal to 5. Now to help us plot our function, we also need to determine the extrema for this cubic equation. And so, to determine the points where our extrema are located, we need to take a derivative of our function and set it equal to 0, and then solve for T. Now, since we will be taking a derivative with respect to T, we're going to be using two derivatives rules, our power rule and our constant rule. So for the power rule, we call that the derivative with respect to X of the quantity of X-rays to the N is equal to N multiplied by X raises to the N minus 1. And for the constant rule, we call that the derivative with respect to X of a constant, which will label C is going to be equal to 0. So we're gonna use these two rules to calculate the first derivative of S with respect to T. So, we'll have S of T. is equal to, when I take the derivative of TQ using our power rule, we'll get 3 T squad. Minus or the derivative of 3 T squared, that's going to give us 6T, since we'll have 3 multiplied by 2, which will give me a 6, and I'll subtract one from 2, which will give you 1, and T to the 1 is just T. Then then I'll have minus the derivative T, which is just 1. And then plus the derivative of +88 the consonant, so it's derivative is 0. So now what I need to do is set this first derivative equal to 0 and solve this for T. And notice here that I have a quadratic equation, and it's already in the quadratic form, so I can just directly apply our quadratic formula. So that means that T is going to be equal to minus. -6, plus or minus? The square root of the quantity of minus 6 squared. -4, multiplied by 3, multiplied by -1. And that whole quantity is divided by. The quantity of 2 multiplied by 3. And so we can simplify this expression a bit. So this is going to be equal to 6. Plus or minus the square root? Of 36 plus 12. And the net quantity is divided by? 6. So we can simplify this even further. So this is going to be equal to 6, plus or minus. We have 36 plus 12, which is 48, but 48 is 16 multiplied by 3, and so the square root of 16 is going to be 4, and that's gonna still be multiplied by the square root of 3. And then that quantity is divided by 6. And then finally we can simplify this even more just by dividing um both the numerator and denominator by 2, and so we'll get 3 plus or minus. 2 multiplied by the square of 3. And that whole quantity is divided by 3. So we actually have two solutions here. We will have T1. Which is going to be our positive root, so that's gonna be equal to 3, plus 2 multiplied by the square root 3, divided by 3, and this is approximately equal to 2.15. Here we just round it to 2 decimal places. Our second solution, T2, is going to be equal to the negative root. Of our square root function. So we'll have 3 minus 2 multiplied by the square root 3, divided by 3. And this is going to be approximately equal to -0.15. So, T2 is outside of our domain, so we only need to really calculate the value of our function S for T1. So we'll have S of 3 + 2 multiplied by the square root 3, and that whole quantity is divided by 3. This is going to be equal to the quantity of 3 + 2 multiplied by the square root of 3. Divided by 3. The whole quantity is cute. -3, multiplying the quantity. Of 3 + 2 multiplied by the squared 3 and quantity divided by 3, and then that quantity is squared. Minus the quantity of 3 + 2 multiplied by the square root of 3, and quantity divided by 3. Plus 8. And this turns out to be approximately equal to 1.92. When I plug those values into my calculator. So, now that I've identified my extrema, we also need to identify our inflection points. And so here we're going to take another derivative, we'll need the second derivative of our functions, and we're going to set equal to 0 to find our inflection points. So, taking another derivative of S we'll have S of T. It's equal to, and when I apply the power rule to our first derivative, our first term, there is 3 T squared, when I take its derivative, that's gonna give me. 6 T. Then we'll have minus the derivative of 6T, which is just 6, again applying our power rule, and then we'll take the derivative with respect to T of 1, which is going to be 0. And so I need to set the second derivative equal to 0. And so I can solve this equation now for T, so we'll get 6 T is equal to or sorry 6 T is equal to 6. And then T is going to be equal to 1 when I divide both sides by 6. And we've already calculated the value for T equal 1, and that turned out to be 5. So we have an inflection point at T equal to 1. So now that I have my extrement identified, as well as my inflection points, and some points to actually plot on our graph, we'll come down here and plot our graph by hand. So On our axis here. We're gonna start off by plotting the points that we initially found. So we'll have 08 for our first point. We'll have 15 as our second point, we'll have 22 as our third point, and our fourth point will be 35. We'll also plot on the graph R.4 R Xtrema, which was at 2.15 and 1.92. So that comes slightly just down into the right of our point attitude too. And so remembering that we have an inflection point at 115. We're going to draw our graph by hand just by connecting these points. Ensuring that we have an inflection point at 15. And then, once we connect all of our points on our graph, we should get what we see on the screen here. So, this is the answer to the question that we were looking for, and so I hope that this explanation has been useful, and I'll see you in the next video.

Key Concepts

Position Function

Graphing Functions

Calculus and Motion

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