Suppose a stone is thrown vertically upward from the edge of a...

Question

Suppose a stone is thrown vertically upward from the edge of a cliff on Earth with an initial velocity of 64 ft/s from a height of 32 ft above the ground. The height (in feet) of the stone above the ground t seconds after it is thrown is s(t) = -16t² + 64t + 32.
On what intervals is the speed increasing?

On what intervals is the speed increasing?

Accepted Answer

Welcome back for once another video. Determine the time intervals at which the speed of a ball is increasing if the ball is launched straight up from the top of a building with an initial speed of 80 ft per second and a height of 150. Feet above the ground, the height and feet of the ball above the ground 2 seconds after it is launched is given by the equation H of T equals -16 T2 + 80 T + 50. So we're essentially given the position function. We can visualize it if we want. This is definitely not necessary, but we can say that if we plot it, well, we're going to have H on the vertical axis and time T on the horizontal axis, and this is a parabola. That opens down, right, because we have a negative coefficient and if time is 0, well, height is 50. So we can essentially visualize this parabola as follows. This is definitely not necessary, but now we're going to break it down and in some cases, we can use the sketch to understand what's happening, right? So we're given our a H of T function. It tells us the vertical position relative to time. We have -16 T2 + 80 T + 50, and our goal is to analyze the velocity function. We have to recall that velocity is the first derivative of the position function. So let's go ahead and differentiate the given function. We're going to apply the power rule. So for T2d, we're going to get 2 T as the derivative. So overall, 2T multiplied by -16 gives us -30 to T, and the derivative of T is 1. So we get plus 80, and the derivative of 50 is 0. So essentially, from here, we're going to determine the acceleration function. Which is the first derivative of the velocity function or the second derivative of the position function. In this case, we want to differentiate C, which gives us 1, and we can see that we have a constant acceleration of -32, right? So what we want to understand is that speed is increasing when velocity and acceleration vectors are pointing in the same direction. Notice that acceleration is negative for any T, right? It's independent of time, so acceleration is less than 0 for all time values, and this means that the speed is only increasing if velocity is negative as well, right? Because we want to make sure that acceleration and velocity are pointing in the same direction. So let's solve B of T less than 0. And this means that 32 + 80 must be less than 0. And of course, this equation can be factored out. So what we're going to do is simply factor out. 16, or we can just factor out negative 16, and then in parenthesis, we're going to have 2T minus 5 being less than 0. And now from here, we can essentially divide both sides by -16, right? So if we do that, we're going to say that 2 T minus 5 must be greater than 0. Why is it greater? Well, essentially because we're dividing by a negative value. And from here, we can just add 5 to both sides, so TT is greater than 5, and time is greater than 5 divided by 2 or 2.5 seconds. So what we're going to do is simply understand that we start. Going up and there's the highest point within the parabola where velocity is zero and after that point. We are going down, right? Speed is increasing. So essentially this point is 2.5 and then of course we want to find the point where the object or the ball in this case hits the ground. So we have our lower bound of time, right? We're going to say that time is greater than 2.5 seconds and then we want to identify the final. Value of time at which this pole hits the grounds we're going to set H of T equals 0, and we want to solve -16 T2 + 80 T + 50 equals 0. So we're setting the position equals 0, and to solve this equation, well, essentially we can apply the quadratic formula. Now, before we do that, we can essentially simplify the given expression, right? So we can divide both sides by -2. We're going to get 8 T squad minus 40 T plus. It will be -25, right? Because we have to flip all of those signs. So -25 equals 0. And from here, we are going to have two solutions based on the quadratic formula. So we're taking negative B, our B is -40, negative of -40. Plus or minus. Square root of the discriminant, which is B2 minus 4AC, so -40 squad minus 4 multiplied by 8 multiplied by -25. All of that is going to be divided by 2A, so 2 multiplied by 8. And what we're going to do is simplify and evaluate two time values. So time 1 is going to be equal to 40. We're going to take the positive solution first, right? So essentially the positive radical value. And we want to evaluate that. So we have 1600 minus 32 multiplied by -25. This gives us a value of 2400, and then we are dividing by 2A, right? So 2 multiplied by 8 gives us 16. Now from here we're going to simplify 2400 can be written as 400 multiplied by 6, right? So square root of 400 is 20, and we can write 40 plus 20 square root of 6, because 400 multiplied by 6 gives us 2400, and then we are dividing by 16. So what we're going to do is evaluate this value. Notice that we can divide both sides by 2 and we are going to get. 20 plus 102 6 divided by 8, but of course we can even divide by 4. So instead we're going to get 10 plus 20 divided by 4 is 5. Then we have square root of 6, and in the denominator 16 divided by 4 is 4. Now for the second solution we're going to have a negative sign, and if we subtract 10 minus 5 square root of 6, this is going to give us a negative value, right? So we are only interested in positive time values. So there's no necessity to solve for the second time value. We can just say that the the speed of the pool is increasing when time is within the open interval. Because at 2.5 seconds, the ball is station rewrite, it stops for a moment, so we're going to say 2.5 seconds, comma. And then our final value is 0 10 + 5 square root of 6 divided by 4 seconds, which, which is our finally answer. So let's label it and thank you for watching.

Key Concepts

Velocity and Speed

Acceleration

Critical Points and Intervals

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